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Mamont248 [21]
3 years ago
5

A What does the data tell us? Using the storm data and the computer model, check all statements that are true.

Chemistry
2 answers:
ioda3 years ago
3 0
I think the answer is a. and d.

hope this helps!!
dem82 [27]3 years ago
3 0

It's A, B, and D. This question was also in my homework.


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6. This part of Earth's atmosphere blocks the sun's harmful rays. (4 points)
Mrrafil [7]
Stratosphere:) hope this helped
7 0
2 years ago
Read 2 more answers
There is an electrolytic cell in which Mn2+ is reduced to Mn and Sn is oxidized to Sn2+.
leonid [27]

Answer:

See explanation

Explanation:

Anode;

Sn(s) ------> Sn^2+(aq) + 2e

Cathode;

Mn^2+(aq) + 2e ------> Mn(s)

The minimum voltage required to drive the reaction is the cell voltage. The cell voltage is obtained from;

E°cell= E°cathode - E°anode

E°cell= -1.19 - (-0.14)

E°cell= -1.05 V

4 0
3 years ago
Hydrogen gas (a potential future fuel) can be formed by the reaction of methane with water according to the following equation:
rusak2 [61]

Answer:

60.42% is the percent yield of the reaction.

Explanation:

Moles of methane gas at 734 Torr and a temperature of 25 °C.

Volume of methane gas = V = 26.0 L

Pressure of the methane gas = P = 734 Torr = 0.9542 atm

Temperature of the methane gas = T = 25 °C = 298.15 K

Moles of methane gas = n

PV=nRT

n=\frac{PV}{RT}=\frac{0.9542 atm\times 26.0L}{0.0821 atm L/mol K\times 298.15 K}=1.0135 mol

Moles of water vapors at 700 Torr and a temperature of 125 °C.

Volume of water vapor = V' = 23.0 L

Pressure of water vapor = P' = 700 Torr = 0.9100 atm

Temperature of  water vapor = T' = 125 °C = 398.15 K

Moles of water vapor gas = n'

P'V'=n'RT'

n'=\frac{PV}{RT}=\frac{0.9100 atm\times 23.0L}{0.0821 atm L/mol K\times 398.15 K}=0.6402 mol

CH_4(g)+H_2O(g)\rightarrow CO(g)+3H_2(g)

According to reaction , 1 mol of methane reacts with 1 mol of water vapor. As we can see that moles of water vapors are in lessor amount which means it is a limiting reagent and formation of hydrogen gas will depend upon moles of water vapors.

According to reaction 1 mol of water vapor gives 3 moles of hydrogen gas.

Then 0.6402 moles of water vapor will give:

\frac{3}{1}\times 0.6402 mol=1.9208 mol of hydrogen gas

Moles of hydrogen gas obtained theoretically = 1.9208 mol

The reaction produces 26.0 L of hydrogen gas measured at STP.

At STP, 1 mole of gas occupies 22.4 L of volume.

Then 26 L of volume of gas will be occupied by:

\frac{1}{22.4 L}\times 26 L= 1.1607 mol

Moles of hydrogen gas obtained experimentally = 1.1607 mol

Percentage yield of hydrogen gas of the reaction:

\frac{Experimental}{Theoretical}\times 100

\%=\frac{ 1.1607 mol}{1.9208 mol}\times 100=60.42\%

60.42% is the percent yield of the reaction.

8 0
3 years ago
What is the volume of 2 mol of chlorine gas at STP?<br> 2.0 L<br> 11.2 L<br> 22.4 L<br> 44.8 L
nirvana33 [79]

Answer:

44.8 L

Explanation:

Using the ideal gas law equation:

PV = nRT

Where;

P = pressure (atm)

V = volume (L)

n = number of moles (mol)

R = gas law constant (0.0821 Latm/molK)

T = temperature (K)

At Standard temperature and pressure (STP);

P = 1 atm

T = 273K

Hence, when n = 2moles, the volume of the gas is:

Using PV = nRT

1 × V = 2 × 0.0821 × 273

V = 44.83

V = 44.8 L

7 0
2 years ago
To a 25.00 mL volumetric flask, a lab technician adds a 0.150 g sample of a weak monoprotic acid, HA , and dilutes to the mark w
Elis [28]

<u>Answer:</u> The number of moles of weak acid is 4.24\times 10^{-3} moles.

<u>Explanation:</u>

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

\text{Molarity of the solution}=\frac{\text{Moles of solute}}\text{Volume of solution (in L)}}

We are given:

Volume of solution = 43.81 mL = 0.04381 L      (Conversion factor: 1L = 1000 mL)

Molarity of the solution = 0.0969 moles/ L

Putting values in above equation, we get:

0.0969mol/L=\frac{\text{Moles of KOH}}{0.04381}\\\\\text{Moles of KOH}=4.24\times 10^{-3}mol

The chemical reaction of weak monoprotic acid and KOH follows the equation:

HA+KOH\rightarrow KA+H_2O

By Stoichiometry of the reaction:

1 mole of KOH reacts with 1 mole of weak monoprotic acid.

So, 4.24\times 10^{-3}mol of KOH will react with = \frac{1}{1}\times 4.24\times 10^{-3}=4.24\times 10^{-3}mol of weak monoprotic acid.

Hence, the number of moles of weak acid is 4.24\times 10^{-3} moles.

6 0
3 years ago
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