The law of conservation of mass dictates that the total mass of reactants must be equal to the total mass of the products. Thus:
mass(MgO) = mass(Mg) + mass(O)
mass(MgO) = 24 + 16
mass(MgO) = 40 g

The third option is correct.

4 0

3 years ago

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The enthalpy of solution (∆H) of KOH is -57.6 kJ/mol. If 3.66 g KOH is dissolved in enough water to make a 150.0 mL solution, wh

Wewaii [24]

When 3.66 g of KOH (∆Hsol = -57.6 kJ/mol) is dissolved in 150.0 mL of solution, it causes a temperature change of 5.87 °C.

The enthalpy of solution of KOH is -57.6 kJ/mol. We can calculate the heat released by the solution (Qr) of 3.66 g of KOH considering that the molar mass of KOH is 56.11 g/mol.

$3.66g \times \frac{1mol}{56.11g} \times \frac{(-57.6kJ)}{mol} = -3.76 kJ$

According to the law of conservation of energy, the sum of the heat released by the solution of KOH (Qr) and the heat absorbed by the solution (Qa) is zero.

$Qr+Qa = 0\\\\Qa = -Qr = 3.76 kJ$

150.0 mL of solution with a density of 1.02 g/mL were prepared. The mass (m) of the solution is:

$150.0 mL \times \frac{1.02g}{mL} = 153 g$

Given the specific heat capacity of the solution (c) is 4.184 J/g･°C, we can calculate the change in the temperature (ΔT) of the solution using the following expression.

$Qa = c \times m \times \Delta T\\\\\Delta T = \frac{Qa}{c \times m} = \frac{3.76 \times 10^{3}J }{\frac{4.184J}{g.\° C } \times 153g} = 5.87 \° C$

When 3.66 g of KOH (∆Hsol = -57.6 kJ/mol) is dissolved in 150.0 mL of solution, it causes a temperature change of 5.87 °C.

Learn more: brainly.com/question/4400908

7 0

2 years ago

At 393 K, the pressure of a sample of nitrogen is 1.07 atm. What will the pressure be at 478 K, assuming constant volume?

True [87]

Answer:

About 1.301 atm

Explanation:

The formula that you should is PV=nRT, where P stands for pressure, V stands for volume, n stands for the number of moles, R stands for the universal gas constant, and T stands for temperature in Kelvin. Since the volume, number of moles, and universal gas constant don't change, you don't need to worry about them.

1.07V=393nR

PV=498nR

P=1.301 atm. Hope this helps!

7 0

3 years ago

sing any data you can find in the ALEKS Data resource, calculate the equilibrium constant K at 25.0°C for the following reaction

prisoha [69]

Answer:

2.76 × 10⁻¹¹

Explanation:

I don’t have access to the ALEKS Data resource, so I used a different source. The number may be different from yours.

1. Calculate the free energy of formation of CCl₄

C(s)+ 2Cl₂(g)→ CCl₄(g)

ΔG°/ mol·L⁻¹: 0 0 -65.3

ΔᵣG° = ΔG°f(products) - ΔG°f(reactants) = -65.3 kJ·mol⁻¹

2. Calculate K

$\text{The relationship between $\Delta G^{\circ}$ and K is}\\\Delta G^{\circ} = -RT \ln K$

T = (25.0 + 273.15) K = 298.15 K

$\begin{array}{rcl}-65 300 & = & -8.314 \times 298.15 \ln K \\65300& = & 2479 \ln K\\26.34 & = & \ln K\\K& = & e^{26.34}\\&= & \mathbf{2.76 \times 10}^{\mathbf{11}}\\\end{array}$

3 0

3 years ago