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2 years ago

A student sets up the following equation...

Chemistry

1 answer:

Sphinxa [80]2 years ago

3 0

Answer:

The answer to your question is given below

Explanation:

From the question given above, the following data were obtained:

Concentration (g/mL) = 0.04 g/mL

Concentration (g/L) =?

We can convert g/mL to g/L by doing the following:

1 g/mL = 1000 g/L

Therefore,

0.04 g/mL = 0.04 g/mL × 1000 g/L / 1 g/mL

0.04 g/mL = 40 g/L

0.04 g/mL is equivalent to 40 g/L.

Thus, the concentration (g/L) is 40 g/L

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This is your answer:

-13 degrees C

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Please help! Four different people are using sound wave technology. Which describes the form of technology most likely being use

Andru [333]

Answer:

The correct answer is option D

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2 years ago

Consider the following equation: SiO2 (s) + 3C (graphite) --> SiC (s) + 2CO (g) ΔH rxn = 624.6 kJ / mol rxn. Using the follow

DerKrebs [107]

<u>Answer:</u> The enthalpy of the formation of $SiC(s)$ is coming out to be -65.3 kJ/mol

<u>Explanation:</u>

Enthalpy change is defined as the difference in enthalpies of all the product and the reactants each multiplied with their respective number of moles. It is represented as $\Delta H^o$

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

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

For the given chemical reaction:

$SiO_2(s)+3C\text{ (graphite)}(s)\rightarrow SiC(s)+2CO(g)$

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

$\Delta H^o_{rxn}=[(1\times \Delta H^o_f_{(SiC(s))})+(2\times \Delta H^o_f_{(CO(g))})]-[(1\times \Delta H^o_f_{(SiO_2(s))})+(3\times \Delta H^o_f_{(C(s))})]$

We are given:

$\Delta H^o_f_{(CO(g))}=-110.5kJ/mol\\\Delta H^o_f_{(SiO_2(s))}=-910.9kJ/mol\\\Delta H^o_f_{(C(s))}=0kJ/mol\\\Delta H^o_{rxn}=624.6kJ$

Putting values in above equation, we get:

$624.6=[(1\times \Delta H^o_f_{(SiC(s))})+(2\times (-110.5))]-[(1\times (-910.9))+(3\times (0))]\\\\\Delta H^o_f_{(SiC(s))}=-65.3kJ/mol$

Hence, the enthalpy of the formation of $SiC(s)$ is coming out to be -65.3 kJ/mol.

8 0

3 years ago

The ka of benzoic acid is 6.30 ⋅ 10-5. The ph of a buffer prepared by combining 50.0 ml of 1.00 m potassium benzoate and 50.0 ml

Free_Kalibri [48]

<h3><u>Answer;</u></h3>

pH = 4.20

<h3><u>Explanation;</u></h3>

pKa = -log(6.30 × 10^-5)

pKa = 4.20

Moles of Benzoic acid = volume × molarity

= 0.050L × 1.00M

= 0.050moles benzoic acid

Moles of the salt = 0.050L × 1.00M

= 0.050 moles salt

Therefore;

0.050mols / 0.1 L = 0.50M

Thus;

pH = 4.20 + log(0.50/0.50)

pH = 4.20

6 0

3 years ago