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Levart [38]
3 years ago
10

2/10

Chemistry
1 answer:
emmainna [20.7K]3 years ago
6 0

Answer:

pH=2.34

Explanation:

HBr  -> H  + Br

The dissociation it's complete, for that reason the concentration of the products is the same of HBr

[H+]=[Br-]=0.00234 M

pH= - log (0.00234)=2.34

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10 points get it right ADVPH
Sav [38]

Answer:

Advanced Pharmaceutics Inc

Explanation:

Advanced Pharmaceutics Inc

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8 0
2 years ago
Calculate the number of hydrogen atoms in a 110.0 sample of tetraborane(B4H10) . Be sure your answer has a unit symbol if necess
Contact [7]

Answer:

1.242 \times 10^{25}\text{ atoms H}

Explanation:

You must convert the mass of B₄H₁₀ to moles of B₄H₁₀, then to molecules of B₄H₁₀, and finally to atoms of H.

1. Moles of B₄H₁₀

\text{Moles of B$_{4}$H}_{10} = \text{110.0 g B$_{4}$H}_{10} \times \dfrac{\text{1 mol B$_{4}$H}_{10}}{\text{53.32 g B$_{4}$H}_{10}} = \text{2.063 mol B$_{4}$H}_{10}

2. Molecules of B₄H₁₀

\text{No. of molecules} = \text{2.063 mol B$_{4}$H}_{10} \times \dfrac{6.022 \times 10^{23}\text{ molecules B$_{4}$H}_{10}}{\text{1 mol B$_{4}$H}_{10}}\\\\=1.242 \times 10^{24}\text{ molecules B$_{4}$H}_{10}

3. Atoms of H

\text{Atoms of H} = 1.242 \times 10^{24}\text{ molecules B$_{4}$H}_{10} \times \dfrac{\text{10 atoms H}}{\text{1 molecule B$_{4}$H}_{10}}\\\\= \mathbf{1.242 \times 10^{25}}\textbf{ atoms H}

8 0
3 years ago
Plllllllllllllllllllllllllllllllll hellllllllllllllllppppppppppppppp mmmmmmmmmmmeeeeeeee​
kykrilka [37]

Answer:b yes b yes c no c yes d yes d no

Explanation: wanna trade r o b l o x acc

7 0
2 years ago
How much heat is lost when 575 g of molten iron at 1825 K becomes solid at 1181 K and cools to 293 K?
meriva
Answer is: -963,8 kJ.
Q₁ = m(Fe) · C · ΔT₁.
C - specific heat capacity of liquid iron, C(Fe) = 0,82 J/g°<span>C.
</span>m(Fe) = 575 g.
ΔT₁ = 1181 - 1825 = -644°C.
Q₁ = -859306,5 J = -859,3 kJ.
Q₂ = m(Fe) · C · ΔT₂.
ΔT₂ = 293 - 1181 = -888°C.
C - specific heat capacity, C(Fe) = 0,44 J/g°C.
Q₂ = -224664 J = -224,66 kJ.
Q₃ =- heat of fusion, ΔH = 209 J/g.
Q₃ = 120175 J = 120,17 kJ.
Q = Q₁ + Q₂ + Q₃ = -963,8 kJ.

7 0
2 years ago
Read 2 more answers
Calculate the energy (in kJ) required to heat 10.1 g of liquid water from 55 oC to 100 oC and change it to steam at 100 oC. The
Maksim231197 [3]

Answer:

           \large\boxed{\large\boxed{24.6kJ}}

Explanation:

<u>1. Energy to heat the liquid water from 55ºC to 100ºC</u>

     Q=m\times C\times \Delta T

  • m = 10.1g
  • C = 4.18g/JºC
  • ΔT = 100ºC - 55ºC = 45ºC

     Q=10.1g\times 4.18J/g\ºC\times 45\ºC=1,899.81J

<u>2. Energy to change the liquid to steam at 100ºC</u>

      L=\lambda \times n

  • λ = 40.6kJ/mol
  • n = 10.1g / 18.015g/mol = 0.5606mol

      L=40.6kJ/mol\times 0.5604mol=22.76214kJ=22,762.14J

<u>3. Total energy</u>

       1,899.81J+22,762.14J=24,661.95J\approx24,662J\approx24.6kJ

7 0
3 years ago
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