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djyliett [7]
3 years ago
8

Nault 25000L 250 our = 25000L 15040 8. How many grams of CaCl2 are needed to make 150.0 mL of a 0.500 M CF solution? (Note: CaCl

2 is a soluble salt. The molar mass of CaCl2 is 110.98 g/mol.)
Chemistry
1 answer:
yuradex [85]3 years ago
3 0

Answer:

You need 8,324 g of CaCl₂ yo make this solution

Explanation:

Molarity is a way to express concentration in a solution, in units of moles of solute per liter of solution.

To know the grams of CaCl₂ it is necessary to know, first, the moles of this substance with the desired volume and concentration , thus:

0,1500 L × \frac{0,500 mol}{L} = 0,075 CaCl₂ moles

Now, with the molar mass of CaCl₂ you will obtain the necessary grams, thus:

0,075 CaCl₂ moles  × \frac{110,98 g}{mol} = 8,324 g of CaCl₂

So, you need <em>8,324 g of CaCl₂</em> to make 150,0 mL of a 0,500M solution

I hope it helps!

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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
Convert 1.248×1010 g to each of the following units.
Leona [35]

Answer:

a) 1.248 x 10⁷ kg

b) 1.248 x 10⁴ Mg

c) 1.248 x 10¹³ mg

d) 1.248 x 10⁴ ton

Explanation:

a) Since 1000 g = 1 kg we can convert grams to kg by multiplyig any given quantity in grams by the conversion factor ( 1 kg / 1000 g):

1.248 x 10¹⁰ g * (1 kg / 1000 g) = 1.248 x 10⁷ kg

b) Since 1 Mg = 1 x 10⁶ g, the conversion factor will be ( 1 Mg / 1 x 10⁶ g):

1.248 x 10¹⁰ g * (  1 Mg / 1 x 10⁶ g) = 1.248 x 10⁴ Mg

c) Since 1 mg = 1 x 10⁻³ g, the conversion factor will be ( 1 mg / 1 x 10⁻³ g):

1.248 x 10¹⁰ g ( 1 mg / 1 x 10⁻³ g) = 1.248 x 10¹³ mg

d) Since 1 metric ton = 1000 kg and 1000 g = 1 kg, we can use these conversions factors: ( 1 kg / 1000 g) and (1 ton / 1000 kg):

1.248 x 10¹⁰ g * ( 1 kg / 1000 g) * ( 1 ton / 1000 kg) = 1.248 x 10⁴ ton

8 0
2 years ago
How many atoms of hydrogen-1 are in a 1.007-g sample?
Greeley [361]
I think it's 2 I tried looking it up because I was not sure.
8 0
3 years ago
Calculate the equilibrium constant k for the isomerization of glucose-1-phosphate to fructose-6-phosphate at 298 k. express your
k0ka [10]
We cannot solve this problem without using empirical data. These reactions have already been experimented by scientists. The standard Gibb's free energy, ΔG°, (occurring in standard temperature of 298 Kelvin) are already reported in various literature. These are the known ΔG° for the appropriate reactions.

<span>glucose-1-phosphate⟶glucose-6-phosphate          ΔG∘=−7.28 kJ/mol
fructose-6-phosphate⟶glucose-6-phosphate          ΔG∘=−1.67 kJ/mol
</span>
Therefore, the reaction is a two-step process wherein glucose-6-phosphate is the intermediate product.

glucose-1-phosphate⟶glucose-6-phosphate⟶fructose-6-phosphate 

In this case, you simply add the ΔG°. However, since we need the reverse of the second reaction to end up with the terminal product, fructose-6-phosphate, you'll have to take the opposite sign of ΔG°.

ΔG°,total = −7.28 kJ/mol  + 1.67 kJ/mol = -5.61 kJ/mol

Then, the equation to relate ΔG° to the equilibrium constant K is

ΔG° = -RTlnK, where R is the gas constant equal to 0.008317 kJ/mol-K.
-5.61 kJ./mol = -(0.008317 kJ/mol-K)(298 K)(lnK)
lnK = 2.2635
K = e^2.2635
K = 9.62


6 0
3 years ago
Read 2 more answers
What element does gold identity as
ICE Princess25 [194]
Gold is a chemical element with symbol Au (from Latin: aurum) and atomic number 79, making it one of the higher atomic number elements that occur naturally. In its purest form, it is a bright, slightly reddish yellow, dense, soft, malleable, and ductile metal.
8 0
2 years ago
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