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2 years ago

Suppose that we combined 75 g of magnesium with 45 g of nitrogen and heat them up. Predict the product of

Chemistry

1 answer:

juin [17]2 years ago

6 0

Answer:

Mg is the limiting reactant

The theoretical yield is 105.2 g of Mg3N2

Explanation:

3Mg(s) + N2(g) -----> Mg3N2(s)

Number of moles of Mg = 75g/24g/mol = 3.125 moles

3 moles of Mg yields 1 mole of Mg3N2

3.125 moles of Mg yields 3.125 × 1/3 = 1.042 moles of Mg3N2

Number of moles of N2 = 45g/28g/mol = 1.607 moles

1 mole of N2 yields 1 mole of Mg3N2

1.607 moles of N2 yields 1.607 moles Mg3N2

Hence, Mg is the limiting reactant

The theoretical yield of Mg3N2 = 1.042 moles of Mg3N2 × 101 g/mol = 105.2 g of Mg3N2

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At T = 250 °C the reaction PCl5(g) PCl3(g) + Cl2(g) has an equilibrium constant in terms of pressures Kp = 2.15. (a) Suppose the

Ganezh [65]

Answer:

To the right

Explanation:

Step 1: Given data

Partial pressure of PCl₅ (pPCl₅) = 0.548 atm

Partial pressure of PCl₃ (pCl₃) = 0.780 atm

Partial pressure of Cl₂ (pCl₂) = 0.780 atm

Step 2: Write the balanced equation

PCl₅(g) ⇄ PCl₃(g) + Cl₂(g)

Step 3: Calculate the pressure reaction quotient

$Q_p = \frac{pPCl_3 \times pCl_2 }{pPCl_5} = \frac{0.780 \times 0.780 }{0.548} =1.11$

Step 4: Determine whether the reaction proceeds to the right or to the left as equilibrium is approached

Since <em>Qp < Kp</em>, the reaction will proceed to the right to attain the equilibrium.

7 0

3 years ago

Suppose of barium acetate is dissolved in of a aqueous solution of ammonium sulfate. Calculate the final molarity of barium cati

ollegr [7]

Explanation:

Let us assume that the given data is as follows.

mass of barium acetate = 2.19 g

volume = 150 ml = 0.150 L (as 1 L = 1000 ml)

concentration of the aqueous solution = 0.10 M

Therefore, the reaction equation will be as follows.

$Ba(C_{2}H_{3}O_{2})_{2} \rightarrow Ba^{2+} + 2C_{2}H_{3}O^{-}_{2}$

Hence, moles of $C_{2}H_{3}O^{-}_{2}$ = $2 \times Ba(C_{2}H_{3}O_{2})_{2}$ .......... (1)

As, No. of moles = $\frac{mass}{\text{molar mass}}$

Hence, moles of $Ba(C_{2}H_{3}O_{2})_{2}$ will be calculated as follows.

No. of moles = $\frac{mass}{\text{molar mass}}$

= $\frac{2.19 g}{255.415 g/mol}$ (molar mass of $Ba(C_{2}H_{3}O_{2})_{2}$ is 255.415 g/mol)

= $8.57 \times 10^{-3}$

Moles of $C_{2}H_{3}O^{-}_{2}$ = $2 \times 8.57 \times 10^{-3}$

= 0.01715 mol

Hence, final molarity will be as follows.

Molarity = $\frac{\text{no. of moles}}{volume}$

= $\frac{0.01715 mol}{0.150 L}$

= 0.114 M

Thus, we can conclude that final molarity of barium cation in the solution is 0.114 M.

5 0

3 years ago

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kvv77 [185]

Yeah I can help I'm an A student in science

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3 years ago

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-Dominant- [34]

Answer:

appropriately shaped receptors

Explanation:

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2 years ago

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3 years ago

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