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Blababa [14]
3 years ago
11

How to calculate from moles to grams? 4.0 moles of Cu(CN)2

Chemistry
1 answer:
andreyandreev [35.5K]3 years ago
3 0

Answer: The mass of given amount of copper (II) cyanide is 462.4 g

Explanation:

To calculate the number of moles, we use the equation:

We are given:

Moles of copper (II) cyanide = 4 moles

Molar mass of copper (II) cyanide = 115.6 g/mol

Putting values in above equation, we get:

Hence, the mass of given amount of copper (II) cyanide is 462.4 g

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Write a program that allows the user to enter a number of quarters, dimes, and nickels and then outputs the monetary value of th
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Answer:

// Program is written in C++ Programming Language

// Comments are used for explanatory purpose

// Program Starts here

#include<iostream>

using namespace std;

int main ()

{

// Declare Variables

int quarter, dimes, nickel, cent;

// Enter values for each

cout<<"Quarter: ";

cin>>quarter;

cout<<"Dimes: ";

cin>>dimes;

cout<<"Nickels: ";

cin>>nickel;

/*

In the United States, these coins have the following values

Quarter = 25 cents

Done = 10 cents

Nickel = 5 cent

Total cent is calculated below

*/

cent = 25 * quarter + 10 * dimes + 5 * nickel;

// Print Total

cout<<"The coins are worth "<<cent<<" cents";

return 0;

}

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3 years ago
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What is the water vapor pressure at 120 degrees f.
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28.815

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Calculate the molarity of a solution containing 15.2 grams of NaCl dissolved in 2.5 L of solution.
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8 0
3 years ago
During an experiment, 95 grams of calcium carbonate reacted with an excess amount of hydrochloric acid. If the percent yield of
almond37 [142]

Answer:

Actual yield: 86.5 grams.

Explanation:

How many moles of formula units in 95 grams of calcium carbonate \rm CaCO_3?

Refer to a modern periodic table for relative atomic mass data:

  • Ca: 40.078;
  • C: 12.011;
  • O: 15.999.

Formula mass of \rm CaCO_3:

M(\mathrm{CaCO_3})  = \underbrace{1\times 40.078}_{\rm Ca} + \underbrace{1\times 12.011}_{\rm C} + \underbrace{3\times 15.999}_{\rm O} = \rm 100.086\;g\cdot mol^{-1}.

\displaystyle n(\mathrm{CaCO_3}) = \frac{m(\mathrm{CaCO_3})}{M(\mathrm{CaCO_3})} = \rm \frac{95\;g}{100.086\;g\cdot mol^{-1}} = 0.949184\;mol.

How many moles of \rm CaCl_2 will be produced?

The coefficient in front of \rm CaCO_3 in the chemical equation is the same as that in front of \rm CaCl_2. That is:

\displaystyle \frac{n(\rm CaCl_2)}{n(\rm CaCO_3)} = 1.

\displaystyle n(\mathrm{CaCl_2}) = n(\mathrm{CaCO_3})\cdot \frac{n(\rm CaCl_2)}{n(\rm CaCO_3)} = n(\mathrm{CaCO_3}) = \rm 0.949184\;mol.

What's the theoretical yield of calcium chloride? In other words, what's the mass of \rm 0.949184\;mol of \rm CaCl_2?

Again, refer to a periodic table for relative atomic data:

  • Ca: 40.078;
  • Cl: 35.45.

M(\mathrm{CaCl_2}) = \underbrace{1\times 40.078}_{\rm Ca} + \underbrace{2\times 35.45}_{\rm Cl} = \rm 110.978\;g\cdot mol^{-1}.

\begin{aligned}m(\mathrm{CaCl_2}) &= n(\mathrm{CaCl_2})\cdot M(\mathrm{CaCl_2})\\ &= \rm 0.949184\;mol\times 110.978\;g\cdot mol^{-1}\\ &= \rm 105.339\; g\end{aligned}.

What's the actual yield of calcium chloride?

\displaystyle \text{Percentage Yield} = \frac{\text{Actual Yield}}{\text{Theoretical Yield}}\times 100\%.

\displaystyle \begin{aligned}\text{Actual Yield} &= \text{Theoretical Yield}\cdot \frac{\text{Percentage Yield}}{100\%}\\ &=\rm 105.339\; g \times \frac{82.15\%}{100\%}\\&= \rm 86.5\;g \end{aligned}.

8 0
3 years ago
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