What percentage of oxygen is attributed fossil fuel combustion

Coal can be used to generate hydrogen gas (a potential fuel) by this endothermic reaction. If this reaction mixture is at equili

VladimirAG [237]

The question ask to predict the effect where as coal can be used to generate hydrogen gas by this endothermic reason and the best answer among your choices based on the element you give is letter C. raising the temperature of the reaction mixture. I hope my answer is correct

8 0

3 years ago

How many grams are 3.01 × 1023 molecules of CuSO4?

zvonat [6]

Answer is: 79.8 grams of copper(II) sulfate.
N(CuSO₄) = 3.01·10²³; number of molecules.
n(CuSO₄) = N(CuSO₄) ÷ Na.
n(CuSO₄) = 3.01·10²³ ÷ 6.02·10²³ 1/mol.
n(CuSO₄) = 0.5 mol; amount of substance.
m(CuSO₄) = n(CuSO₄) · M(CuSO₄).
m(CuSO₄) = 0.5 mol · 159.6 g/mol.
m(CuSO₄) = 79.8 g; mass of substance.
M - molar mass.

4 0

3 years ago

If you mixed three Dixie cups’ contents together containing 0.05L of 1.0M lemonade, 0.05L of 2.5M lemonade, and 0.05L of 0.5M le

viva [34]

Molarity of the resulting solution will be 1.33 M.

<u>Explanation:</u>

First we have to find the number of moles for each of the solution using the formula, moles = molarity × volume

For cup 1 = 1 M ×0.05 L = 0.05 moles

For cup 2 = 2.5 M × 0.05 L= 0.125 moles

For cup 3 = 0.5 M × 0.05 L = 0.025 moles

Total moles = 0.05 + 0.125 + 0.025 = 0.2 moles

We have to find the total volume as, 0.05 + 0.05 + 0.05 = 0.15 L

Now we have to find the molarity as, moles / volume = 0.2 moles/ 0.15 L = 1.33 M

3 0

3 years ago

When 7.80 mL of 0.500 M AgNO3 is added to 6.25 mL of 0.300 M NH4Cl, how many grams of AgCl are formed?

irina1246 [14]

Answer:

The answer to your question is 0.269 grams of AgCl

Explanation:

Data

[AgNO₃] = 0.50 M

Vol AgNO₃ = 7.80 ml

[NH₄Cl] = 0.30 M

Vol NH₄Cl = 6.25 ml

mass of AgCL

Balanced reaction

AgNO₃(aq) + NH₄Cl(aq) ⇒ AgCl (s) + NH₄NO₃ (aq)

Process

1.- Calculate the moles of AgNO₃

Molarity = moles / volume

moles = Molarity x volume

moles = 0.50 x 0.0078

moles = 0.0039

2.- Calculate the moles of NH₄Cl

moles = 0.30 x 0.0063

moles = 0.00188

3.- Calculate the limiting reactant

The proportion of AgNO₃(aq) to NH₄Cl(aq) is 1 :1, then, we conclude that the limiting reactant is NH₄Cl(aq), because there are less amount of this reactant in the experiment.

4.- Calculate the moles of AgCl

1 mol of NH₄Cl ---------------- 1 mol of AgCl

0.00188 mol of NH₄Cl ------------- x

x = (0.00188 x 1) /1

x = 0.00188 moles of AgCl

5.- Calculate the grams of AgCl

molecular mass of AgCl = 108 + 35.5 = 143.5 g

143.5 grams of AgCl -------------- 1 mol

x -------------------------------------------0.00188 moles of AgCl

x = (0.00188 x 143.5) / 1

x = 0.269 grams of AgCl

8 0

3 years ago

Which pair shares the same empirical formula? C2H4 and C6H6 CH2 and CH4 CH3 and C2H6 CH and C2H4

hoa [83]

Answer:

CH3 and C2H6

Explanation:

4 0

3 years ago

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