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zepelin [54]
4 years ago
6

100 points PLZ hurry!!! Balance the chemical reaction PLZ.

Chemistry
2 answers:
mario62 [17]4 years ago
8 0
Ba(OH)2 Ba-2 power + 2 OH H2 + SO 4

All numbers are exponents.
lora16 [44]4 years ago
4 0

Hello!

After balancing the equation the equation should look like this:

Ba(OH)2 Ba-2 power + 2 OH H2 + SO 4

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The answer is 20g N2
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How is this model useful?
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A) It shows how electrons are distributed in the Shells of an Iron atom.

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A solution prepared by mixing 10 ml of 1 m hcl and 10 ml of 1.2 m naoh has a ph of
blagie [28]

Answer: pH of resulting solution will be 13

Explanation:

pH is the measure of acidity or alkalinity of a solution.

Moles of H^+ ion = Molarity\times {\text {Volume in L}}=1M\times 0.01L=0.01mol

Moles of OH^- ion = Molarity\times {\text {Volume in L}}=1.2M\times 0.01L=0.012mol

HCl+NaOH\rightarrow NaCl+H_2O

For neutralization:

1 mole of H^+ ion will react with 1 mole of OH^- ion

0.01 mol of H^+  ion will react with =\frac{1}{1}\times 0.01mole of OH^- ion

Thus (0.012-0.01)= 0.002 moles of OH^- are left in 20 ml or 0.02 L of solution.

[OH^-]=\frac{0.002}{0.02L}=0.1M

pOH=-log[OH^-]

pOH=-log[0.1]=1

pH+pOH=14

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Thus the pH of resulting solution will be 13

7 0
3 years ago
What’s the chemical formula for lithium carbonate
nordsb [41]

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8 0
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Read 2 more answers
Air at 25°C with a dew point of 10 °C enters a humidifier. The air leaving the unit has an absolute humidity of 0.07 kg of water
madreJ [45]

Explanation:

The given data is as follows.

         Temperature of dry bulb of air = 25^{o}C

          Dew point = 10^{o}C = (10 + 273) K = 283 K

At the dew point temperature, the first drop of water condenses out of air and then,

        Partial pressure of water vapor (P_{a}) = vapor pressure of water at a given temperature (P^{s}_{a})

Using Antoine's equation we get the following.

            ln (P^{s}_{a}) = 16.26205 - \frac{3799.887}{T(K) - 46.854}

            ln (P^{s}_{a}) = 16.26205 - \frac{3799.887}{283 - 46.854}

                               = 0.17079

                   P^{s}_{a} = 1.18624 kPa

As total pressure (P_{t}) = atmospheric pressure = 760 mm Hg

                                   = 101..325 kPa

The absolute humidity of inlet air = \frac{P^{s}_{a}}{P_{t} - P^{s}_{a}} \times \frac{18 kg H_{2}O}{29 \text{kg dry air}}

                  \frac{1.18624}{101.325 - 1.18624} \times \frac{18 kg H_{2}O}{29 \text{kg dry air}}

                 = 0.00735 kg H_{2}O/ kg dry air

Hence, air leaving the humidifier has a has an absolute humidity (%) of 0.07 kg H_{2}O/ kg dry air.

Therefore, amount of water evaporated for every 1 kg dry air entering the humidifier is as follows.

                 0.07 kg - 0.00735 kg

              = 0.06265 kg H_{2}O for every 1 kg dry air

Hence, calculate the amount of water evaporated for every 100 kg of dry air as follows.

                0.06265 kg \times 100

                  = 6.265 kg

Thus, we can conclude that kg of water the must be evaporated into the air for every 100 kg of dry air entering the unit is 6.265 kg.

3 0
3 years ago
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