Grams of BW
i think thats irtu9rgirg
Independent variable in an investigation is the variable is does not depends on any other variable and on which we can have the control. Hence, the grams of BW is the independent variable here.
What is independent variables?In an experiment the parameters which changes by control or depending on other changes are called variables. There are two kinds variables namely dependant and independent variables.
The dependant variables are those variable which depends upon other variables and whose changes are studying with respect to certain parameters.
Independent variables does not depends on other variable and can be controlled by the researcher. Here, we are studying the % of developing cold with respect to the amounts of BW. The changing variable is BW amount on which the percentage of cold depends.
Therefore, the independent variable is grams of BW and dependant variable is % development of cold.
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How many sulfur atoms are present in 100 grams of this compound? Report your answer to three significant figures.
Answer:
1.88 × 10²⁴ atoms
Explanation:
Step 1: Given data
Mass of sulfur: 100 g
Step 2: Calculate the moles corresponding to 100 g of sulfur
The molar mass of sulfur is 32.07 g/mol. The moles corresponding to 100 g of sulfur are:
100 g × (1 mol/32.07 g) = 3.12 mol
Step 3: Calculate the number of atoms in 3.12 moles of sulfur
We will use Avogadro's number: there are 6.02 × 10²³ atoms of sulfur in 1 mole of sulfur.
3.12 mol × (6.02 × 10²³ atoms/1 mol) = 1.88 × 10²⁴ atoms
What are some applications of flame tests in industry? Describe at least three, one of them MUST include
firework industry.
Answer:
Flame tests use simple equipment, making them ideal for fieldwork. Geologists use the flame test to identify the presence of metals. Forensic scientists can use flame tests at crime scenes for quick analysis of elements present. Miners use the test for analysis of samples, particularly when prospecting. Flame tests provide a good teaching tool for chemistry students learning about emission spectra.
Unlike more sophisticated spectrographic equipment, a flame test requires only a gas burner, a hydrochloric acid solution and nichrome wire to hold the sample. The process is simple: dip the nichrome wire in the acid solution and hold it in the flame to remove any impurities, then affix the sample and hold it in the flame. The emitted colors show what metal ions the sample contains. For example, Copper emits a deep blue, Sodium bright orange and Lead a grey-white color. A table of elements and their characteristic colors helps with identification.
Explanation:
I find the pH of a solution whose hydrogen ion concentration 10^-3mol/l also deduce the pOH of the solution
II
Answer:
pOH = 11
pH = 3
Explanation:
Given data:
Hydrogen ion concentration = 10⁻³ mol/L
pOH of solution = ?
pH of solution = ?
Solution:
pH = -log[H⁺]
Now we will put the values.
pH = -log[ 10⁻³]
pH = -(-3) log 10
pH = 3
pOH of solution:
pOH + pH = 14
pOH = 14 - pH
pOH = 14 - 3
pOH = 11
Which energy source would be the best source of energy for rescue workers?
List the sources from worst to best.
Combustion Engine
Solar Cell
Nuclear Power Plant
Hydroelectric Power Plant
Human Powered Generator
Wind Turbine
Fuel Burning Power Plant
Answer:
Solar Cell
Explanation:
Claim 1: The sun (solar cells) is the best energy source for the rescue team.
what might happen to variables in a science experiment that would lead to unusable results?
Answer:
ejeb094
Explanation:
nnb3neneie9eei rje
Please I need help with this
Answer: put 45 mm
Explanation:
What is it called when two or more atoms combine and are held together 2 poin
by a chemical bond
Answer:
Covalent bonds
Explanation:
i'm pretty sure it's fusion but i may be wrong.
explain the reason each step of the separation is performed with three portions of the solvent rather than with a single poriton of solvent
Answer:
Several extractions is more effective than a single extraction.
Explanation:
When extraction is carried out multiple times, for instance, in this case, the extraction was carried out with three portions of the solvent rather than with just a single portion of the solvent, the amount of material left in the residue will be lower, because the extraction is more complete.
Several extractions with smaller volumes of solvent are more effective than a single extraction with a large volume of solvent.
what is the overall charge of an atom with 17 protons 17 neutrons and 20 electrons
Answer:
Overall charge = -3
Explanation:
The atom of every chemical element is its smallest indivisible part. However, this atom further consists of subatomic particles namely: proton, electron, and neutron. The proton and electron are the positively charged and negatively charged particle respectively.
In a neutral atom, the amount of proton and electrons in that atom equates. However, the amount of electron and proton present in an atom determines the charge of that atom. For example, in this question, an atom is said to contain 17 protons, 17 neutrons and 20 electrons.
Since the negatively charged electrons (20) are more than the positively charged protons (17) by 3, hence, the net charge is -3.
A student measured the masses of four different-sized blocks. The student determined that each block had a mass of 50 grams.
(There is a small block, a little bit bigger block, a big block and the biggles block)
Which block has the least density?
Answer:..
Explanation:
A first-order decomposition reaction has a rate constant of 0.00140 yr−1. How long does it take for [reactant] to reach 12.5% of its original value? Be sure to report your answer to the correct number of significant figures.
Reactants take 504.87 yr to reach 12.5% of their original value in first-order decomposition reaction.
Equation for the first-order decomposition reaction:-[tex]A_{t} =A_{0} e^{-kt}[/tex]....(1)
Here, [tex]A_{t}[/tex] is the final concentration, t is the time, [tex]A_{0}[/tex] is the initial concentration, and k is the rate constant.
Given:-
[tex]A_{t} =0.125A_{0}[/tex]
k= [tex]0.00140yr^{-1}[/tex]
Substitute the above value in equation (1) as follows:-
[tex]0.125A_{0} =A_{0} e^{-kt} \\0.125A_{0} =A_{0} e^{-k\times0.00140 yr^{-1} }\\ln(0.125)/(-0.00140)=t\\t=504.87 year[/tex]
So, 504.87 yr does it take for the reactant to reach 12.5% of its original value.
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WhatGiven the particle diagram:
Which type of matter is represented by the particle diagram?
(1) an element
(2) a compound
(3) a homogeneous mixture
(4) a heterogeneous mixture
Answer:
A compound
Explanation:the mixture can vary
Volume of HCl used 25.0mL 4 l
Initial burette reading 0.50mL
Final burette reading 25.60mL
Concentration of KOH 1.0M
the molarity
HCl solution
Calculate
Answer:
1.0 M
Explanation:
Reaction equation;
KOH(aq) + HCl(aq) -----> KCl(aq) + H2O(l)
Concentration of acid CA = ?
Concentration of base CB = 1.0 M
Volume of base VB = 25.60 - 0.50 = 25.1 ml
Volume of acid VB = 25.0 ml
Number of moles of acid NA = 1
Number of moles of base NB =2
CAVA/CBVB =NA/NB
CAVANB = CBVBNA
CA = CBVBNA/VANB
CA = 1 * 25.1 * 1/25.0 *1
CA = 1.0 M
Pyridine, C5H5N, is a bad-smelling liquid that is a weak base in water. Its pKb is 8.77. What is the pH of a 0.20 M aqueous solution of this compound
Answer:
pH = 9.26
Explanation:
Let's make the reaction of this weak base:
C₅H₅N + H₂O ⇄ C₅H₅NH⁺ + OH⁻ Kb
We do not know the Kb, but we can discover it from pKb
Kb = 10^⁻pKb → 10⁻⁸'⁷⁷ = 1.70×10⁻⁹
So, let's make the expression for Kb
Kb = [OH⁻] . [C₅H₅NH⁺] / [C₅H₅N]
In the equilibrum, we would have 0.20 moles from the begining - x (react)
So we would proudce x moles of OH⁻ and x moles of C₅H₅NH⁺.
In conclussion:
1.70×10⁻⁹ = x . x / (0.20 - x)
To avoid the quadratic formula we can miss the x from the substraction
1.70×10⁻⁹ = x² / 0.20
1.70×10⁻⁹ . 0.20 = x²
We avoid the x from the (0.20-x), because the base is so concentrated, and Kb is very small.
x = √(1.70×10⁻⁹ . 0.20) → 1.84×10⁻⁵
That's the value of [OH⁻], so we can calculate pOH
- log [OH⁻] = pOH → - log 1.84×10⁻⁵ = 4.73
Then pH = 14 - pOH
14 - 4.73 = 9.26
The pH of a 0.20 M aqueous solution of pyridine is approximately 9.26.
Given,
pKb = 8.77
Concentration = 0.20M
The pKb is related to the pKa by the equation: pKa + pKb = 14.
pKa = 14 - 8.77 = 5.23.
Since pyridine is a base, it can be treated as the conjugate base of its corresponding acid. The acid is the pyridinium cation, which is formed by the protonation of pyridine.
Now, the equilibrium between the pyridine (base) and the pyridinium ion (acid):
C₅H₅N + H₂O ⇌ C₅H₅NH⁺ + OH⁻
The equilibrium constant expression for this reaction can be expressed as:
Kb = [tex]\frac{([C_5H_5NH^+][OH^-])}{[C_5H_5N]}[/tex]
At equilibrium, the concentration of OH⁻ will be equal to the concentration of [C₅H₅NH⁺], as pyridine is a weak base and will not significantly dissociate in water.
Kb = [tex]\frac{[OH^-]^2}{[C_5H_5N]}[/tex]
Kb = [tex]10^{(-8.77)[/tex]
Kb = [tex]\frac{[OH^-]^2}{[C_5H_5N]}[/tex]
[tex]10^{(-8.77)[/tex] = [tex]\frac{[OH^-]^2}{(0.20 M)}[/tex]
[OH⁻]² = [tex]10^{(-8.77)[/tex] ₓ 0.20 M
[OH⁻] = [tex]\sqrt{10^{-8.77} \times 0.20M}[/tex]
[OH⁻] ≈ 1.42 x 10⁻⁵ M.
The pOH (the negative logarithm of the hydroxide ion concentration):
pOH = [tex]-log_{10}([OH^-])[/tex]
pOH = [tex]-log_{10}[/tex](1.42 x 10⁻⁵)
pOH ≈ 4.85
Since pH + pOH = 14, the pH:
pH = 14 - pOH
pH = 14 - 4.85
pH ≈ 9.26
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Automobile air bags inflate following a serious impact. The impact triggers the following chemical reaction.
2NaN3(s)->2Na(s)+3N2(g)
If an automobile air bag has a volume of 11.9 L, what mass of NaN3 (in g) is required to fully inflate the air bag upon impact? Assume STP conditions.
Answer:
Explanation:
To determine the molar volume of the gas according to the equation at stp;
1 mole = 22.4 dm³
2 moles = 44.8 dm³
To determine the mass of NaN₃ inflated according to the equation
2NaN₃ (where Na = 23g and N = 14) = (2 × 23) + 2(14 × 3)
= 130 g
Hence, if 130g of NaN₃ is required to inflate 44.8 dm³ airbag upon impact
what mass of NaN₃ is required to fully inflate the air bag upon impact;
130g ⇒ 44.8 dm³
? ⇒ 11.9 dm³ (dm³ is same as L)
? = 130 × 11.9/44.8
? = 34.5g
34.5g of NaN₃ is required to fully inflate 11.9 L of air bag upon impact
Your task is to create a buffered solution. You are provided with 0.10 M solutions of formic acid and sodium formate. Formic acid has a pKa of 3.75. 2. Create approximately 20 mL of buffer solution with a pH of 4.25.
Answer:
15.2mL of the 0.10M sodium formate solution and 4.8mL of the 0.10M formic acid solution.
Explanation:
To find the pH of a buffer based on the concentration of the acid and conjugate base we must use Henderson-Hasselbalch equation:
pH = pKa + log [A⁻] / [HA]
Where [A⁻] could be taken as moles of the sodium formate and [HA] moles of the formic acid
4.25 = 3.75 + log [A⁻] / [HA]
0.5 = log [A⁻] / [HA]
3.162 = [A⁻] / [HA] (1)
As both solutions are 0.10M and you want to create 20mL of the buffer, the moles are:
0.10M * 20x10⁻³L =
2x10⁻³moles = [A⁻] + [HA] (2)
Replacing (2) in (1):
3.162 = 2x10⁻³moles - [HA] / [HA]
3.162 [HA] = 2x10⁻³moles - [HA]
4.162[HA] = 2x10⁻³moles
[HA] = 4.805x10⁻⁴ moles
[A⁻] = 2x10⁻³moles - 4.805x10⁻⁴ moles = 1.5195x10⁻³moles
That means, to create the buffer you must add:
[A⁻] = 1.5195x10⁻³moles * (1L / 0.10mol) = 0.0152L =
15.2mL of the 0.10M sodium formate solution[HA] = 4.805x10⁻⁴ moles * (1L / 0.10mol) = 0.0048L =
4.8mL of the 0.10M formic acid solutionHow are resonance structures for a molecule or polyatomic ion represented?
As a progression of structures.
The parts of each structure that vary are circled.
The parts of each structure that vary are highlighted.
With a double-headed arrow in between each structure.
The answer is number 4
How many moles of HNO3 are present in 697 grams of HNO3?
Answer:
Explanation:
given mass = 697 g
molar mass = mass of H × mass of N × 3 × mass of O
= 1 × 14 × 3 × 16 = 672 g
no. of moles = given mass / molar mass = 697 / 672 = 1.03 moles
Hope this helps
plz mark as brainliest!!!!!!
Quick electron emissions are called
Which temperature is warmer 0° F or 0°C
Answer:
0 degree C
Explanation:
0 degree C = 32 degree F
0 degree F = -17.7778 degree C
Answer:
0 degrees C would be warmer
Explanation:
What is the mass in grams of 8 moles of Cu?
Answer:
mass = 508 g
Explanation:
Given data:
Number of moles of Cu = 8 mol
Mass in gram = ?
Solution:
Number of moles = mass/molar mass
Molar mass of Cu is 63.5 g/mol.
Now we will put the values in formula.
8 mol = mass /63.5 g/mol
mass = 63.5 g/mol ×8 mol
mass = 508 g
what element has a higher ionization energy carbon or silicon
which objects would have a greater gravitational force between them, Objects A and B, or Objects B and C? HELPPP
A piece of metal has a mass of 0.650 kilograms, has a width of 0.136 meters, and has a length of 0.0451 meters.Part A: If the metal’s volume is 291 cm3, what is the height of the metal in centimeters? (The width & length values given above are in a different unit!)
Part B: What is the density of this piece of metal?
Answer:
height = 4.74 cm
density = 2.23 g/ cm³
Explanation:
Mass of metal = 0.650 kg (650 g)
Width = 0.136 m
Length = 0.0451 m
Volume of metal = 291 cm³
Height in cm = ?
density of metal =?
Solution:
Width = 0.136 m (0.136 m×100 cm/1m = 13.6 cm)
Length = 0.0451 m (0.0451 m×100 cm/1m = 4.51 cm)
First of all we will calculate the height:
Volume = height× width× length
291 cm³ = h × 13.6 cm × 4.51 cm
291 cm³ = h × 61.34 cm²
h = 291 cm³ / 61.34 cm²
h = 4.74 cm
Density:
d = m/v
d = 650 g/291 cm³
d = 2.23 g/ cm³
Can you guys answer question 4 on new substance for Chemistry tysm
The answer is a bevasue it then becomes a chemical compound
Answer:
a
Explanation:
this would result in a compound and compounds are chemical changes so i think im right....
For the combustion of methane presented in Example 5.4, the chemical reaction is CH4 +2O2 →CO2 +2H2O Suppose that methane flows into a burner at 30 gmol/s, while oxygen flows into the same burner at 75 gmol/s. If all the meth- ane is burned and a single output stream leaves the burner, what is the mole fraction of CO2 in that output stream? Hint 1: Does the fact that all the methane is burned mean that all the oxygen is burned also? Hint 2: Find the molar flow rate of each component gas in the outlet gas ("flue gas").
Answer:
[tex]x_{CO_2}^{out} =0.25[/tex]
Explanation:
Hello.
In this case, for the reactive scheme, it is very convenient to write each species' mole balance as shown below:
[tex]CH_4:f_{CH_4}^{out}=f_{CH_4}^{in}-\epsilon \\\\O_2:f_{O_2}^{out}=f_{O_2}^{in}-2\epsilon\\\\CO_2:f_{CO_2}^{out}=\epsilon\\\\H_2O:f_{H_2O}^{out}=2\epsilon[/tex]
Whereas [tex]\epsilon[/tex] accounts for the reaction extent. However, as all the methane is consumed, from the methane balance:
[tex]0=f_{CH_4}^{in}-\epsilon \\\\\epsilon=30gmol/s[/tex]
Thus, we can compute the rest of the outlet mole flows since not all the oxygen is consumed as it is in excess:
[tex]f_{O_2}^{out}=f_{O_2}^{in}-2\epsilon=75gmol/s-2*30gmol/s=15gmol/s\\\\f_{CO_2}^{out}=15gmol/s\\\\f_{H_2O}^{out}=2*15gmol/s=30gmol/s[/tex]
It means that the mole fraction of carbon dioxide in that output is:
[tex]x_{CO_2}^{out}=\frac{15}{15+15+30} =0.25[/tex]
Best regards.
True or false this model represents an ion
Answer:
False
Explanation:
From the question given above, the following data were obtained:
Electron number = 6
Proton number = 6
Neutron number = 6
From the question given above, we can see clearly that the model has the same number of protons and electrons. Hence the model is not an ion.
This can further be explained if we determine the charge.
Electron number = 6
Proton number = 6
Charge =?
Charge = Proton – Electron
Charge = 6 – 6
Charge = 0
Since the charge is zero, the model is not an ion.
A 10.0 cm 3sample of copper has a mass of 89.6 g. What is the density of copper?
Answer:
19.3 g/cm3
Explanation:
A 10.0 cm3 sample of copper has a mass of 89.6 g. What is the density of copper? 19.3 g/cm3.
The energy required to ionize boron is 801 kJ/mol. You may want to reference (Pages 93 - 98) Section 2.5 while completing this problem. Part A What minimum frequency of light is required to ionize boron
Answer:
The frequency is [tex]f = 2,01 * 10^{15} \ Hz [/tex]
Explanation:
From the question we are told that
The energy required to ionize boron is [tex]E_b = 801 KJ/mol[/tex]
Generally the ionization energy of boron pre atom is mathematically represented as
[tex]E_a = \frac{E_b}{N_A}[/tex]
Here [tex]N_A[/tex] is the Avogadro's constant with value [tex]N_A = 6.022*10^{23}[/tex]
So
[tex]E_a = \frac{801}{6.022*10^{23}}[/tex]
=> [tex]E_a = 1.330 *10^{-18} \ J/atom [/tex]
Generally the energy required to liberate one electron from an atom is equivalent to the ionization energy per atom and this mathematically represented as
[tex]E = hf = E_a[/tex]
=> [tex] hf = E_a[/tex]
Here h is the Planks constant with value [tex]h = 6.626 *10^{-34}[/tex]
So
[tex]f = \frac{1.330 *10^{-18}}{ 6.626 *10^{-34}}[/tex]
=> [tex]f = 2,01 * 10^{15} \ Hz [/tex]
How many atoms are in 64 g sulfur (S)?
O A. 6.02 x 1023 atoms
OB. 646.02 x 1023) atoms
O C. 32(6.02 x 1023) atoms
D. 2(6.02 x 1023) atoms
Answer:
D) 2( 6.02 ×10^23)
Explanation:
using n=m
M
where n = moles
m=mass of substance
M=molar mass .
n( sulfur ) =64
32
= 2
using N =nL
where N is number of entities
n = moles
L= 6.02×20^23
N( of atoms of sulfur) =2(6.02×10^23)
32(6.02 x 10²³) number of atoms are in 64 g sulfur (S) is:
C. 32(6.02 x 10²³) atoms
To determine the number of atoms in 64 grams of sulfur (S), you need to use Avogadro's number, which is approximately 6.02 x 10²³ atoms/mol. This number represents the number of atoms or molecules in one mole of a substance.
1. Find the molar mass of sulfur (S):
The molar mass of sulfur (S) is the mass of one mole of sulfur atoms, and you can find it on the periodic table. The molar mass of sulfur is approximately 32.06 g/mol.
2. Calculate the number of moles of sulfur:
To calculate the number of moles, divide the given mass (64 g) by the molar mass of sulfur (32.06 g/mol):
Number of moles = Mass of sulfur / Molar mass of sulfur
Number of moles = 64 g / 32.06 g/mol
Number of moles = 1.997 moles
3. Use Avogadro's number to find the number of atoms:
Now, multiply the number of moles by Avogadro's number:
Number of atoms = Number of moles × Avogadro's number
Number of atoms = 1.997 moles × 6.02 x 10²³ atoms/mol
Number of atoms = 1.202 x 10²⁴ atoms
So, there are approximately 1.202 x 10²⁴ atoms in 64 grams of sulfur (S). The closest option to this value is:
C. 32(6.02 x 10²³) atoms (which is equivalent to 1.924 x 10²⁴ atoms)
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