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3 years ago

Identify the correct name for each compound. NaOH: CaSO4 NHACN: Al2(SO4)3

Chemistry

1 answer:

klemol [59]3 years ago

5 0

Answer:

Sodium hydroxide

Calcium sulfate

Ammonium cyanide

Aluminum sulfate

Explanation:

Just did it

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What is generally true about rate over the course of a reaction?. Rate decreases because there are more collisions as the reacti

Vlad1618 [11]

The best answer among the choices listed is the third option. The rate over the course of a reaction <span>decreases because the concentration of reactants decreases. As the concentration of the reactants decreases, there is less particles to form the products therefore less rate.</span>

8 0

3 years ago

Read 2 more answers

For the reaction below, Kp = 1.16 at 800.°C. CaCO3(s) equilibrium reaction arrow CaO(s) + CO2(g) If a 25.0-g sample of CaCO3 is

goblinko [34]

Answer:

76.0%

Explanation:

Let's consider the following reaction.

CaCO₃(s) ⇄ CaO(s) + CO₂(g)

At equilibrium, the equilibrium constant Kp is:

Kp = 1.16 = pCO₂ ⇒ pCO₂ = 1.16 atm

We can calculate the moles of CO₂ at equilibrium using the ideal gas equation.

$P.V=n.R.T\\n=\frac{P.V}{R.T} =\frac{1.16atm\times 14.4 L}{(0.08206atm.L/mol.K)\times 1073K} =0.190mol$

From the balanced equation, we know that 1 mole of CO₂ is produced by 1 mole of CaCO₃. Taking into account that the molar mass of CaCO₃ is 100.09 g/mol, the mass of CaCO₃ that reacted is:

$0.190molCO_{2}.\frac{1molCaCO_{3}}{1molCO_{2}} .\frac{100.09gCaCO_{3}}{1molCaCO_{3}} =19.0gCaCO_{3}$

The percentage by mass of the CaCO₃ that reacted to reach equilibrium is:

$\frac{19.0g}{25.0g} \times 100\%=76.0\%$

5 0

2 years ago

A balance was tested against a standard calibration mass with a certified value of 200.002 g and produced the following readings

wlad13 [49]

Answer:

Precise but not accurate.

Explanation:

We can tell the performance of the balance is precise, because the repeated measurements give values close to one another.

However, the performance of the balance is not accurate, as the mean value of the repeated measurements (195.587) is not close to the value considered as true (in this case the standard calibration mass with a certified value of 200.002 g).

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3 years ago

1.) I am in period two and have an atomic mass of eleven. Who am I?

zavuch27 [327]

1.)Boron

2.)Cadmium has 48 electrons not 121 Mercury has 80 And Copernicum so they all have no 121 electrons

3.)Hydrogen

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3 years ago

a 20.5g sample of cleaning detergent contains 8.61g of NH40H.CALCULATE the percentage composition of nitrogen in the cleaning de

AysviL [449]

Answer:

The mass percentage composition of nitrogen in the sample of the cleaning detergent is approximately 16.78%

Explanation:

The given mass of the sample of the cleaning detergent, m₁ = 20.5 g

The mass of the ammonium hydroxide, NH₄OH in the detergent, m₂ = 8.61 g

The molar mass of NH₄OH = 35.04 g/mol

The molar mass of nitrogen, N = 14.01 g/mol

Therefore, the mass, m₃ of nitrogen, N, in 8.61 g of ammonium hydroxide, NH₄OH, is found as follows;

m₃ = (14.01/35.04) × 8.61 g = (402,087/118,800) g ≈ 3.44 g

The mass of nitrogen, N, in the ammonium hydroxide, NH₄OH, contained in the 20.5 g sample of the cleaning agent, m₃ ≈ 3.44 grams

The percentage composition of nitrogen in the sample of the cleaning detergent, %N is given as follows;

$\% Composition = \dfrac{Mass \ of \ component}{Total \ mass \ of \ cleaning \ detergent} \times 100$

Therefore;

%N ≈ ((3.44 g)/(20.5 g)) × 100 ≈ 16.78 %

The percentage composition of nitrogen, %N ≈ 16.78%.

6 0

2 years ago