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2 years ago

Please help me

Chemistry

1 answer:

klio [65]2 years ago

6 0

Answer:

Vascular plants are successful due to better transportation for water, nutrients and reproduction.

Explanation:

I got it right at school and hope this helps you learn to STUDY.jk

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Chromium(III) oxide can be prepared by heating chromium(IV) oxide in vacuo at high temperature: 4Cr02 —2Cr2O3 +02 The reaction o

kkurt [141]

<u>Answer:</u> The theoretical yield and percent yield of chromium (III) oxide is 434.72 grams and 92.6 % respectively.

<u>Explanation:</u>

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ .....(1)

Given mass of $CrO_2$ = 480.1 g

Molar mass of $CrO_2$ = 84 g/mol

Putting values in equation 1, we get:

$\text{Moles of }CrO_2=\frac{480.1g}{84g/mol}=5.72mol$

For the given chemical equation:

$4CrO_2\rightarrow 2Cr_2O_3+O_2$

By Stoichiometry of the reaction:

4 moles of $CrO_2$ produces 2 moles of chromium (III) oxide

So, 5.72 moles of $CrO_2$ will produce = $\frac{2}{4}\times 5.72=2.86mol$ of chromium (III) oxide

Now, calculating the mass of chromium (III) oxide from equation 1, we get:

Molar mass of chromium (III) oxide = 152 g/mol

Moles of chromium (III) oxide = 2.86 moles

Putting values in equation 1, we get:

$2.86mol=\frac{\text{Mass of chromium (III) oxide}}{152g/mol}\\\\\text{Mass of chromium (III) oxide}=(2.86mol\times 152g/mol)=434.72g$

To calculate the percentage yield of chromium (III) oxide, we use the equation:

$\%\text{ yield}=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100$

Experimental yield of chromium (III) oxide = 402.4 g

Theoretical yield of chromium (III) oxide = 434.72 g

Putting values in above equation, we get:

$\%\text{ yield of chromium (III) oxide}=\frac{402.4g}{434.72g}\times 100\\\\\% \text{yield of chromium (III) oxide}=\%$

Hence, the theoretical yield and percent yield of chromium (III) oxide is 434.72 grams and 92.6 % respectively.

7 0

2 years ago

What is the pressure in atm exerted by 2.48 moles of a gas in a 250.0 mL container at 58 degrees celsius

Dvinal [7]

Since the question manages to include moles, pressure, volume, and temperature, then it is evident that in order to find the answer we will have to use the Ideal Gas Equation: PV = nRT (where P = pressure; V = volume; n = number of moles; R = the Universal Constant [0.082 L·atm/mol·K]; and temperature.

First, in order to work out the questions, there is a need to convert the volume to Litres and the temperature to Kelvin based on the equation:
250 mL = 0.250 L
58 °C = 331 K

Also, based on the equation P = nRT ÷ V

⇒ P = (2.48 mol)(0.082 L · atm/mol · K)(331 K) ÷ 0.250 L
⇒ P = (67.31 L · atm) ÷ 0.250 L
⇒ P = 269.25 atm

Thus the pressure exerted by the gas in the container is 269.25 atm.

7 0

2 years ago

(~40 Points, Please answer and explain)

Helen [10]

Answers:

1. CO₂ < Ar < N₂ < He;

2. Cl₂ < CO₂ < Ar < N₂ < H₂

Step-by-step explanation:

Graham’s Law applies to the diffusion of gases:

The rate of diffusion (r) of a gas is inversely proportional to the square root of its molar mass (M).

$r = \frac{1 }{\sqrt{M}}$

If you have two gases, the ratio of their rates of diffusion is

$\frac{r_{2}}{r_{1}} = \sqrt{\frac{M_{1}}{M_{2}}}$

1. Order of diffusion rates

According to Graham's Law, the lightest gases will have the highest diffusion rates and the heavier gases the slowest.

The molecular masses of the gases are:

Ar 39.95; CO₂ 44.01; He 4.00; N₂ 28.02

Putting them in order,we get

44.01 > 39.95 > 28.02 > 4.00

CO₂ > Ar > N₂ > He

Thus, the relative rates of diffusion are

CO₂ < Ar < N₂ < He

2. Order of molecular speeds

A postulate of the Kinetic Molecular Theory is that at a given temperature, the average kinetic energy of the molecules is directly proportional to the Kelvin temperature.

KE = ½ mv² ∝ T

mv² ∝ T Divide each side by m

v² ∝ T/m

If T is constant.

v² ∝ 1/m Take the square root of each side

v ∝ 1/√m

This is an inverse relationship, so the molecules with the smallest molecular mass should have the highest average speeds.

The molecular masses of the gases are:

N₂ 28.02; H₂ 2.016; Cl₂ 70.91; CO₂ 44.01; Ar 39.95

Putting them in order. we get

70.91 > 44.01 > 39.95 > 28.02 > 2.016

Cl₂ > CO₂ > Ar > N₂ > H₂

Thus, the relative molecular speeds are

Cl₂ < CO₂ < Ar < N₂ < H₂

4 0

2 years ago

Calculate the standard cell potential at 25 ∘c for the reaction x(s)+2y+(aq)→x2+(aq)+2y(s) where δh∘ = -793 kj and δs∘ = -319 j/

QveST [7]

First we will calculate free energy change:
ΔG₀ = ΔH₀ - (T * ΔS₀)
= - 793 kJ - (298 * - 0.319 kJ/K) = - 698 kJ
We know the relation between free energy change and cell potential is:
ΔG₀ = - n F E⁰ where
F = Faraday's constant = 96485 C/mol
n = 2 (given by equation that the electrons involved is 2)
ΔG₀ = - 2 x 96485 x E⁰
- 698 kJ = - 2 x 96485 x E⁰
E⁰ = (698 x 1000) / (2 x 96485) = 3.62 volts

5 0

2 years ago

Which gas variable has to do with the amount of collisions of gas particles has?

Firlakuza [10]

The answer is temperature

3 0

2 years ago