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Leona [35]
2 years ago
12

a 1.00litre vessel contains 0.215 mole of nitrogen gas and 0.0118 mole of hydrogen gas at 25°C. determine the partial pressure o

f each component and the total pressure of the vessel​
Chemistry
1 answer:
stepladder [879]2 years ago
8 0

Answer:

THANK YOU SA BRAINLIEST

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Calculate the concentration of hydronium and hydroxide ions in a 0.050 M solution of nitric acid.
Mazyrski [523]

Answer:

[H₃O⁺] = 0.05 M & [OH⁻] = 2.0 x 10⁻¹³.

Explanation:

  • HNO₃ is completely ionized in water as:

<em>HNO₃ + H₂O → H₃O⁺ + NO₃⁻.</em>

  • The concentration of hydronium ion is equal to the concentration of HNO₃:

[H₃O⁺] = 0.05 M.

∵ [H₃O⁺][OH⁻] = 10⁻¹⁴.

<em>∴ [OH⁻] = 10⁻¹⁴/[H₃O⁺] </em>= 10⁻¹⁴/0.05 = <em>2.0 x 10⁻¹³.</em>

3 0
2 years ago
What type of reaction is shown below? <br>a) Addition reaction <br>b) Esterification​
Gemiola [76]

Answer:

a) Addition reaction, is your answer

6 0
3 years ago
Use bond energies from Table 10.3 in the textbook to estimate the enthalpy change (ΔH) for the following reaction. C2H2(g)+H2(g)
Digiron [165]

Answer: =176.6kJmol^{-1}

Explanation:Bond energy of H-H is 436.4 kJ/mole

Bond energy of  C-H is 414 kJ/mol

Bond energy of C=C is 620 kJ/mol

Bond energy of C≡C is 835 kJ/mol

\Delta H= {\text {sum of bond energies of reactants}}-  {\text {sum of bond energies of products}}

\Delta H= {1B.E(C≡C)+2B.E(C-H) +1B.E(H-H)} - {1B.E(C=C)+4B.E(C-H)}

\Delta H= {1B.E(835kJmole^{-1})+2B.E(414kJmole^{-1}) +1B.E(436.4kJmole^{-1})} -  {1B.E(620kJmole^{-1})+4B.E(414kjmole^{-1})}

=176.6kJmol^{-1}







7 0
3 years ago
I need help with 2. I’ll give brainliest
noname [10]

Answer:

Lines of latitude run from east to west.

Explanation:

8 0
3 years ago
How many bromine atoms are present in 37.9 g of CH2Br2?
VARVARA [1.3K]

Answer:

The answer to your question is: 6.55 x 10 ²³ atoms of Br

Explanation:

CH2Br2 = 37.9 g

MW CH2Br2 = (12 x 1) + (2 x 1) + (80 x 2) = 174 g

                   174 g of CH2Br2 ------------------  160 g of Br2

                   37.9 g of CH2Br2   ---------------     x

                x = 37.9 x 160/174 = 34.85 g of Br

                      1 mol of Br -----------------   160 g Br2

                         x              ----------------    174 g Be2

               x = 174 x 1 /160 = 1.088 mol of Br2

                1 mol of Br -----------------  6.023 x 10 ²³ atoms

            1.088 mol of Br -------------    x

                    x = 1.088 x 6.023 x 10 ²³ / 1 = 6.55 x 10 ²³ atoms

4 0
3 years ago
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