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2 years ago

Identify the element from the following substances

Chemistry

1 answer:

Orlov [11]2 years ago

8 0

Answer:

sodium chloride

Explanation:

an element is a pure substance that cannot be split into

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Naming compounding<br> N2O

solniwko [45]

Answer:

Nitrogenous oxide/ nitrogen oxide

Explanation:

N2O is nitrogen oxide

6 0

2 years ago

6.5 Moles of Al reacts with 7.2 Moles of H2O What is the limiting reactant and calculate the Theoretical Yield?

SVEN [57.7K]

Answer:

H₂O is the limiting reactant

Theoretical yield of 240 g Al₂O₃ and 14 g H₂

Explanation:

Find how many moles of one reactant is needed to completely react with the other.

6.5 mol Al × (3 mol H₂O / 2 mol Al) = 9.75 mol H₂O

We need 9.75 mol of H₂O to completely react with 6.5 mol of Al. But we only have 7.2 mol of H₂O. Therefore, H₂O is the limiting reactant.

Now find the theoretical yield:

7.2 mol H₂O × (1 mol Al₂O₃ / 3 mol H₂O) × (102 g Al₂O₃ / mol Al₂O₃) ≈ 240 g Al₂O₃

7.2 mol H₂O × (3 mol H₂ / 3 mol H₂O) × (2 g H₂ / mol H₂) ≈ 14 g H₂

Since the data was given to two significant figures, we must round our answer to two significant figures as well.

4 0

3 years ago

Which of the following correctly orders the types of radiation from the LONGEST wavelength to the SHORTEST wavelength?

gulaghasi [49]

Answer:

For the wavelenth of light spectrum, I believe the order would be red visible light, infrared, microwaves and radio waves.

5 0

3 years ago

1. Hydrogen, Helium, Lithium?

Reika [66]

Answer:

what happened to it, have the gone extinct

6 0

3 years ago

Consider the reaction 4PH3(g) → P4(g) + 6H2(g) At a particular point during the reaction, molecular hydrogen is being formed at

Svetllana [295]

Answer:

The rate at which $P_4$ is being produced is 0.0228 M/s.

The rate at which $PH_3$ is being consumed is 0.0912 M/s.

Explanation:

$4PH_3\rightarrow P_4(g)+6H_2(g)$

Rate of the reaction : R

$R=\frac{-1}{4}\frac{d[PH_3]}{dt}=\frac{1}{6}\frac{d[H_2]}{dt}=\frac{1}{1}\frac{d[P_4]}{dt}$

The rate at which hydrogen is being formed = $\frac{d[H_2]}{dt}=0.137 M/s$

$R=\frac{1}{6}\frac{d[H_2]}{dt}$

$R=\frac{1}{6}\times 0.137 M/s=0.0228 M/s$

The rate at which $P_4$ is being produced:

$R=\frac{1}{1}\frac{d[P_4]}{dt}$

$0.0228 M/s=\frac{1}{1}\frac{d[P_4]}{dt}$

The rate at which $PH_3$ is being consumed :

$R=\frac{-1}{4}\frac{d[PH_3]}{dt}$

$0.0228 M/s\times 4=\frac{-1}{1}\frac{d[PH_3]}{dt}$

$\frac{-1}{1}\frac{d[PH_3]}{dt}=0.912 M/s$

6 0

3 years ago