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Mrrafil [7]
2 years ago
11

If 500ml of H₂ gas at 600mmHg and 400ml of CO₂ gas at 700mmHg are mixed in a 1 lit vessel, find the total pressure of mixture of

gases at constant temperature..​
Chemistry
1 answer:
seraphim [82]2 years ago
6 0

Answer:

Total pressure increased

Explanation:

When gas C is added in the vessel then number of mole increases and number of collision depends on the number of molecules present in the vessel and on adding gas C ,mole also increases hence  number of collision increases therefore pressure also increases because number of collision increases.

Total pressure increases.

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What is the Molecular Mass of CuSo4.5H2O​
serious [3.7K]
The answer to this question is 159.609 g/mol
8 0
4 years ago
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Which of these equations is balanced? a. H2SO4 + 2Al → Al2(SO4)3 + H2 b. 2KCl + Pb(NO3)2 → 2KNO3 + PbCl2
Lapatulllka [165]
The answer is B, you just check if it is the same on the left and right side

A:
Left side - Right side
2xH - 2xH
1xS - 3xS
4xO - 12xO
2xAl - 2xAl

Therefore A is not correct

B:
Left side - right side
2xK - 2xK
1xCl - 1xCl
1xPb - 1xPb
2xN - 2xN
6xO - 6xO

B is therefore correct as both sides add up
6 0
4 years ago
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Calculate the pH of a solution that has a [OH−] of 2.6 × 10^−6 M.
TiliK225 [7]

Answer:

A. 8.4

Explanation:

[OH⁻] = 2.6 × 10⁻⁶               Take the negative log of each side

-log[OH⁻] = pOH = 5.59     Apply the pH/pOH relation

pH + pOH = 14.00               Insert the value of pOH

pH + 5.59 = 14.00               Subtract 5.59 from each side

pH = 14.00 -5.59 = 8.41

7 0
3 years ago
Oh, no! You just spilled 85.00 mL of 1.500 M sulfuric acid on your lab bench and need to clean it up immediately! Right next to
vredina [299]

Explanation:

We will balance equation which describes the reaction between sulfuric acid and sodium bicarbonate: as follows.

   H_2SO_4(aq) + 2NaHCO_3(s) \rightarrow Na2SO_4(aq) + 2H_2O(l) + 2CO_2(g)

Next we will calculate how many moles of H_2SO_4 are present in 85.00 mL of 1.500 M sulfuric acid.

As,       Molarity = \frac{\text{moles of solute}}{\text{liters of solution
}}

            1.500 M = \frac{n}{0.08500 L
}

                    n = 0.1275 mol H_2SO_4

Now set up and solve a stoichiometric conversion from moles of H_2SO_4  to grams of NaHCO_3. As, the molar mass of NaHCO_3 is 84.01 g/mol.

 0.1275 mol H_2SO_4 \times (\frac{2 mol NaHCO_3}{1 mol H_2SO_4}) \times (\frac{84.01 g NaHCO_3}{1 mol NaHCO_3})

                 = 21.42 g NaHCO_3

So unfortunately, 15.00 grams of sodium bicarbonate will "not" be sufficient to completely neutralize the acid. You would need an additional 6.42 grams to complete the task.

4 0
3 years ago
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