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3 years ago

What is the number of moles of NaOH in 16.5 mL of 0.750 Molar NaOH?

Chemistry

1 answer:

Alchen [17]3 years ago

5 0

Answer:

0.0165 (L) * 0.750 (mol/L) = 0.0124 mol NaOH in 16.5 mL

Explanation:

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A 50 L cylinder is filled with argon gas to a pressure of 10130.0 kPa at 300°C. How many moles of argon gas does the cylinder co

Vlad1618 [11]

To answer this question, we will use the general gas law which states that:
PV = nRT where:
P is the pressure of the gas = 10130.0 kPa
V is the volume of the gas = 50 liters
n is the number of moles that we want to calculate
R is the gas constant = 8.314 L∙kPa/K∙mol
T is the temperature = 300+273 = 573 degree kelvin

Substitute with the givens in the equation to get the number of moles as follows:
10130 * 50 = n * 8.314 * 573
506500 = 4763.922 n
n = 506500 / 4763.922
n = 106.3199 moles

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4 years ago

What is the volume of as sample that has a mass of 40.0 g and a density of 4.30 g/mL?

kati45 [8]

Explanation:

density = mass/volume so volume = mass/density = 40/4.30

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3 years ago

What is the solubility in moles/liter for silver chloride at 25 oC given a Ksp value of 1.6 x 10-10. Write using scientific nota

trapecia [35]

Answer:

Explanation:

AgCl ⇄ Ag⁺ + Cl⁻

m m m

If x mole of AgCl be dissolved in one litre .

[ Ag⁺ ] [ Cl⁻ ] = 1.6 x 10⁻¹⁰

m² = 1.6 x 10⁻¹⁰

m = 1.26 x 10⁻⁵ moles

So solubility of AgCl is 1.26 x 10⁻⁵ moles / L

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3 years ago

H2CO3(aq) → CO2(aq) + H2O(l) After strenuous exercise, you breathe heavily, exhaling and removing CO2 from your body. The remova

Aleksandr-060686 [28]

Answer:

did you find the answer

Explanation:

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6 0

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]

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

Explanation:

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

6 0

3 years ago