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3 years ago

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If you mix equal concentrations of reactants and products, will the reaction proceed to the right or the left? HF(aq)+NO3−(aq)⇌H

NO3(aq)+F−(aq) Match the items in the left column to the appropriate blanks in the sentences on the right. ResetHelp HFH F blank is a stronger acid than blank, and blank is a stronger base than blank. Because proton transfer occurs from the stronger blank to the stronger blank, the reaction proceeds to the blank. is a stronger acid than HNO3H N O Subscript3 blank is a stronger acid than blank, and blank is a stronger base than blank. Because proton transfer occurs from the stronger blank to the stronger blank, the reaction proceeds to the blank.,

1 answer:

kaheart [24]3 years ago

4 0

Answer:

hypompast

Explanation:

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Determine the mass of 5.2 x 10 power of 21 molecules of propanol C3H7OH(l), on grams.

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where $N_A = 6.02 \times 10^{23}\; \text{mol}^{-1}$ the Avogadro's constant that relates the number of particles to their number, in the unit moles $\text{mol}$ .

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$M(\text{C}_3\text{H}_7\text{OH}) = \underbrace{3 \times 12.01}_{\text{carbon}} + \underbrace{8 \times 1.008}_{\text{hydrogen}} + \underbrace{1\times 16.00}_{\text{oxygen}} = 60.09 \; \text{g} \cdot \text{mol}^{-1}$

$8.6 \times 10^{-3} \; \text{mol}$ of propanol molecules would thus have a mass of $8.6 \times 10^{-3} \; \text{mol} \times 60.09 \; \text{g} \cdot \text{mol}^{-1} = 0.52 \; \text{g}$

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3 years ago

Why might a geologist ask an analytical chemist to help identify the minerals in a rock.

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2 years ago

Calculate the percent ionization of propionic acid (C2H5COOH) in solutions of each of the following concentrations (Ka is given

tekilochka [14]

Answer:

a) = 0.704%

b) = 1.30%

c) = 2.60%

Explanation:

Given that:

$K_a$ = $1.34*10^{-5$

For Part A; where Concentration of A = 0.270 M

Percentage Ionization(∝) $\alpha = \sqrt{\frac{K_a}{C} }$

$\alpha = \sqrt{\frac{1.34*10^{-5}}{0.270} }$

$\alpha = \sqrt{4.9629*10^{-5}}$

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percentage% (∝) = $7.044*10^{-3}*100$

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For Part B; where Concentration of B = $7.84*10^{-2$ M

$\alpha = \sqrt{\frac{1.34*10^{-5}}{7.84*10^{-2}} }$

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$\alpha = 0.0130\\$

percentage% (∝) = 0.0130 × 100%

= 1.30%

For Part C; where Concentration of C= $1.92*10^{-2} M$

$\alpha = \sqrt{\frac{1.34*10^{-5}}{1.97*10^{-2}} }$

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$\alpha =0.02608$

percentage% (∝) = 0.02608 × 100%

= 2.60%

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3 years ago