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givi [52]
2 years ago
9

When dissolved beryllium chloride reacts with dissolved silver nitrate in water, aqueous beryllium nitrate and silver

Chemistry
1 answer:
lisabon 2012 [21]2 years ago
4 0

Answer:

The balanced chemical equation is :

BeCl_{2} + 2AgNO_{3}\rightarrow Be(NO_{3})_{2} + 2AgCl

Explanation:

The chemical equation in which number of atoms in reactants is equal to products is called balanced equation.

Formula of , Beryllium Chloride :[BeCl_{2}

Beryllium Nitrate : BeNO_{3}

Silver Chloride : AgCl

Silver Nitrate  :AgNO_{3}

When beryllium chloride reacts with dissolved silver nitrate in water , following reaction occur :

 BeCl_{2} + 2AgNO_{3}\rightarrow Be(NO_{3})_{2} + 2AgCl

The number of atoms in reactant as well as in products are balanced :

Be = 1

Ag = 2

N =2

O = 6

Cl = 2

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Which two scenarios illustrate the relationship between pressure and volume as described by Boyle’s law?
Airida [17]

The correct answer is option 2 and 3.

The two scenarios that illustrate the relationship between pressure and volume as described by Boyle’s law are as follows:

2. The volume of an underwater bubble increases as it rises and the pressure decreases.

3. The pressure increases in an inflated plastic bag when the bag is stepped on.

According to Boyle's law, pressure of a gas is inversely proportional to its volume at constant temperature. This means that pressure rises as the volume increases and vice versa.

3 0
2 years ago
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Oxalic acid is a diprotic acid. calculate the percent of oxalic acid (h2c2o4) in a solid given that a 0.7984-g sample of that so
vlada-n [284]
This method of quantitative determination of percent purity is titrimetric reactions. These reactions most commonly involve neutralization reactions between an acid and a base. Then, we look at the neutralization reaction:

H₂C₂O₄ + 2 NaOH ⇒ Na₂C₂O₄ + 2 H₂O

So, we do the stoichiometric calculations. The important data we should know is the molar mass of oxalic acid which is equal to 90 g/mol.

(0.2283 mol/L NaOH * 0.3798 L * 1 mol H₂C₂O₄/ 2mol NaOH * 90 g/mol H₂C₂O₄) ÷ 0.7984 g *100%
= 488%

This is impossible. The purity can't be more than 100%. Looking at our calculations and the balance reaction, all steps were done correctly. So, I think there is some typographical error in the given. The mass of the sample should be 7.984 g. Then, the answer would be 48.87% purity.
5 0
2 years ago
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Write the reduction reaction of glucose to form sorbitol. List and explain the side effects caused by too much sorbitol consumpt
Lemur [1.5K]

Answer:

The product of reduction of glucose is sorbitol

The side effects caused by too much sorbitol consumption include: Diarrhea, Nausea, stomach discomfort

Explanation:

Please find attached the reaction of glucose with NADPH to produce sorbitol

5 0
2 years ago
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Given the following balanced reaction of hydrochloric acid and oxygen gas forming chlorine gas and water, how many grams of hydr
nadya68 [22]
The ratio of moles of reactants to moles of products can be seen from the coefficients in a balanced equation. In our case  4 moles of hydrochloric acid reacts with one mole of oxygen to produce two moles of chlorine and water.  So, <span> the ratio of moles of hydrochloric acid to moles of chlorine is 2:1. To determine the number moles, divide the mass by the mass of one mole. </span>
<span>Cl2 = 2 * 35.45 = 70.9 grams </span>
<span>Number of moles = 335 ÷ 70.9 </span>
<span>This is approximately 4.72 moles. The number of moles of hydrochloric acid is twice this number. </span>

<span>Mass of one mole = 1 + 35.46 = 36.45 grams </span>
<span>Total mass = 2 * (335 ÷ 70.9) * 36.45 </span>
<span>This is approximately 344.45 grams. 
Correct answer A.</span>
6 0
2 years ago
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14. What is the pH of a 0.24 M solution of sodium propionate, NaC3H502, at 25°C? (For
Murrr4er [49]

Answer:

9.1

Explanation:

Step 1: Calculate the basic dissociation constant of propionate ion (Kb)

Sodium propionate is a strong electrolyte that dissociates according to the following equation.

NaC₃H₅O₂ ⇒ Na⁺ + C₃H₅O₂⁻

Propionate is the conjugate base of propionic acid according to the following equation.

C₃H₅O₂⁻ + H₂O ⇄ HC₃H₅O₂ + OH⁻

We can calculate Kb for propionate using the following expression.

Ka × Kb = Kw

Kb = Kw/Ka = 1.0 × 10⁻¹⁴/1.3 × 10⁻⁵ = 7.7 × 10⁻¹⁰

Step 2: Calculate the concentration of OH⁻

The concentration of the base (Cb) is 0.24 M. We can calculate [OH⁻] using the following expression.

[OH⁻] = √(Kb × Cb) = √(7.7 × 10⁻¹⁰ × 0.24) = 1.4 × 10⁻⁵ M

Step 3: Calculate the concentration of H⁺

We will use the following expression.

Kw = [H⁺] × [OH⁻]

[H⁺] = Kw/[OH⁻] = 1.0 × 10⁻¹⁴/1.4 × 10⁻⁵ = 7.1 × 10⁻¹⁰ M

Step 4: Calculate the pH of the solution

We will use the definition of pH.

pH = -log [H⁺] = -log 7.1 × 10⁻¹⁰ = 9.1

5 0
2 years ago
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