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snow_lady [41]
3 years ago
5

What is the mass of ethyl alcohol that fills a 250.0-mL graduated cylinder? The density of ethyl alcohol is 0.789 g/mL.

Chemistry
2 answers:
solong [7]3 years ago
4 0

Answer:

\boxed {\tt 197.25 \ grams }

Explanation:

The density formula is:

d=\frac{m}{v}

Rearrange the formula for m, the mass. Multiply both sides of the equation by v.

d*v=\frac{m}{v} *v

d*v=m

Mass can be found by multiplying the density and volume. The density of ethyl alcohol is 0.789 grams per milliliter. The alcohol fills a 250 milliliter graduated cylinder, so the volume is 250 milliliters.

d= 0.789 \ g/mL \\v= 250 mL

Substitute the values into the formula.

0.789 \ g/mL * 250 \ mL=m

Multiply. Note that the milliliters, or mL will cancel each other out.

0.789 \ g * 250=m

197.25 \ g=m

The mass of the ethyl alcohol is 197.25 grams.

UNO [17]3 years ago
3 0

Answer: 157.8 g

Explanation: I hope this help my child UwU

What is the weight of the ethyl alcohol that exactly fills a 200.0 mL container the density of ethyl alcohol is 0.789 g ml?

Density review

A

B

What is the weight of the ethyl alcohol that exactly fills a 200.0 mL container? The density of ethyl alcohol is 0.789 g/mL.

g = (0.789 g/mL) (200.0 mL) = 158 g

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<u>Answer:</u> The standard heat for the given reaction is -138.82 kJ

<u>Explanation:</u>

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The equation used to calculate enthalpy change is of a reaction is:

\Delta H^o_{rxn}=\sum [n\times \Delta H_f_{(product)}]-\sum [n\times \Delta H_f_{(reactant)}]

For the given chemical reaction:

4CH_3NH_2(g)+2H_2O(l)\rightarrow 3CH_4(g)+CO_2(g)+4NH_3(g)

The equation for the enthalpy change of the above reaction is:

\Delta H_{rxn}=[(3\times \Delta H_f_{(CH_4(g))})+(1\times \Delta H_f_{(CO_2(g))})+(4\times \Delta H_f_{(NH_3(g))})]-[(4\times \Delta H_f_{(CH_3NH_2(g))})+(2\times \Delta H_f_{(H_2O(l))})]

We are given:

\Delta H_f_{(H_2O(l))}=-285.8kJ/mol\\\Delta H_f_{(NH_3(g))}=-46.1kJ/mol\\\Delta H_f_{(CH_4(g))}=-74.8kJ/mol\\\Delta H_f_{(CO_2(g))}=-393.5kJ/mol\\\Delta H_f_{(CH_3NH_2(g))}=-22.97kJ/mol

Putting values in above equation, we get:

\Delta H_{rxn}=[(3\times (-74.8))+(1\times (-393.5))+(4\times (-46.1))]-[(4\times (-22.97))+(2\times (-285.8))]\\\\\Delta H_{rxn}=-138.82kJ

Hence, the standard heat for the given reaction is -138.82 kJ

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