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2 years ago

2a. Write down the names of the new substances formed when each of the following substance reacts with dilute hydrochloric acid.

Chemistry

2 answers:

Naddika [18.5K]2 years ago

7 0

Answer:

i) Dilute hydrochloric acid will react with Ammonia to form ammonia salt.

ii) dilute hydrochloric acid will react with soduim hydroxide to form sodium chloride and water

iii) dilute hydrochloric acid will react with calcuim carbonate to form Calcium chloride, Carbon dioxide and water.

Explanation:

CHEMICAL EQUATIONS :

I) HCL + NH3 = NH4Cl

ii) HCL+ NaOH = NaCl + H2O

iii) HCL + CaCo = CaCl2 + CO2 + H2O

vampirchik [111]2 years ago

4 0

Sodium hydroxide I think it is right

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What is the concentration in milligrams per milliter of a solution containing 23.5 meq sodium chlorise per milliliter? mw nacl =

ValentinkaMS [17]

1374.75 is the concentration in milligrams per ml of a solution containing 23.5 meq sodium chloride per milliliter.

Concentration in chemistry is calculated by dividing a constituent's abundance by the mixture's total volume.

It is calculated in mg/ml.

The unit of measurement frequently used for electrolytes is the milliequivalent (mEq). This value compares an element's chemical activity, or combining power, to that of 1 mg of hydrogen.

Formula for calculating concentration in mg/ml is

Conc. (mg/ml) = M(eq) /ml × Molecular weight / Valency

Given

M(eq) NaCl/ ml = 23.5

Molecular weight pf NaCl = 58.5 g/mol

Valency = 1

Putting the values into the formula

Conc. (mg/ml) = 23.5 ×58.5/1

= 1374.75 mg/ml

Hence, 1374.75 is the concentration in milligrams per ml of a solution containing 23.5 meq sodium chloride per milliliter.

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7 0

2 years ago

Nitric oxide, NO, is made from the oxidation of NH3, and the reaction is represented by the equation: 4NH3 + 5O2 → 4NO + 6H2O Wh

tamaranim1 [39]

Answer:

16.16g of O2.

Explanation:

We'll begin by writing the balanced equation for the reaction. This is illustrated below:

4NH3 + 5O2 → 4NO + 6H2O

Next, we shall determine the mass of NH3 and O2 that reacted from the balanced equation. This is illustrated below:

Molar mass of NH3 = 14 + (3x1) = 17g/mol

Mass of NH3 from the balanced equation = 4 x 17 = 68g

Molar Mass of O2 = 16x2 = 32g/mol

Mass of O2 from the balanced equation = 5 x 32 = 160g.

From the balanced equation above,

68g of NH3 reacted with 160g of O2.

Now, we can calculate the mass of O2 that will be required to react completely with 6.87 g of NH3. This is illustrated below:

From the balanced equation above,

68g of NH3 reacted with 160g of O2

Therefore, 6.87g of NH3 will react with = (6.87 x 160)/68 = 16.16g of O2.

Therefore, 16.16g of O2 is needed for the reaction.

4 0

3 years ago

A chemical reaction can theoretically produce 1375.5 grams of product , but in actuality 112.9 grams are product . Which is the

zalisa [80]

Answer:

8.20 % → Percent yield reaction

Explanation:

To find the percent yield of reaction we apply this:

(Produced yield / Theoretical yield) . 100 = %

Produced yield = 112.9 g

Theoretical yield = 1375.5 g

We replace → (112.9g / 1375.5 g) . 100

8.20 % → Percent yield reaction

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3 years ago

C10H22 > C8H18+C2H4 if only ethane can be sold what is the atom econmy of this process

insens350 [35]

Answer:

The atom economy of ethane in this process is 19.72 %.

What is atom economy?

The conversion efficiency of a chemical reaction in terms of all the atoms involved and the desired products produced is known as atom economy (atom efficiency/percentage).

Explanation:

C₁₀H₂₂ → C₈H₁₈ + C₂H₄

Molecular weight of C₁₀H₂₂ = 142.28

Molecular weight of C₈H₁₈ = 114.228
Molecular weight of C₂H₄ = 28.053

% Atom economy = $\frac {Molecular weight of C2H4} {Molecular weight of C10H22}$

= $\frac{28.053}{142.28}*100$

= 19.716 %

≈ 19.72 %

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5 0

1 year ago

Freon−12 (CF2Cl2), widely used as a refrigerant and aerosol propellant, is a dangerous air pollutant. In the troposphere, it tra

nevsk [136]

Answer:

141g of CCl₄

Explanation:

First, we have to write the balanced equation.

CCl₄(g) + 2 HF(g) ⇄ CF₂Cl₂(g) + 2 HCl(g)

We can calculate how many moles of CF₂Cl₂ using the ideal gas equation.

V = 14.9 dm³ = 14.9 L

T = 21°C + 273.15 = 294.15 K

P = 1.48 atm

R = 0.08206 atm.L/mol.K

$P.V=n.R.T\\n=\frac{P.V}{R.T} =\frac{1.48atm.14.9L}{0.08206\frac{atm.L}{mol.K}.294.15K }=0.914mol$

We can use proportions to find the mass of CCl₄ required to obtain 0.914 moles of CF₂Cl₂. According to the balanced equation, 1 mol of CF₂Cl₂ is produced when 1 mol of CCl₄ reacts. And the molar mass of CCl₄ is 154 g/mol.

$0.914molCF_{2}Cl_{2}.\frac{1molCCl_{4}}{1molCF_{2}Cl_{2}} .\frac{154gCCl_{4}}{1molCCl_{4}} =141gCCl_{4}$

7 0

3 years ago