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3 years ago

If 4520 kj of heat is needed to boil a sample of water, what is the mass of water

Chemistry

1 answer:

Cerrena [4.2K]3 years ago

5 0

Answer:

1,085g of water

Explanation:

If we have the value 4520kj is because the question is related to Energy and heat capacity. In this case, the law and equation that we use is the following:

Q= m*C*Δt where;

Q in the heat, in this case: 4520kj

m is the mas

Δt= is the difference between final-initial temperature (change of temperature), in this exercise we don´t have temperatura change.

In order to determine the mass, I will have the same equation but finding m

m= Q/C*Δt without m=Q/C

So: m= 4,520J/4.18J/g°C

m= 1,0813 g

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Answer:

0.252 milimoles

Explanation:

To convert mass of a substance to moles it is necessary to use the molar mass of the substance.

The formula of morphine is C₁₇H₁₉NO₃, thus, its molar mass is:

C: 17*12.01g/mol = 204.17g/mol

H: 19*1.01g/mol = 19.19g/mol

N: 1*14g/mol = 14g/mol

O: 3*16g/mol = 48g/mol.

204.17 + 19.19 + 14 + 16 = 285.36g/mol

Thus, moles of 71.891 mg = 0.071891g:

0.071891g × (1mol / 285.36g) = 2.5193x10⁻⁴ moles

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2.5193x10⁻⁴ moles = 0.252 milimoles

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4 years ago

Given the [H+] = 3.89x10^-7M, what is the pOH?

Orlov [11]

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Explanation:

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3 years ago

A gas occupies a volume of 2.45 L at a pressure of 1.03 atm. What volume will the

Gemiola [76]

Answer:

2.58 L

Explanation:

Please see the step-by-step solution in the picture attached below.

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3 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]

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

Explanation:

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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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4 years ago

What is the molarity of a KOH solution if it requires 20 milliters of 2.0m HCL to exactcly neutalize 20 milliters of the KOH sol

soldier1979 [14.2K]

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3 years ago

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