Answer:
2.70 g of glucose.
Explanation:
The following data were obtained from the question:
Molarity of the glucose solution = 0.1 M
Volume of solution = 0.15 L
Molar mass of glucose = 180.16 g/mol
Mass of glucose =.?
Next, we shall determine the number of mole of glucose in the solution. This can be obtained as follow:
Molarity of the glucose solution = 0.1 M
Volume of solution = 0.15 L
Mole glucose =?
Molarity = mole /Volume
0.1 = Mole of glucose /0.15
Cross multiply
Mole of glucose = 0.1 × 0.15
Mole of glucose = 0.015 mole
Finally, we shall determine the mass of glucose in the solution as follow:
Mole of glucose = 0.015 mole
Molar mass of glucose = 180.16 g/mol
Mass of glucose =.?
Mole = mass /molar mass
0.015 = mass of glucose /180.16
Cross multiply
Mass of glucose = 0.015 × 180.16
Mass of glucose = 2.70 g
Therefore, the solution contains 2.70 g of glucose.
The grams that the solution must contain is :
- 2.70 g of glucose.
Mole ConceptGiven:
Molarity of the glucose solution = 0.1 M
Volume of solution = 0.15 L
Molar mass of glucose = 180.16 g/mole
Molarity of the glucose solution = 0.1 M
Volume of solution = 0.15 L
Molarity = mole /Volume
0.1 = Mole of glucose /0.15
Mole of glucose = 0.1 × 0.15
Mole of glucose = 0.015 mole
Mole of glucose = 0.015 mole
Molar mass of glucose = 180.16 g/mol
Mass of glucose =.?
Mole = mass /molar mass
0.015 = mass of glucose /180.16
Mass of glucose = 0.015 × 180.16
Mass of glucose = 2.70 g
The solution contains 2.70 g of glucose.
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trioxocarbonate iv acid
Answer:
Trioxocarbonate (iv) are salts derived from trioxocarbonate (iv) acid when it reacts with metals and metallic oxides. All trioxocarbonate (iv) salts of Sodium (Na), Potassium (K) and Ammonium(NH4+) are soluble while all others are insoluble.
How many moles are in 1.29 x 1024 hydrogen atoms in HF
Answer:
No. Moles = 2.142 mole
When a solution of ammonium sulfate is added to a
solution of lead(II) nitrate, Pb(NO3)2, a white
precipitate, lead(II) sulfate, forms. Write a balanced
net ionic equation for this reaction. Both ammonium
sulfate and lead(II) nitrate, Pb(NO3)2 exist as
dissociated ions in aqueous solution.
Answer:
Explanation:
(NH4)2SO4+Pb(NO3)2...............PbSO4+2NH4NO3
The balanced net ionic equation for the reaction between ammonium sulfate (NH₄)2SO₄ and lead(II) nitrate Pb(NO₃)₂, resulting in the formation of lead(II) sulfate (PbSO₄), can be written as follows:
[tex]2NH_{4+} + SO_{4}^2- + Pb^{2+} + NO_{3-} + PbSO_{4} (s) + 2NH_{4+} + 2NO_{3-}[/tex]
In this equation, the ammonium cation (NH₄₊) and nitrate anion (NO₃₋) are spectator ions, meaning they appear on both sides of the equation and do not participate in the actual chemical reaction. Therefore, they are not included in the net ionic equation.
The net ionic equation represents only the species that undergo a change, which in this case is the sulfate anion (SO₄²⁻) and the lead(II) cation (Pb²⁺), leading to the formation of the white precipitate, lead(II) sulfate (PbSO₄).
Therefore, the balanced net ionic equation can be wriiten as:
[tex]2NH_{4+} + SO_{4}^2- + Pb^{2+} + NO_{3-} + PbSO_{4} (s) + 2NH_{4+} + 2NO_{3-}[/tex]
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What is O?
(A)Metal
(B)Metalloid
(C)Nonmetal
(D)None of the above
Answer:nonmetal
Explanation: gggg
CH4+O2—-CO2+H2O what is the best classification for the unbalanced equations reaction and why
What is the representative particle for boron?
The molecule is the representative particle of molecular compounds. It is also the representative particle of diatomic elements.
I hope this helps you, have a beautiful day ✨
hlo good evening everyone
Answer:
hi
Explanation:
Hello good evening' but for me it's a good night
Explanation:
Anyway, thank you for the points have a great day!
-1
1 point
Which is not a force that causes erosion and deposition?
carbonation
1
glaciers
waves
2
water
3
Which of the following is an example of sp3 d2 hybridization?
A. C₂H6
B. PC15
C. 1F7
D. SFO
A chemistry graduate student is given 250.mL of a 1.60M chlorous acid HClO2 solution. Chlorous acid is a weak acid with =Ka×1.1 x 10^2. What mass of NaClO2 should the student dissolve in the HClO2 solution to turn it into a buffer with pH =1.45?
Answer:
28.04g of NaClO2 must be dissolved
Explanation:
When NaClO2 and HClO2 are mixed in a solution, a buffer is produced. The pH of this buffer is obtained using H-H equation:
pH = pKa + log [NaClO2] / [HClO2]
Where pKa is -log Ka = 1.9586
And [NaClO2] and [HClO2] could be taken as moles of each compound.
Moles of HClO2 are:
250mL = 0.250L * (1.60mol / L) = 0.40 moles HClO2.
Replacing:
pH = pKa + log [NaClO2] / [HClO2]
1.45 = 1.9586 + log [NaClO2] / [0.40mol]
0.31mol = [NaClO2]
To convert these moles to mass we need to use molar mass of NaClO2: 90.44g/mol:
0.31mol * (90.44g / mol) =
28.04g of NaClO2 must be dissolved
Calculate the pH at 25 degrees celsius of a 0.39 M solution of pyridinium chloride (c5h5nhcl) . Note that pyridine (c5h5n) is a weak base with a pkb of 8.77 .
Round your answer to 1 decimal place.
Answer:
[tex]pH=2.3[/tex]
Explanation:
Hello!
In this case, since pyridinium chloride has a pKb of 8.77 which is a Kb of 1.70x10⁻⁹ and therefore a Ka of 5.89x10⁻⁵ which means it tends to be acidic, we write its ionization via:
[tex]C_5H_5NHCl(aq)+H_2O(l)\rightleftharpoons C_5H_5NCl^-(aq)+H_3O^+(aq)[/tex]
Because it is a Bronsted base which donates one hydrogen ion to the water to produce hydronium. Thus, we write the equilibrium expression with the aqueous species only:
[tex]Ka=\frac{[C_5H_5NCl^-][H_3O^+]}{[C_5H_5NHCl]}[/tex]
In terms of the reaction extent [tex]x[/tex], we write:
[tex]5.89x10^{-5}=\frac{x*x}{0.39-x}[/tex]
Thus, solving for [tex]x[/tex] we obtain:
[tex]x_1=-0.0048M\\\\x_2=0.0048M[/tex]
Clearly the solution is 0.0048 M because to negative values are not allowed, therefore, since it equals the concentration of hydronium which defines the pH, we write:
[tex]pH=-log([H_3O^+])=-log(0.0048)\\\\pH=2.3[/tex]
Best regards!
A mixture of nitrogen and xenon gases contains nitrogen at a partial pressure of 417 mm Hg and xenon at a partial pressure of 427 mm Hg. What is the mole fraction of each gas in the mixture
Answer:
0.4941, 0.5059
Explanation:
[tex]P_N[/tex] = Partial pressure of nitrogen = 417 mm Hg
[tex]P_{Xe}[/tex] = Partial pressure of xenon = 427 mm Hg
Total pressure in the system is given by
[tex]P=P_N+P_{Xe}\\\Rightarrow P=417+427=844\ \text{mm Hg}[/tex]
Mole fraction is given by
[tex]X_N=\dfrac{P_N}{P}\\\Rightarrow X_N=\dfrac{417}{844}\\\Rightarrow X_{N}=0.4941[/tex]
For xenon
[tex]X_{Xe}=1-0.4941=0.5059[/tex]
or
[tex]X_{Xe}=\dfrac{P_{Xe}}{P}\\\Rightarrow X_{Xe}=\dfrac{427}{844}\\\Rightarrow X_{Xe}=0.5059[/tex]
So, mole fraction of nitrogen is 0.4941 and xenon is 0.5059.
Given 700 ml of oxygen at 7 ºC and 80.0 cm Hg pressure, what volume does it take at 27 ºC and 50.0 cm Hg pressure?
Answer:
I think that it might be 2.7
Explanation:
A 11.1-g sample of granite initially at 76.0°C is immersed into 22.0 g of water initially at 22.0°C. What is the final temperature of both substances when they reach thermal equilibrium? (For water, Cs=4.18J/g⋅∘C and for granite, Cs=0.790J/g⋅∘C.)
Answer:
[tex]T_f=26.7\°C[/tex]
Explanation:
Hello.
In this case, when two substances at different temperature are placed in contact in an isolated container, we can say that the heat lost by the hot substance is gained by the cold substance. In such a way, since granite is at 76.0 °C and water at 22.0 °C we infer granite is hot and water is cold, so we write:
[tex]Q_{granite}=-Q_{water}[/tex]
In terms of mass, specific heat and change in temperature, we write:
[tex]m_{granite}C_{granite}(T_f-T_{granite})=-m_{water}C_{water}(T_f-T_{water})[/tex]
Thus, since the temperature is the same for both substance, we can solve for it as shown below:
[tex]T_f=\frac{m_{granite}C_{granite}T_{granite}+m_{water}C_{water}T_{water}}{m_{granite}C_{granite}+m_{water}C_{water}}[/tex]
By plugging in each variable, we obtain:
[tex]T_f=\frac{11.1g*0.790\frac{J}{g\°C} *76.0\°C+22.0g*4.18\frac{J}{g\°C} *22.0\°C}{11.1g*0.790\frac{J}{g\°C} +22.0g*4.18\frac{J}{g\°C}}\\\\T_f=26.7\°C[/tex]
Best regards!
Which structure could a scientist look for in a plant that would identify it as a club moss rather than a liverwort?
O phloem
O spores
O rhizoids
O flowers
Answer:
what e said
Explanation:
how many moles of nitroge are there in 50.0 g of nitrogen?
1.785714286 moles
The number of moles (n) for nitrogen is: [ n=50.0÷28.0 ] = 1.785714286 moles.
When do chemical reactions occur?
Answer:
When the molecular composition changes.
Explanation:
A physical change is a change in size or appearance that can usually be reversed; a chemical change cannot be reversed easily and changes the atomic composition of something.
Networks can be made of bonded ____________________ or ____________________.
The pH of a solution prepared by mixing 50.0 mL of 0.125 M KOH and 50.0 mL of 0.125 M HCl is ________.
Answer:
7
Explanation:
This question is a neutralization reaction.
50 ml of 0.125 M of KOH = (50 x 0.125) = 6.25 ml
In this same way, 50 ml of 0.125M of HCl = 50x0.125
= 6.25 ml
Then KOH is going to neutralise the HCl fully suvh that the pH of the medium is going to be 7
(H+) + (OH-) = H2O
Then Kw = [H=][OH-]
= [H+]² = 10^-14
Such that [H+] = 10^-7
The pH would be = -log [H+]
= -log[10^-7]
When inputted on a calculator this gives us 7
= 7
This is our answer
.
As you move from left to right, the atomic number of each element increases
by
Answer:
Ionization Energy increases from left to right on the periodic table. Atomic radius decreases moving left to right on the periodic table.
Atomic size gradually decreases from left to right across a period of elements. This is because, within a period or family of elements, all electrons are added to the same shell.
compare the size of I, I+ and I-
umm explanation pls so i answee
Water can be an atom.
True
False
true because I asked my sister
Answer:
false
Explanation:
yep
When carbon dioxide is dissolved in an aqueous solution of sodium hydroxide, the mixture reacts to yield aqueous sodium carbonate and liquid water. Determine the net ionic equation for this process.
Answer:
CO + 2NaOH(aq) ---------- Na2CO3 + H2O
When carbon dioxide is dissolved in an aqueous solution of sodium hydroxide, the mixture reacts to yield aqueous sodium carbonate and liquid water. The net ionic equation is CO + [tex]2OH^{-}[/tex] → [tex]CO^{2-} _{3}[/tex]
What is the net ionic equation?The molecular formula for a process known as the net ionic equation only includes the species which are really involved in the reaction. In double displacement reactions, and redox reactions, including acid-base neutralization processes, the net ionic equation would be frequently employed.
When carbon dioxide is dissolved in an aqueous solution of sodium hydroxide, the mixture reacts to yield aqueous sodium carbonate and liquid water.
CO + 2NaOH → [tex]Na_{2} CO_{3} + H_{2}O[/tex]
The ionic equation can be written as:
CO + [tex]2Na^{+} + 2OH^{-}[/tex] → [tex]2Na^{+} + CO^{2-} _{3}[/tex]
Hence, the net ionic equation is CO + [tex]2OH^{-}[/tex] → [tex]CO^{2-} _{3}[/tex]
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What isotope does the model show? ASAP
1. lithium-6
2. lithium-7
3. beryllium-7
4. beryllium-4
Answer:
Lithium 7
Explanation:
An isotope is based off the number of neutrons present 3+4=7
Look on the periodic table to find the element number
Calculate the mass percentage of NaClNaCl in a solution containing 1.50 gg of NaClNaCl in 50.0 gg of water.
Answer:
%m/m = 2.9 %
Explanation:
The mass percentage (m/m %) of a solution is calculated as follows:
m/m % = mass of solute/ mass of solution x 100
The solution is composed by : solute + solvent. In this case, the solute is NaCl and the solvent is water. We have:
mass of solute = 1.50 g
mass of solvent = 50.0 g
mass of solution = mass of solute + mass of solvent = 1.50 g = 50.0 g = 51.5 g
Thus, the mass percentage of NaCl in the solution is :
% m/m = (1.50 g)/(51.5 g) x 100 = 2.91 % ≅ 2.9 %
need help thx if you take the time to help me out
Answer:
A is the only neutral atom there.
Explanation:
It consists of 6 atoms 6 neutrons and 6 electrons which means it is neutral. I am currently in chemistry.
Production of 6.5 grams of C2H2 requires consumption of how many grams of H2O
Answer:
C
Explanation:
Grams 3/4
Answer:
9g
Explanation:
number of moles × the moleculars mass
who wants these points
Answer:
GRACIAS
Explanation:
Answer:
Thnx
Explanation:
what the equipment of diffusion?
Answer:
liquid
a semi permeable membrane
oxygen
Explanation:
plz give me brainiest
How many grams are in 1.50 moles of KMnO4?
A. 165.06
B. 660.24
C. 178.41
D. 237.06
E. 484.29
Explanation:
mole=mass/Molar mass
mass= mole×molar mass
m= 1.5 × (39+55+64)
m= 1.5 × 158
m= 237g
There are 237.06 grams in 1.50 moles of KMnO4.
HOW TO CALCULATE MASS:
The mass of a substance can be calculated by multiplying the number of moles in the substance by its molar mass. That is;mass of KMnO4 = no. of moles × molar mass of KMnO4Molar mass of KMnO4 = 39 + 55 + 16(4)= 158g/mol
mass of KMnO4 = 158g/mol × 1.5molMass of KMnO4 = 237gTherefore, there are 237.06 grams in 1.50 moles of KMnO4.Learn more at: https://brainly.com/question/15743584?referrer=searchResults
