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3 years ago

4 Weathering Assessment

Chemistry

1 answer:

Leni [432]3 years ago

6 0

Answer:

This is an example of a physical change because the ice cubes began to melt.

Explanation:

The above is the right answer to the question about the dissolution of the whole mixture mentioned in the excerpts above.

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Under certain conditions the rate of this reaction is zero order in dinitrogen monoxide with a rate constant of ·0.0047Ms−1: 2N2

leva [86]

Answer:

250 mmol are left after 26.60 secs ≅ 26.6 seconds

Explanation:

zero order kinetics formula is:

[A] = [A₀] - kt

where [A] = amount left; [A₀] = amount remaining; k = rate constant; t = time

concentration [A₀] = mole/volume = 0.5 mole/2.0L = 0.25

[A] = 0.25 mole/2.0L = 0.125 M

k = 0.0047 Ms⁻¹

t = {[A]-[A₀]}/-k = (0.125 - 0.25)/(-0.0047) = 26.60 seconds = 26.6 seconds

the amount is reduced by half after 26.6 seconds

6 0

3 years ago

There are more electrons than protons in a negatively charged atom. (2 points) True False

Mkey [24]

Answer:

it's false i think

Explanation:

5 0

3 years ago

Read 2 more answers

A gas takes up a volume of 35 L, and has a pressure of 4.8 atm. What is the new pressure of

joja [24]

Answer:

7.3 atm

Explanation:

- Use the formula P1V1 = P2V2

- Rearrange formula and then plug in values.

- Hope this helped! Let me know if you need more help or a further explanation.

3 0

3 years ago

According to the oxygen-hemoglobin dissociation curve, PO2 in the lungs of 100 mm Hg results in Hb being 98% saturated. At high

Ad libitum [116K]

Answer:

Hb would be 78.4% saturated.

Explanation:

This problem can be solved by using simple unitary method.

At 100 mm Hg pressure of oxygen, Hb is saturated by 98%

So, at 1 mm Hg pressure of oxygen, Hb is saturated by $\frac{98}{100}$ %

Hence, at 80 mm Hg pressure of oxygen, Hb is saturated by $\frac{98\times 80}{100}$ % or 78.4%

Therefore, at 80 mm Hg pressure of oxygen in the lungs, Hb would be 78.4% saturated.

4 0

3 years ago

Methanol, CH3OH, is a useful fuel that can be made as follows: CO(g) + 2H2(g) → CH3OH(l) A reaction mixture used 12.0 g of H2 an

algol [13]

Answer:

A = Theoretical yield = 82.24 g

B = Amount of excess reactant left = 1.72 g

Explanation:

Given data:

Mass of H₂ = 12 g

Mass of CO = 74.5 g

Theoretical yield of CH₃OH = ?

Amount of excess reactant left = ?

Solution:

First of all we will write the balance chemical equation:

CO + 2H₂ → CH₃OH

Number of moles of H₂ = mass / molar mass

Number of moles of H₂ = 12 g/ 2 g/mol = 6 mol

Number of moles of CO = mass / molar mass

Number of moles of CO = 74.5 g/ 29 g/mol = 2.57 mol

Now we compare the moles of CH₃OH with CO and H₂ from balance chemical equation:

CO : CH₃OH

1 : 1

2.57 : 2.57

H₂ : CH₃OH

2 : 1

6 : 1/2 × 6 = 3 mol

The number of moles of CH₃OH produce by CO are less so it will limiting reactant.

mass of CH₃OH = number of moles × molar mass

mass of CH₃OH = 2.57 mol × 32 g/mol

mass of CH₃OH = 82.24 g

Excess amount of H₂:

CO : H₂

1 : 2

2.57 : 2×2.57 = 5.14 mol

The moles of H₂ that react with CO are 5.14. While the total number of moles of H₂ available are 6 moles. So,

The number of moles of H₂ remain untreated = 6 mol - 5.14 mol = 0.86 mol

Mass of H₂ remain untreated = 0.86 mol × 2 g/mol

Mass of H₂ remain untreated = 1.72 g

6 0

3 years ago