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2 years ago

96 K = _____ -177°C 0°C 96°C 369°C

Chemistry

2 answers:

velikii [3]2 years ago

6 0

ºC = K - 273

ºC = 96 - 273

= -177ºC

answer A

hope this helps!

Inessa05 [86]2 years ago

3 0

Answer: Option (A) is the correct answer.

Explanation:

It is known that to convert temperature from degree celsius to kelvin we have to add 273 into the given temperature.

Whereas when we have to convert temperature from kelvin to degree celsius then we have to subtract 273.

So, convert 96 K into degree celsius as follows.

$(96 - 273)^{o}C$

= $- 177^{o}C$

Thus, we can conclude that 96 K equals $- 177^{o}C$ .

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3 years ago

Read 2 more answers

What is the molarity of a solution of 58.7 grams of MgCl2 in 359 ml of solution?

jekas [21]

Answer:

1.72 M

Explanation:

Molarity is the molar concentration of a solution. It can be calculated using the formula a follows:

Molarity = number of moles (n? ÷ volume (V)

According to the information provided in this question, the solution has 58.7 grams of MgCl2 in 359 ml of solution.

Using mole = mass/molar mass

Molar mass of MgCl2 = 24 + 35.5(2)

= 24 + 71

= 95g/mol

mole = 58.7g ÷ 95g/mol

mole = 0.618mol

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2 years ago

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2 years ago

Calculate the standard enthalpy change for the reaction at 25 ∘ 25 ∘ C. Standard enthalpy of formation values can be found in th

WINSTONCH [101]

<u>Answer:</u> The standard enthalpy change of the reaction is coming out to be -16.3 kJ

<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 $\Delta H$

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

$\Delta H_{rxn}=\sum [n\times \Delta H_f(product)]-\sum [n\times \Delta H_f(reactant)]$

For the given chemical reaction:

$Mg(OH)_2(s)+2HCl(g)\rightarrow MgCl_2(s)+2H_2O(g)$

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

$\Delta H_{rxn}=[(1\times \Delta H_f_{(MgCl_2(s))})+(2\times \Delta H_f_{(H_2O(g))})]-[(1\times \Delta H_f_{(Mg(OH)_2(s))})+(2\times \Delta H_f_{(HCl(g))})]$

We are given:

$\Delta H_f_{(Mg(OH)_2(s))}=-924.5kJ/mol\\\Delta H_f_{(HCl(g))}=-92.30kJ/mol\\\Delta H_f_{(MgCl_2(s))}=-641.8kJ/mol\\\Delta H_f_{(H_2O(g))}=-241.8kJ/mol$

Putting values in above equation, we get:

$\Delta H_{rxn}=[(1\times (-641.8))+(2\times (-241.8))]-[(1\times (-924.5))+(2\times (-92.30))]\\\\\Delta H_{rxn}=-16.3kJ$

Hence, the standard enthalpy change of the reaction is coming out to be -16.3 kJ

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2 years ago