Can someone help this is a final exam its really important

Determine the concentration of a solution made by dissolving 44.0 g of calcium chloride (CaCl2) in 0.30 L of solution. SHOW YOUR

ycow [4]

Answer:

[CaCl₂] = 1.32 M

Explanation:

We know the volume of solution → 0.30 L

We know the mass of solute → 44 g of CaCl₂

Let's convert the mass of solute to moles.

44 g . 1 mol / 110.98 g = 0.396 moles

Molarity (mol/L) → 0.396 mol / 0.3 L = 1.32 M

6 0

3 years ago

Ethyl butyrate, CH3CH2CH2CO2CH2CH3, is an artificial fruit flavor commonly used in the food industry for such flavors as orange

SIZIF [17.4K]

Answer:

A. 10.0 grams of ethyl butyrate would be synthesized.

B. 57.5% was the percent yield.

C. 7.80 grams of ethyl butyrate would be produced from 7.60 g of butanoic acid.

Explanation:

$CH_3CH_2CH_2CO_2H(l)+CH_2CH_3OH(l)+H^+\rightarrow CH_3CH_2CH_2CO_2CH_2CH_3(l)+H_2O(l)$

A

Moles of butanoic acid = $\frac{7.60 g}{88 g/mol}=0.08636 mol$

According to reaction ,1 mole of butanoic acid gives 1 mol of ethyl butyrate,then 0.08636 mol of butanoic acid will give :

$\frac{1}{1}\times 0.08636 mol=0.08636 mol$ of ethyl butyrate

Mass of 0.08636 moles of ethyl butyrate =

0.08636 mol × 116 g/mol = 10.0 g

Theoretical yield = 10.0 g

Experimental yield = ?

Percentage yield of the reaction = 100%

$Yield\%=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100$

$100\%=\frac{\text{Experimental yield}}{10.0 g}\times 100$

Experimental yield = 10.0 g

10.0 grams of ethyl butyrate would be synthesized.

B

Theoretical yield of ethyl butyrate = 10.0 g

Experimental yield ethyl butyrate = 5.75 g

Percentage yield of the reaction = ?

$Yield\%=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100$

$=\frac{5.75 g}{10.0 g}\times 100=57.5\%$

57.5% was the percent yield.

C

Moles of butanoic acid = $\frac{7.60 g}{88 g/mol}=0.08636 mol$

According to reaction ,1 mole of butanoic acid gives 1 mol of ethyl butyrate,then 0.08636 mol of butanoic acid will give :

$\frac{1}{1}\times 0.08636 mol=0.08636 mol$ of ethyl butyrate

Mass of 0.08636 moles of ethyl butyrate =

0.08636 mol × 116 g/mol = 10.0 g

Theoretical yield = 10.0 g

Experimental yield = ?

Percentage yield of the reaction = 78.0%

$Yield\%=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100$

$78.0\%=\frac{\text{Experimental yield}}{10.0 g}\times 100$

Experimental yield = 7.80 g

7.80 grams of ethyl butyrate would be produced from 7.60 g of butanoic acid.

8 0

3 years ago

Identify the reversible chemical reaction below. A. reactants=>products B. Products <=> reactants C. products => rea

Korvikt [17]

Answer:

Its B

Explanation:

Products <=> reactants; products can go back to reactant and reactants can form products, depending on the equilibrium position.

The equilibrium position, if it is on the left, and much of the product is formed. If the equilibrium position is not the right, then its the vice versa The equilibrium position is determined by ;

Temperature

If the reaction is endothermic, it will be favoured by increase in temperature and equilibrium position will shift to the right ( reactants )

If the reaction is exothermic, its the vice versa

NOTE: Only temperature affects the equilibrium position

5 0

3 years ago

2.2 Chromium has four naturally-occurring isotopes: 4.34% of 50Cr, with an atomic weight of 49.9460 amu; 83.79% of 52Cr, with an

shutvik [7]

Answer:

Average atomic mass = 51.9963 amu

Explanation:

Given data:

Abundance of Cr⁵⁰ with atomic mass= 4.34% , 49.9460 amu

Abundance of Cr⁵² with atomic mass = 83.79%, 51.9405 amu

Abundance of Cr⁵³ with atomic mass =9.50%, 52.9407 amu

Abundance of Cr⁵⁴ with atomic mass = 2.37%, 53.9389 amu

Average atomic mass = 51.9963 amu

Solution:

Average atomic mass = (abundance of 1st isotope × its atomic mass) +(abundance of 2nd isotope × its atomic mass +....n) / 100

Average atomic mass = (4.34×49.9460)+(83.79×51.9405) +(9.50×52.9407)+ (2.37×53.9389) / 100

Average atomic mass = 216.7656 + 4352.0945 + 502.9367 +127.8352 / 100

Average atomic mass = 5199.632 / 100

Average atomic mass = 51.9963 amu

4 0

3 years ago

What are two ways carbon returns from animal into the water

soldi70 [24.7K]

Split and merge into it. While they are alive, carbon returns from animals into water through waste products from respiration and defecation/urination. Another way when they are dead is from decaying remains. While they are alive, carbon returns from animals into water through waste products from respiration and defecation/urination.

Good enough?

8 0

3 years ago