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7nadin3 [17]
2 years ago
11

H2(g) + Co(g) _CH3OH balance the equation ​

Chemistry
1 answer:
Tom [10]2 years ago
8 0

Answer:

CO + 2H2 = CH3OH

Explanation:

1. Label Each Compound With a Variable

  aCO + bH2 = cCH3OH

2. Create a System of Equations, One Per Element

  C: 1a + 0b = 1c

  O: 1a + 0b = 1c

  H: 0a + 2b = 4c

3. Solve For All Variables (using substitution, gauss elimination, or a calculator)

  a = 1

  b = 2

  c = 1

4. Substitute Coefficients and Verify Result

  CO + 2H2 = CH3OH

      L R

  C: 1 1 ✔️

  O: 1 1 ✔️

  H: 4 4 ✔️

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The surface tension of water is 7.28 ✕ 10−2 J/m2 at 20°C. Predict whether the surface tension of heptane would be higher or lowe
stira [4]

Answer:

Lower  

Explanation:

Surface tension occurs because molecules at the surface do not have molecules above them, so they cohere more strongly to their neighbours.

The stronger cohesive forces make it more difficult to move an object through the surface than when it is beneath the surface.

The attractive forces in water are strong because of hydrogen bonding.

A hexane molecule is nonpolar, so the only attractions are the weak London dispersion forces.

The cohesive forces at the surface are much lower than those in water, so the surface tension of hexane is lower than that of water at the sane temperature.

3 0
3 years ago
Alvin is heating a solution in his chemistry lab. The temperature of the solution is at 14°C. He turns on the burner and finds t
dem82 [27]

Answer:

(3 20) is the answer

Explanation:

3 0
3 years ago
The dilution factor D for an unseeded mixture of wastewater is 0.05. The DO of the mixture is initially 8.0 mg/L, and after 5 da
enyata [817]

Answer:

(i) 5-day BOD of the waste is 120 mg/l.

(ii) The ultimate carbonaceous BOD (Lo) is 20 mg/l.

Explanation:

The dilution factor D is 0.05.

The initial DO is 8.0 mg/L and the DO after 5 days is 2.0 mg/L.

The BOD of the waste for an unseeded mixture is

BOD_5=(DO_5-DO_0)/D=(8-2)/0.05=6/0.05=120mg/l

The ultimate carbonaceous BOD (Lo) can be calculated as

L = L_o*10 ^{- k_1t} \\L_o=L/10^{- k_1t}=2/10^{- 0.2*5}=2/10^{- 1}=2*10=20mg/l

6 0
3 years ago
In order for nuclear fusion to occur, the core of the sun must be hot enough to overcome the _________ of the nuclei. select one
andreyandreev [35.5K]
The answer is option a, that is " <span>electromagnetic repulsions</span><span>".
The sun generates energy by nuclear fusion, and converts a part of its mass into energy and nuclear fusion is the source of all energy that is released by the sun. The two things which are required for the process of nuclear fusion are high temperature and the high densities.

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7 0
4 years ago
How many grams of N2 gas are present in a 20.0 L tank of N2 that is at 290 K and 2.75 atm?
Rasek [7]

Considering the ideal gas law, 64.68 grams of N₂ gas are present in a 20.0 L tank of N₂ that is at 290 K and 2.75 atm.

<h3>Ideal gas law</h3>

Ideal gases are a simplification of real gases that is done to study them more easily. It is considered to be formed by point particles, do not interact with each other and move randomly. It is also considered that the molecules of an ideal gas, in themselves, do not occupy any volume.

The pressure, P, the temperature, T, and the volume, V, of an ideal gas, are related by a simple formula called the ideal gas law:

P×V = n×R×T

where:

  • P is the gas pressure.
  • V is the volume that occupies.
  • T is its temperature.
  • R is the ideal gas constant. The universal constant of ideal gases R has the same value for all gaseous substances.
  • n is the number of moles of the gas.  

<h3>Mass of N₂</h3>

In this case, you know:

  • P= 2.75 atm
  • V= 20 L
  • T= 290 K
  • R= 0.082 \frac{atmL}{molK}
  • n= ?

Replacing in the ideal gas law:

2.75 atm× 20 L = n×0.082 \frac{atmL}{molK}× 290 K

Solving:

n= (2.75 atm× 20 L)÷ (0.082 \frac{atmL}{molK}× 290 K)

<u><em>n= 2.31 moles</em></u>

Being the molar mass of N₂ 28 g/mol, then the mass of 2.31 moles is calculated as:

mass= 28 \frac{g}{mole}× 2.31 moles

Solving:

<u><em>mass= 64.68 grams</em></u>

Finally, 64.68 grams of N₂ gas are present in a 20.0 L tank of N₂ that is at 290 K and 2.75 atm.

Learn more about the ideal gas law:

brainly.com/question/4147359

#SPJ1

7 0
2 years ago
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