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4 years ago

Using the standard enthalpies of formation, what is the standard enthalpy of reaction?

Chemistry

1 answer:

bulgar [2K]4 years ago

8 0

Answer:

Standard enthalpy for the given reaction is -41.166 kJ

Explanation:

Standard enthalpy of a reaction = $\sum [n_{i}\times \Delta H_{f}(product)_{i}]-\sum [n_{j}\times \Delta H_{f}(reactant)_{j}]$

where $n_{i}$ and $n_{j}$ are number of moles of i-th product and j-th reactant in balanced reaction respectively.

Hence Standard enthalpy for the given reaction = $[(1\times \Delta H_{f}(CO_{2})_{g})]+[(1\times \Delta H_{f}(H_{2})_{g})]-[(1\times \Delta H_{f}(CO)_{g})]-[(1\times \Delta H_{f}(H_{2}O)_{g})]$

So, Standard enthalpy for the given reaction = $[(1\times -393.509]+[(1\times 0]-[(1\times -110.525]-[(1\times -241.818]$ kJ = -41.166 kJ

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Which of the following is an example of pseudoscience?

Westkost [7]

Psychic experiences it the answer out of the 4

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4 years ago

Calculate the molarity of a solution consisting of 65.5 g of K2S0 4 in 5.00 L of solution.

tia_tia [17]

Answer:

Molarity is 0.075 M. 

Explanation:

Moles:

${ \tt{ = \frac{65.5}{RFM} }}$

RFM of potassium sulphate :

${ \tt{ = (39 \times 2) + 32 + (16 \times 4)}} \\ = 174 \: g$

substitute:

${ \tt{moles = \frac{65.5}{174} = 0.376 \: moles}}$

In volume of 5.00 l:

${ \tt{5.00 \: l = 0.376 \: moles}} \\ { \tt{1 \: l = ( \frac{0.376}{5.00} ) \: moles}} \\ { \tt{molarity = 0.075 \: mol \: l {}^{ - 1} }}$

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3 years ago

In order to prepare very dilute solutions, a lab technician chooses to perform a series of dilutions instead of measuring a very

NeTakaya

Answer: $5.70\times 10^{-6}M$

Explanation:

Molarity of a solution is defined as the number of moles of solute dissolved per Liter of the solution.

$Molarity=\frac{n\times 1000}{V_s}$

where,

n= moles of solute

Given : 0.360 g of $KNO_3$ is dissolved in 500 ml of solution.

$Moles=\frac{\text{Given mass}}{\text{Molar mass}}=\frac{0.360g}{101g/mol}=3.56\times 10^{-3}mole$

$V_s$ = volume of solution = 500 ml

$Molarity=\frac{3.56\times 10^{-3}\times 1000}{500}=7.12\times 10^{-3}M$

According to the neutralization law,

$M_1V_1=M_2V_2$

where,

$M_1$ = molarity of stock solution = $7.12\times 10^{-3}M$

$V_1$ = volume of stock solution = 10.0 ml

$M_2$ = molarity of diluted solution = ?

$V_2$ = volume of diluted solution = 500.0 ml

$7.12\times 10^{-3}M\times 10.0=M_2\times 500.0$

$M_2=1.42\times 10^{-4}M$

b) On further dilution

$M_1$ = molarity of stock solution = $1.42\times 10^{-4}M$

$V_1$ = volume of stock solution = 10.0 ml

$M_2$ = molarity of diluted solution = ?

$V_2$ = volume of diluted solution = 250.0 ml

$1.42\times 10^{-4}M\times 10.0=M_2\times 250.0$

$M_2=5.70\times 10^{-6}M$

Thus the final concentration of the $KNO_3$ solution is $5.70\times 10^{-6}M$

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3 years ago

"0.35 g of hydrogen chloride (hcl is dissolved in water to make 3.0 l of solution. what is the ph of the resulting hydrochloric

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Firstly calculating number of moles of hcl disssolved. The answer would be in the pic above. Concentration of hydrogen ions can be calculated by using that mole amount dividing by 3litres of solution to get 0.003196. Hence the ph of solution will be -log10(h+ concentration) which is equal to 2.495

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3 years ago

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valentinak56 [21]

C is the correct answer ( calcium)

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3 years ago