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2 years ago

A chemical reaction was used to produce 2.95 moles of copper(II) bicarbonate, Cu(HCO3)2.

Chemistry

1 answer:

BARSIC [14]2 years ago

8 0

Answer:

About 547 grams.

Explanation:

We want to determine the mass of copper (II) bicarbonate produced when a reaction produces 2.95 moles of copper (II) bicarbonate.

To do so, we can use the initial value and convert it to grams using the molar mass.

Find the molar mass of copper (II) bicarbonate by summing the molar mass of each individual atom:

$\displaystyle \begin{aligned} \text{MM}_\text{Cu(HCO$_3$)$_2$} &= (63.55 + 2(1.01)+2(12.01)+6(16.00))\text{ g/mol} \\ \\ &=185.59\text{ g/mol} \end{aligned}$

Dimensional Analysis:

$\displaystyle 2.95\text{ mol Cu(HCO$_3$)$_2$}\cdot \frac{185.59 \text{ g Cu(HCO$_3$)$_2$}}{1 \text{ mol Cu(HCO$_3$)$_2$}} \Rightarrow 547 \text{ g Cu(HCO$_3$)$_2$ }$

In conclusion, about 547 grams of copper (II) bicarbonate is produced.

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A compound is 52.0% zinc 9.6% carbon and 38.4% oxygen. Calculate the empirical formula of the compound.

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Answer:

So a compound is 52% Zinc(Zn), 9.6% Carbon(C), and 38.4% Oxygen (O). Let’s first start off by assuming that we have 100 g of this compound. This means that we have 52 g of Zinc, 9.6 g of Carbon, and 38.4 g of Oxygen.Zinc = 65.38 g/molCarbon = 12 g/molOxygen = 16 g/molThis means we have:52 g of Zn(1 mol Zn/65.38 g of Zn) ≈0.8 mol of Zn.9.6 g of C(1 mol C/12 g of C) = 0.8 mol of C38.4 g of O(1 mol of O/16 g of O) = 2.4 mol of O.

Explanation:

What we want to do next is divide each element by the common factor of all of them, which is 0.8. In most cases, you divide each element by the element with the least amount of moles. After we divide each by 0.8, you’ll notice you have 1 Zn, 1 C, and 3 O. This gives you the empirical formula of ZnCO3, or Zinc Carbonate.

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1 year ago

The colour of light depends on it's wave length.Red light have a wave length in the order of 7.8*10 raise to power of -7metre, e

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Calculate the percent ionization of nitrous acid in a solution that is 0.249 M in nitrous acid. The acid dissociation constant o

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Answer:

4.26 %

Explanation:

There is some info missing. I think this is the original question.

<em>Calculate the percent ionization of nitrous acid in a solution that is 0.249 M in nitrous acid. The acid dissociation constant of nitrous acid is 4.50 × 10 ⁻⁴.</em>

<em />

Step 1: Given data

Initial concentration of the acid (Ca): 0.249 M

Acid dissociation constant (Ka): 4.50 × 10 ⁻⁴

Step 2: Write the ionization reaction for nitrous acid

HNO₂(aq) ⇒ H⁺(aq) + NO₂⁻(aq)

Step 3: Calculate the concentration of nitrite in the equilibrium ([A⁻])

We will use the following expression.

$[A^{-} ] = \sqrt{Ca \times Ka } = \sqrt{0.249 \times 4.50 \times 10^{-4} } = 0.0106 M$

Step 4: Calculate the percent ionization of nitrous acid

We will use the following expression.

$\alpha = \frac{[A^{-} ]}{[HA]} \times 100\% = \frac{0.0106M}{0.249} \times 100\% = 4.26\%$

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