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ruslelena [56]
2 years ago
15

1.Calculate the mass of compound required to prepare 100.00 mL of a 1.40x10-2M CuCl2.Show your work.

Chemistry
1 answer:
Verdich [7]2 years ago
5 0

The concentration of the diluted solution is 0.0625 M.

<h3>What is concentration?</h3>

The term concentration refers to the amount of substance present in solution.

1) To find the mass of the compound;

mass/134 g/mol = 1.40x10-2M × 100/1000

mass = 1.40x10-2M × 100/1000 × 134 g/mol

mass = 0.19 g

2) To find the molarity of the solution;

2.50 g/40 = 100/1000 × M

M = 2.50 g/40 × 1000/100

M = 0.625 M

3) Using;

C1V1 = C2V2

0.625 M × 10.00mL = C2 × 100.00 mL

C2 = 0.625 M × 10.00mL/ 100.00 mL

C2 = 0.0625 M

Learn more about concentration: brainly.com/question/3045247

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A rectangle has a volume of 395cm3. It has a mass of 147g. What is its density? *
LekaFEV [45]

The density of a rectangle : ρ = 0.372 g/cm³

<h3>Further explanation</h3>

Given

The volume of rectangle : 395 cm³

Mass : 147 g

Required

The density

Solution

Density is a quantity derived from the mass and volume  

Density is the ratio of mass per unit volume  

Density formula:  

\large {\boxed {\bold {\rho ~ = ~ \frac {m} {V}}}}

ρ = density  

m = mass  

v = volume

Input the value :

ρ = 147 g : 395 cm³

ρ = 0.372 g/cm³

6 0
3 years ago
How did scientists solve the nitrogen problem?
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Unfortunately, we have not fully solved the 'nitrogen problem'. To do this, we must halve the amount of nitrogen we dump into the environment by mid-century or our ecosystems will face epidemics of toxic tides, lifeless rivers, and dead oceans. And that to do that will require, among other things, almost doubling the efficiency of nitrogen use on the world’s farms.

3 0
3 years ago
A 7.0 g sample of a hydrocarbon (a molecule that has only hydrogen and carbon) is subject to combustion analysis. The mass of CO
Akimi4 [234]

Answer: The empirical formula for the given compound is CH_2

Explanation:

The chemical equation for the combustion of compound having carbon and hydrogen follows:

C_xH_y+O_2\rightarrow CO_2+H_2O

where, 'x' and 'y' are the subscripts of carbon and hydrogen respectively.

We are given:

Mass of CO_2=22.0g

We know that:

Molar mass of carbon dioxide = 44 g/mol

For calculating the mass of carbon:

In 44 g of carbon dioxide, 12 g of carbon is contained.

So, in 22.0 g of carbon dioxide, \frac{12}{44}\times 22.0=6g of carbon will be contained.

For calculating the mass of hydrogen:

Mass of hydrogen = Mass of sample - Mass of carbon

Mass of hydrogen = 7.0 g - 6 g

Mass of hydrogen = 1.0 g

To formulate the empirical formula, we need to follow some steps:

Step 1: Converting the given masses into moles.

Moles of Carbon =\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{6g}{12g/mole}=0.5moles

Moles of Hydrogen = \frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{1.0g}{1g/mole}=1.0moles

Step 2: Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 0.5 moles.

For Carbon = \frac{0.5}{0.5}=1

For Hydrogen  = \frac{1.0}{0.5}=2

Step 3: Taking the mole ratio as their subscripts.

The ratio of Fe : C : H = 1 : 2

Hence, the empirical formula for the given compound is C_{1}H_{2}=CH_2

4 0
3 years ago
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