URGENT!! This is timed, PLEASE HELP!
Nitrogen gas can be prepared by passing gaseous ammonia over solid copper (II) oxide at high temperatures, as described by the following balanced equation:
2 NH3(g) + 3 CuO(s) → 1N2(g) + 3 Cu(s) + 3 H2O(g)
How many grams of N2 are formed when 120.51 g of NH3 are reacted with excess CuO?
(Please explain using steps and show the whole process. Make sure the answer is in sig figs)
Answer:
99.24 gm of nitrogen .
Explanation:
molecular weight of ammonia = 17 , molecular weight of nitrogen = 28.
2 NH₃(g) + 3 CuO(s) → 1N₂(g) + 3 Cu(s) + 3 H₂O(g)
2 x 17 gm 28 gm
( 34 gm )
34 gm of ammonia forms 28 gms of nitrogen
1 gm of ammonia forms 28 / 34 gms of nitrogen
120.51 gn of ammonia forms 28 x 120.51 / 34 gms of nitrogen
28 x 120.51 / 34 gms
= 99.24 gms of nitrogen will be formed .
If 25.8 mL of an AgNO3 solution is needed to precipitate all Cl- ions in a 1570 mg of KCl (forming AgCl), what is the molarity of the AgNO3nsolution?
Answer:
M=0.816M
Explanation:
Hello,
In this case, we should consider the following reaction:
[tex]AgNO_3+KCl\rightarrow KNO_3+AgCl[/tex]
Thus, by knowing the 1:1 molar ratio of silver nitrate and potassium chloride, we can easily compute the moles of silver nitrate precipitating the 1570 mg of potassium chloride considering its molar mass of 74.5513 g/mol:
[tex]n_{AgNO_3}=1570mgKCl*\frac{1gKCl}{1000mgKCl} *\frac{1molKCl}{74.5513gKCl}*\frac{1molAgNO_3}{1molKCl} \\\\n_{AgNO_3}=0.021molAgNO_3[/tex]
Then, by using the volume of silver nitrate in liters (0.0258 L), we can directly compute the molarity:
[tex]M=\frac{0.021molAgNO_3}{0.0258L}\\ \\M=0.816M[/tex]
Regards.
In what unit do we usually measure the force of the earth gravity? Acceleration due to gravity is 9.8/s^2
Answer:
in short weight
Explanation:
weight is mass x gravitational pull on an object
Monel metal is a corrosion-resistant copper-nickel alloy used in the electronics industry. A particular alloy with a density of 8.80 g/cm3 and containing 0.090 % Si by mass is used to make a rectangular plate that is 15.0 cm long, 12.5 cm wide, and 3.50 mm thick and has a 2.50-cm-diameter hole drilled through its center such that the height of the hole is 3.50 mm .
The silicon in the plate is a mixture of naturally occurring isotopes. One of the those isotopes is silicon-30, which has an atomic mass of 29.97376 amu. The percent natural abundance, which refers to the atoms of a specific isotope, of silicon-30 is 3.10%.
Part A What is the volume of the plate?Express the volume numerically in cubic centimeters.
Part B How many silicon-30 atoms are found in this plate?
Express your answer numerically using two significant figures.
Answer:
Based on the given question, the dimensions of the plate is 15 cm in length, 12.5 cm in width, and 3.50 mm in thickness (0.350 cm). Now the volume of the plate will be,
V = 15 cm × 12.5 cm × 0.350 cm = 65.62 cm³
A hole of diameter 2.50 cm is drilled through the center of the plate, at the height of 3.50 mm or 0.350 cm. Now the volume of the hole is π(r)²h,
= 22/7 × (1.25 cm)² × 0.350 cm = 1.72 cm³
Thus, the volume of the plate will be determined by subtracting the volume of plate with the volume of hole, which will be,
65.62 cm³ - 1.72 cm³ = 63.9 cm³
The density of the alloy is 8.80 g/cm³, therefore, the mass of the alloy can be determined by using the formula, mass = density * volume
mass = 8.80 g/cm³ × 63.9 cm³ = 562.32 grams
Of the total alloy, 0.090 percent is Si, that is,
(0.090/100) × 562.32 g = 0.506 grams of Si
The natural abundance of the element is not determined by mass but by the number of atoms it possess. For this Avogadro's number and atomic mass of Si is used. Now the number of atoms of Si present is,
(0.506 g) (1 mol/28.0855 g) (6.023 × 10²³ atoms /mol) = 1.08 × 10²² Si atoms
Of these Si atoms, 3.10 percent are Si-30 so,
= (3.10 / 100) × (1.08 × 10²² atoms) / 1000 = 3.34 × 10²⁰ atoms of Si-30. or 3.4 × 10²⁰ atoms
When you look at an ant up close, using a convex lens, what do you see?
Answer:
You would be able to see the ants clearly with the unique body parts.
Explanation:
Convex lens is also known as Converging lens. It helps in converging rays of light to become a principal focus which is usually very clear and visible to the eyes. The converging lens when used to view the ant makes the ant appear very visible and the individual is able to see all the unique body parts of the ant. This type of lens is used in individuals who are longsighted.
Which of the following is a chemical property of iron? It
Answer:
is capable of combining with oxygen to form iron oxide
The three‑dimensional structure of a generic molecule is given. Identify the axial and equatorial atoms in the three‑dimensional structure. What is the shape of this molecule?
Answer:
Explanation:
CHECK THE ATTACHMENT FOR THE COMPLETE QUESTION AND THE DETAILED EXPLANATION
NOTE:
Equatorial atoms are referred to atoms that are attached to carbons in the cyclohexane ring which is found at the equator of the ring.
Axial atoms are atoms that exist in a bond which is parallel to the axis of the ring in cyclohexane
For the aqueous reaction dihydroxyacetone phosphate↽−−⇀glyceraldehyde−3−phosphate dihydroxyacetone phosphate↽−−⇀glyceraldehyde−3−phosphate the standard change in Gibbs free energy is ΔG°′=7.53 kJ/molΔG°′=7.53 kJ/mol . Calculate ΔGΔG for this reaction at 298 K298 K when [dihydroxyacetone phosphate]=0.100 M[dihydroxyacetone phosphate]=0.100 M and [glyceraldehyde-3-phosphate]=0.00200 M[glyceraldehyde-3-phosphate]=0.00200 M .
Answer:
ΔG = -2.17 kJ/mol
Explanation:
ΔG of a reaction at any moment could be obtained thus:
ΔG = ΔG° + RT ln Q
Where ΔG° is standard change in free energy of a particular reaction (7.53kJ/mol for the reaction of the problem, R is gas constant (8.314×10⁻³kJ/molK), T is absolute temperature (298K) and Q is reaction quotient of the reaction.
For the reaction:
dihydroxyacetone phosphate ⇄ glyceraldehyde−3−phosphate
Q is defined as:
Q = [glyceraldehyde−3−phosphate] / [dihydroxyacetone phosphate]
Replacing values in ΔG formula:
ΔG = 7.53kJ/mol + 8.314×10⁻³kJ/molK × 298.15K ln [0.00200M] / [0.100M]
ΔG = -2.17 kJ/mol
An electrochemical cell is constructed with a zinc metal anode in contact with a 0.052 M solution of zinc nitrate and a silver cathode in contact with a 0.0042 M solution of silver(I) nitrate. What is the value of Q to use in the Nernst equation for this cell
Answer:
Q = 12.38
Explanation:
The Nernst equation is given as; Ecell = E°cell - (2.303RT/nF) log Q ;where Q is the reaction quotient.
The reaction quotient, Q in a reaction, is the product of the concentrations of the products divided by the product of the concentrations of the reactants.
In an electrochemical cell, Q is the ratio of the concentration of the electrolyte at the anode to that of the electrolyte at the cathode.
Q = [anode]/[cathode]
therefore , Q = 0.052/0.0042 = 12.38
need help and quick answer as fast as possible
One proposed mechanism of the reaction of HBr with O2 is given here. HBr + O2 → HOOBr (slow) HOOBr + HBr → 2HOBr (fast) HOBr + HBr → H2O + Br2 (fast) What is the equation for the overall reaction?
Answer:
4 HBr + O2 → + 2H2O + 2Br2
Explanation:
Based on the following reaction mechanism:
HBr + O2 → HOOBr (slow)
HOOBr + HBr → 2HOBr (fast)
HOBr + HBr → H2O + Br2 (fast)
The equation for the overall reaction is the sum of the three reactions in which intermediaries of reaction (HOBr and HOOBr are canceled). That is 1 + 2 + 2*(3):
HBr + O2 + HOOBr + HBr + 2HOBr + 2HBr → HOOBr + 2HOBr + 2H2O + 2Br2
4 HBr + O2 → + 2H2O + 2Br2Beeing this reaction the equation of the overall reaction.
A sample of carbon dioxide gas at a pressure of 879 mm Hg and a temperature of 65°C, occupies a volume of 14.2 liters. Of the gas is cooled at constant pressure to a temperature of 23°C, the volume of the gas sample will be
Answer:
The correct answer is 12.43 Liters.
Explanation:
Based on the given question, the volume V₁ occupied by the sample of carbon dioxide gas is 14.2 liters at temperature (T₁) 65 degree C or 65+273 K = 338 K.
The gas is cooled at a temperature (T₂) 23 degree C or 273+23 K = 296 K
The volume of the gas (V₂) after cooling can be determined by using the formula,
V₁/T₁ = V₂/T₂
14.2/338 = V₂/296
0.0420 = V₂/296
V₂ = 0.0420 * 296
V₂ = 12.43 Liters.
Which option describes a similarity and a difference between isotopes of an element? A. same atomic number; different number of protons B. same number of protons; different atomic number C. same atomic number; different mass number D. same mass number; different atomic number E. same number of neutrons; same number of protons
Answer:
c
Explanation:
A boy with pneumonia has lungs with a volume of 1.7 L that fill with 0.070 mol of air when he inhales. When he exhales, his lung volume decreases to 1.3 L. Enter the number of moles of gas that remain in his lungs after he exhales. Assume constant temperature and pressure.
Answer:
0.053moles
Explanation:
Hello,
To calculate the number of moles of gas remaining in his after he exhale, we'll have to use Avogadro's law which states that the volume of a given mass of gas is directly proportional to its number of moles provided that temperature and pressure are kept constant. Mathematically,
V = kN, k = V / N
V1 / N1 = V2 / N2= V3 / N3 = Vx / Nx
V1 = 1.7L
N1 = 0.070mol
V2 = 1.3L
N2 = ?
From the above equation,
V1 / N1 = V2 / N2
Make N2 the subject of formula
N2 = (N1 × V2) / V1
N2 = (0.07 × 1.3) / 1.7
N2 = 0.053mol
The number of moles of gas in his lungs when he exhale is 0.053 moles
Suppose an industrial quality-control chemist analyzes a sample from a copper processing plant in the following way. He adds powdered iron to a copper(II) sulfate sample from the plant until no more copper will precipitate. He then washes, dries, and weighs the precipitate, and finds that it has a mass of . Calculate the original concentration of copper(II) sulfate in the sample. Round your answer to significant digits.
Answer:
Concentration of Copper (II) Sulfate in the original sample in mol/L = 0.0035 M
Concentration of Copper (II) Sulfate in the original sample in g/L = 0.56 g/L
Explanation:
Complete Question
Fe(s) + CuSO₄(aq) → Cu(s) + FeSO₄(aq)
Suppose an industrial quality-control chemist analyzes a sample from a copper processing plant in the following way. He adds powdered iron to a 400.mL copper (II) sulfate sample from the plant until no more copper will precipitate. He then washes, dries, and weighs the precipitate, and finds that it has a mass of 89.mg. Calculate the original concentration of copper(II) sulfate in the sample. Round your answer to 2 significant figures.
Solution
Noting that the precipitate is Copper as it is the only solid by-product of this reaction.
89 mg of Copper is produced from this reaction.
We convert this into number of moles for further stoichiometric calculations
Mass of Copper = 89 mg = 0.089 g
Molar mass of Copper = 63.546 amu
Number of moles of Copper produced from the reaction = (0.089/63.546) = 0.0014005602 = 0.001401 mole
From the stoichiometric balance of the reaction,
1 mole of Copper is produced from 1 mole of Copper (II) Sulfate
0.001401 mole of Copper will be produced similarly from 0.001401 mole of Copper (II) Sulfate.
Number of moles of Copper (II) Sulfate in the original sample = 0.001401 mole
Concentration of Copper (II) Sulfate in the original sample in mol/L = (Number of moles) ÷ (Volume in L)
Number of moles = 0.001401 mole
Volume in L = (400/1000) = 0.4 L
Concentration of Copper (II) Sulfate in the original sample in mol/L = (0.001401/0.4) = 0.0035025 mol/L = 0.0035 mol/L to 2 s.f.
Concentration in g/L = (Concentration in mol/L) × (Molar Mass)
Concentration in mol/L = 0.0035025 M
Molar mass of Copper (II) Sulfate = 159.609 g/mol
Concentration of Copper (II) Sulfate in the original sample in g/L = 0.0035025 × 159.609 = 0.559 g/L = 0.56 g/L to 2 s.f
Hope this Helps!!!!
The concentration of the original copper solution is 0.035 M.
The equation of the reaction is;
Fe(s) + CuSO4(aq) -------> FeSO4(aq) + Cu(s)
Number of moles of copper obtained = 89 × 10^-3g/63.5 = 0.0014 moles
Since the reaction is 1:1, the number of moles of copper sulfate that reacted is c.
From the question, we are told that the volume of solution is 400.mL or 0.04L.
Hence, the concentration of the solution is; number of moles /volume
= 0.0014 moles/0.04L = 0.035 M
Learn more: https://brainly.com/question/9352088
Missing parts;
Suppose an industrial quality-control chemist analyzes a sample from a copper processing plant in the following way. He adds powdered iron to a 400.mL copper (II) sulfate sample from the plant until no more copper will precipitate. He then washes, dries, and weighs the precipitate, and finds that it has a mass of 89.mg. Calculate the original concentration of copper(II) sulfate in the sample. Round your answer to 2 significant figures.
Consider the following reaction. I– 2 H2O2 (l) 2 H2O (l) + O2 (g) A solution contains 15 mL 0.1 M KI, 15 mL of DI water and 5 mL of 3% H2O2. After the decomposition of H2O2 is complete, you titrate the solution with 0.1 M AgNO3. If the catalyst, I–, is not consumed in the reaction and is completely recovered, what volume of the 0.1 M AgNO3 is required to reach the end point?
Answer:
Explanation:
The given chemical reaction is:
[tex]2H_2O_{(l)} \to^{I^-}} 2H_2O_{(l)}+O_2_{(g)}[/tex]
From above equation [tex]I^-[/tex] serves as catalyst which is not consumed by the reaction and also it is completely recovered; as a result to that , the full volume of KI will definitely react with AgNO₃.
Given that :
the volume of potassium iodide [tex]V_{KI} = 15 \ ml[/tex]
the molarity of potassium [tex]M_{KI} = 0.1 \ M[/tex]
the volume of distilled water [tex]V_W = 15 \ mL[/tex]
The volume of 3% [tex]H_2O_2 \ \ V_{H_2O_2} = 5 \ mL[/tex]
Molarity of AgNO₃ [tex]M_{AgNO_3} = 0.1 \ M[/tex]
Let take an integral look with the reaction between KI and AgNO₃; we have
[tex]KI + AgNO_3 \to KNO_3 + AgI[/tex]
At the end point; the moles of KI will definitely be equal to the moles of AgNO₃
So;
[tex]M_{KI}V_{KI}= M_{AgNO_3}V_{AgNO_3} \\ \\ V_{AgNO_3} = \dfrac{M_{KI}V_{KI}}{M_{AgNO_3}} \\ \\ \\ V_{AgNO_3} = \dfrac{ 0.1*15}{0.1}[/tex]
[tex]V_{AgNO_3} = 15 \ ml[/tex]
Thus; the volume of 0.1 M AgNO₃ needed to reach the end point is 15 mL
At 25.0 °C the Henry's Law constant for sulfur hexafluoride (SP) gas in water is 2.4x 10 M/atm Calculate the mass in grams of SFo, gas that can be dissolved in S25. ml. of water at 25.0 C and a SF, partial pressure of 1.90 atm Be sure your answer has the correct number of significant digits.
Complete Question
The complete question is shown on the first uploaded image
Answer:
The mass is [tex]m = 0.0349 \ g[/tex]
Explanation:
From the question we are told that
The Henry's Law constant is [tex]k = 2.4 *10^{10} M/atm[/tex]
The volume of water is [tex]V = 525 \ ml = 0.525 \ L[/tex]
The partial pressure is [tex]P = 1.90 \ atm[/tex]
The temperature is [tex]T = 25 ^oC[/tex]
Henry's law is mathematically represented as
[tex]C = P * k[/tex]
Where C is the concentration of sulfur hexafluoride(SP)
substituting value
[tex]C = 1.90 * 2.4*10^{-4}[/tex]
[tex]C = 4.56*10^{-4} \ M[/tex]
The number of moles of SP is mathematically represented as
[tex]n = C * V[/tex]
substituting value
[tex]n = 0.525 * 4.56*10^{-4}[/tex]
[tex]n = 2.39 *10^{-4} \ moles[/tex]
The mass of SP that dissolved is
[tex]m = n * Z[/tex]
Where Z is the molar mass of SP which has a constant value of
[tex]Z = 146 g/mole[/tex]
So
[tex]m = 2.394*10^{-4} * 146[/tex]
[tex]m = 0.0349 \ g[/tex]
A 33.0−g sample of an alloy at 93.00°C is placed into 50.0 g of water at 22.00°C in an insulated coffee-cup calorimeter with a heat capacity of 9.20 J/K. If the final temperature of the system is 31.10°C, what is the specific heat capacity of the alloy? J g·°C
Answer:
THE SPECIFIC HEAT OF THE ALLOY IS 0.9765 J/g K
Explanation:
Mass of alloy = 33 g
Initial temperature of alloy = 93°C
Mass of water = 50 g
Initail temp. of water = 22 °C
Heat capacity of calorimeter = 9.20 J/K
Final temp. = 31.10 °C
specific heat of alloy = unknown
specific heat capacity of water = 4.2 J/g K
Heat = mass * specific heat * change in temperature = m c ΔT
Heat = heat capcity * chage in temperature = Δ H * ΔT
In calorimetry;
Heat lost by the alloy = Heat gained by water + Heat of the calorimeter
mc ΔT = mcΔT + Heat capacity * ΔT
33 * C * ( 93 - 31.10) = 50 * 4.2 * ( 31.10 -22) + 9.20 * ( 31.10 -22)
33 * C * 61.9 = 50 * 4.2 * 9.1 + 9.20 * 9.1
2042.7 C = 1911 + 83,72
C = 1911 + 83.72 / 2042.7
C = 1994.72 /2042.7
C =0.9765 J/g K
The specific heat of the alloy is 0.9765 J/ g K
Identify whether each species functions as a Brønsted-Lowry acid or a Brønsted-Lowry base in this net ionic equation. HF (aq) + SO32- ⇌ F- + HSO3- Brønsted-Lowry _____ Brønsted-Lowry _____ Brønsted-Lowry _____ Brønsted-Lowry _____ In this reaction: The formula for the conjugate _____ of HF is The formula for the conjugate _____ of SO32- is
Explanation:
A Bronsted-Lowry base is a substance that accepts a proton in the form of a hydrogen (H) atom.
On the other hand;
Bronsted-Lowry acid is the substance that donates the proton.
HF (aq) + SO32- ⇌ F- + HSO3-
In the forward reaction;
Bronsted-Lowry acid : HF
Bronsted-Lowry base: SO32-
In the backward reaction;
Bronsted-Lowry acid : HSO3-
Bronsted-Lowry base: F-
The conjugate base of HF is F-
The conjugate acid of SO32- is HSO3-
1. Reaccionan 9.7 Kg de un mineral de níquel al 70% con 8L de una solución de ácido fosfórico al 60% y con una densidad de 1.36g/ml.
Answer:
The reaction produces 201.4 g of hydrogen gas and 12.2 kg of Nickel Phosphate.
Explanation:
English Translation
9.7 Kg of a 70% nickel mineral react with 8L of a 60% phosphoric acid solution and with a density of 1.36g / ml.
Solution
The problem doesn't seen to be complete as it doesn't ask a question in the end. But, we will just calculate the amount of each product expected to cover the grounds.
The balanced chemical reaction between Nickel and Phosphoric acid is given as
3Ni + 2H₃PO₄ → 3H₂ + Ni₃(PO₄)₂
We need to first obtain the limiting reagent, that is, the reagent that is used up during the reaction and is in short supply. This reagent determines the amount of products that will be formed.
Mass of nickel that is present at the start = 70% of 9.7 kg = 6.79 kg
Mass of Phosphoric acid present at the start of the reaction = 60% of (8000 mL × 1.36 g/mL) = 6528 g = 6.528 kg
Converting both of these to number of moles
Number of moles = (mass)/(Molar mass)
For nickel,
Mass = 6.79 kg = 6790 g
Molar mass = 58.6934 g/mol
Number of moles at the start = (6790/58.6934) = 115.7 moles
For Phosphoric acid
Mass = 6528 g
Molar mass = 97.994 g/mol
Number of moles = (6528/97.994) = 66.6 moles
3 moles of Ni reacts with 2 moles of H₃PO₄
From the number of moles present initially, shows that Phosphoric acid is in limited supply and is the limiting reagent.
From the stoichiometric balance of the reaction
2 moles of H₃PO₄ gives 3 moles of H₂
66.6 moles of H₃PO₄ will give (66.6×3/2) of H₂, that is, 99.9 moles of H₂.
Mass of H₂ liberated from the reaction = (Number of moles) × (molar mass) = 99.9 × 2.016 = 201.3984 g = 201.4 g
2 moles of H₃PO₄ gives 1 mole of Ni₃(PO₄)₂
66.6 moles of H₃PO₄ will give (66.6×1/2) of Ni₃(PO₄)₂, that is, 33.3 moles of Ni₃(PO₄)₂.
Mass of Ni₃(PO₄)₂ produced from the reaction = (Number of moles) × (molar mass) = 33.3 × 366.02 = 12,188.466 g = 12.2 kg
Hope this Helps!!!
Aqueous sulfuric acid H2SO4 will react with solid sodium hydroxide NaOH to produce aqueous sodium sulfate Na2SO4 and liquid water H2O. Suppose 62. g of sulfuric acid is mixed with 33.8 g of sodium hydroxide. Calculate the minimum mass of sulfuric acid that could be left over by the chemical reaction. Round your answer to 2 significant digits.
Answer:
Approximately [tex]21\; \rm g[/tex].
Explanation:
[tex]\rm H_2SO_4[/tex] (a diprotic acid) reacts with [tex]\rm NaOH[/tex] (a monoprotic base) at a one-to-two ratio:
[tex]\rm 2\; NaOH\, (s) + H_2SO_4\, (aq) \to Na_2SO_4\; (aq) + 2\; H_2O\, (l)[/tex].
In other words, if [tex]n(\mathrm{NaOH})[/tex] and [tex]n(\mathrm{H_2SO_4})[/tex] represent the number of moles of the two compounds reacted, then:
[tex]\displaystyle \frac{n(\mathrm{H_2SO_4})}{n(\mathrm{NaOH})} = \frac{1}{2}[/tex].
Look up the relative atomic mass data on a modern periodic table:
[tex]\rm H[/tex]: [tex]1.008[/tex].[tex]\rm S[/tex]: [tex]32.06[/tex].[tex]\rm O[/tex]: [tex]15.999[/tex].[tex]\rm Na[/tex]: [tex]22.990[/tex].Calculate the (molar) formula mass of [tex]\rm H_2SO_4[/tex] and [tex]\rm NaOH[/tex]:
[tex]M(\mathrm{H_2SO_4}) = 2 \times 1.008 + 32.06 + 4 \times 15.999 = 98.072\; \rm g \cdot mol^{-1}[/tex].
[tex]M(\mathrm{NaOH}) = 22.990 + 15.999 + 1.008 = 39.997\; \rm g \cdot mol^{-1}[/tex].
Calculate the number of moles of formula units in that [tex]33.8\; \rm g[/tex] of [tex]\rm NaOH[/tex]:
[tex]\begin{aligned}n(\mathrm{NaOH}) &= \frac{m(\mathrm{NaOH})}{M(\mathrm{NaOH})} \\ &= \frac{33.8\; \rm g}{39.997\; \rm g \cdot mol^{-1}} \approx 0.845\; \rm mol\end{aligned}[/tex].
Apply the ratio [tex]\displaystyle \frac{n(\mathrm{H_2SO_4})}{n(\mathrm{NaOH})} = \frac{1}{2}[/tex] to find the (maximum) number of moles of [tex]\rm H_2SO_4[/tex] that would react with the [tex]33.8\; \rm g[/tex] of [tex]\rm NaOH[/tex]:
[tex]\begin{aligned}n(\mathrm{H_2SO_4}) &= \frac{n(\mathrm{H_2SO_4})}{n(\mathrm{NaOH})} \cdot n(\mathrm{NaOH})\\ &= \frac{1}{2} \times 0.845 \approx 0.4225\; \rm mol\end{aligned}[/tex].
Calculate the mass of that [tex]0.4225\; \rm mol[/tex] of [tex]\rm H_2SO_4[/tex]:
[tex]\begin{aligned}m(\mathrm{H_2SO_4}) &= n(\mathrm{H_2SO_4}) \cdot M(\mathrm{H_2SO_4})\\ &= 0.4225 \; \rm mol \times 98.072\; \rm g \cdot mol^{-1} \approx 41.435\; \rm g \end{aligned}[/tex].
When the maximum amount of [tex]\rm H_2SO_4[/tex] is reacted, the minimum would be in excess. Hence, the minimum mass of
[tex]62\; \rm g - 41.435\; \rm g \approx 21\; \rm g[/tex] (rounded to two significant figures.)
Two scientists study data collected during an experiment and reach different conclusions. How would the scientific community address their disagreement?
Please
Answer: D. They would device an experiment that could test the two scientists conclusions.
Explanation:
The results of the scientific study must be verified by peer scientists or members of the scientific community to proof whether the research has been conducted produce a valid evidence.
In the given situation, the two scientists had developed different conclusion for the same experiment. This may mean either of the two may have put up an incorrect conclusion.
The scientific community may address this issue by performing the experiment. Every scientific conclusion is based upon the results of the experimental approach.
Answer:d
Explanation:
g Reduction involves the A) loss of neutrons, gain of electrons, and an increase in oxidation state. B) loss of neutrons. C) increase in oxidation state. D) gain of electrons and an increase in oxidation state. E) gain of electrons.
Answer:
E. Gain of electrons
Explanation:
A reduction reaction is one part of the two concurrent reactions that take place in a redox (reduction-oxidation) reaction.
During reduction, an atom gains electrons from a donor atom, and it's oxidation number becomes smaller.
Option A is wrong because reduction does not increase oxidation state nor are neutrons involved
Option B is wrong because reduction is not a nuclear reaction (does not involve the nucleons)
Option C is wrong because reduction leads to reduction in oxidation state
Option D is wrong leads to a reduction in oxidation state when electrons are gained
Option E is correct because reduction involves gain of electrons
11. Caproic acid, which is responsible for the foul odor of dirty socks, is composed of C, H, and O atoms. Combustion of a 0.225-g sample of this compound produces 0.512 g CO2 and 0.209 g H2O. (a) What is the empirical formula of caproic acid
Answer:
C3H6O
Explanation:
Step 1:
Data obtained from the question include the following:
Mass of the compound = 0.225g
Mass of CO2 = 0.512g
Mass of H2O = 0.209g
Step 2:
Determination of the masses of carbon, hydrogen and oxygen present in the compound.
This is illustrated below:
For Carbon, C:
Molar mass of CO2 = 12 + (2x16) = 44g/mol
Mass of C in CO2 = 12/44 x 0.512 = 0.1396g
For Hydrogen, H:
Molar mass of H2O = (2x1) + 16 = 18g/mol
Mass of H in H2O = 2/18 x 0.209 = 0.0232g
For Oxygen, O:
Mass of O = 0.225 – (0.1396 + 0.0232)
Mass of O = 0.0622g
Step 3:
Determination of the empirical formula for caprioc acid.
This can be obtain as follow:
C = 0.1396g
H = 0.0232g
O = 0.0622g
Divide by their molar mass
C = 0.1396/12 = 0.0116
H = 0.0232/1 = 0.0232
O = 0.0622/16 = 0.0039
Divide by the smallest
C = 0.0116/0.0039 = 3
H = 0.0232/0.0039 = 6
O = 0.0039/0.0039 = 1
Therefore, the empirical formula for caprioc acid is C3H6O
What is the atomic mass of AlNO2?
Answer:
I am not sure, but I think this is the answer 72.987 g/mol
A temperature of 50°F is equal to °C.
Answer:
CONVERT IT:
50°F is equal to 10°C
Answer:
10 degrees Celsius
Explanation:
(50°F − 32) × 5/9 = 10°C
A maple tree could be studied in many fields of science. What aspects of a maple tree might be studied in chemistry?
Answer:
Chemical reactions, kinetics, organic chemistry
Explanation:
You might study the chemical reaction, learn about the differences between products and reactants, about delta H and exothermic and endothermic reactions. You may also study Kinetics by studying the rates of reactions with certain chemicals in a maple's enzymatic processes.
Another thing that you might learn about is organic chemistry. The glucose molecules, carbohydrates, lipids, nucleic acids, all have a structure based on the Carbon atom. You can learn about the specific structures of some chemicals that are involved in photosynthesis and simple hydrocarbons that are involved in photosynthetic/bio-synthetic pathways.
There's probably a lot more - but these are the most basic things I could think of.
The following data show the rate constant of a reaction measured at several different temperatures. Temperature (K) Rate Constant (1/s) 310 0.194 320 0.554 330 1.48 340 3.74 350 8.97 Part APart complete Use an Arrhenius plot to determine the activation barrier for the reaction. Express your answer using three significant figures. Ea
Complete Question
The complete question is shown on the first uploaded image
Answer:
Part A
activation barrier for the reaction [tex]E_a = 84 .0 \ KJ/mol[/tex]
Part B
The frequency plot is [tex]A = 2.4*10^{13} s^{-1}[/tex]
Explanation:
From the question we are told that
at [tex]T_1 = 300 \ K[/tex] [tex]k_1 = 5.70 *10^{-2}[/tex]
and at [tex]T_2 = 310 \ K[/tex] [tex]k_2 = 0.169[/tex]
The Arrhenius plot is mathematically represented as
[tex]ln [\frac{k_2}{k_1} ] = \frac{E_a}{R} [\frac{1}{T_1} - \frac{1}{T_2} ][/tex]
Where [tex]E_a[/tex] is the activation barrier for the reaction
R is the gas constant with a value of [tex]R = 8.314*10^{-3} KJ/mol \cdot K[/tex]
Substituting values
[tex]ln [\frac{0.169}{6*10^-2{}} ] = \frac{E_a}{8.314*10^{-3}} [\frac{1}{300} - \frac{1}{310} ][/tex]
=> [tex]E_a = 84 .0 \ KJ/mol[/tex]
The Arrhenius plot can also be mathematically represented as
[tex]k = A * e^{-\frac{E_a}{RT} }[/tex]
Here we can use any value of k from the data table with there corresponding temperature let take [tex]k_2 \ and \ T_2[/tex]
So substituting values
[tex]0.169 = A e ^{- \frac{84.0}{8.314*10^{-3} * 310} }[/tex]
=> [tex]A = 2.4*10^{13} s^{-1}[/tex]
Gravity on Earth is 9.8 m/s2, and gravity on the Moon is 1.6 m/s2.
so if the mass of an object on earth is 40 kilograms what is the mass on the moon.
and how much does it way
Answer:
Mass is the same but it weights 64 Newtons
Explanation:
First of Mass is the same in any sort of gravity. Now let's calculate weight
W = MG
where W = Weight
M = Mass
G = Gravity
W = (40kg)(1.6)
W = 64
Sorry for the spelling mistakes, hope this helps
Answer:6.61kilo
Explanation: fdfv
Enter your answer in the provided box. To make use of an ionic hydrate for storing solar energy, you place 409.0 kg of sodium sulfate decahydrate on your house roof. Assuming complete reaction and 100% efficiency of heat transfer, how much heat (in kJ) is released to your house at night
Answer:
409.0 kg of sodium sulfate decahydrate will produce 4.49×10⁵ kJ
of heat energy.
Explanation:
CHECK THE COMPLETE QUESTION BELOW
To make use of an ionic hydrate for storing solar energy, you place 409.0 kg of sodium sulfate decahydrate on your house roof. Assuming complete reaction and 100% efficiency of heat transfer, how much heat (in kJ) is released to your house at night? Note that sodium sulfate decahydrate will transfer 354 kJ/mol
EXPLANATION
Here we were asked to calculate the amount of heat will be generated by 409.0 kg of sodium sulfate decahydrate at night assuming there Isa complete reaction and 100% efficiency of heat transfer in the process
The molecular weight of sodium sulfate decahydrate (H₂₀Na₂O₁₄S) is needed here, so it must be firstly calculated.
The molecular weight of sodium sulfate decahydrate (H₂₀Na₂O₁₄S)
( 1*20) + (22.98*2) + (16*14)+ (32*14)= 322.186 g/mol.
Thus 409.0 kg of H₂₀Na₂O₁₄S will have a value which is equivalent to = (409000g)/(322.186 g/mol.)
=1269.453mol of H₂₀Na₂O₁₄S.
But it was stated in the the question that per mole of H₂₀Na₂O₁₄S will transfer 354 kJ heat.
Therefore, 1269.453mol will transfer 1269.453× 354 kJ = 4.49×10⁵ kJ of heat.
Hence, 409.0 kg of sodium sulfate decahydrate will produce
4.49×10⁵ kJ of heat energy.
