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Alexandra [31]
3 years ago
8

What is the pH of a solution with [ht] = 2.3 * 10-3?

Chemistry
1 answer:
Kazeer [188]3 years ago
6 0

Answer:

A. 3.00

sana tama po sagut ko

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In the equation below
yulyashka [42]

Answer:

6.25 moles of N₂ is produced, and 18.8 moles of Cu and H₂O is produced.

Explanation:

We are given the chemical equation:

\displaystyle 2\text{NH$_3$}_\text{(g)} + 3\text{CuO}_\text{(s)} \longrightarrow \text{N$_2$}_\text{(g)}  + 3\text{Cu}_\text{(s)}+3\text{H$_2$O}_\text{(g)}

And we want to determine the amount of products produced when 12.5 moles of NH₃ is reacted with excess CuO.

Compute using stoichiometry. From the equation, we can see the following stoichiometric ratios:

  • The ratio between NH₃ and N₂ is 2:1. (i.e. One mole of N₂ is produced from every two moles of NH₃.)
  • The ratio between NH₃ and Cu is 2:3.
  • The ratio between NH₃ and H₂O is 2:3. (i.e. Three moles of H₂O or Cu is produced frome every two moles of NH₃.)

Dimensional Analysis:

  • The amount of N₂ produced:

\displaystyle 12.5\text{ mol NH$_3$} \cdot \frac{1\text{ mol N$_2$}}{2\text{ mol NH$_3$}} = 6.25\text{ mol N$_2$}

  • The amount of Cu produced:

\displaystyle 12.5\text{ mol NH$_3$} \cdot \frac{3\text{ mol Cu}}{2\text{ mol NH$_3$}} = 18.8\text{ mol Cu}

  • And the amount of H₂O produced:

\displaystyle 12.5\text{ mol NH$_3$} \cdot \frac{3\text{ mol H$_2$O}}{2\text{ mol NH$_3$}} = 18.8\text{ mol H$_2$O}

In conclusion, 6.25 moles of N₂ is produced, and 18.8 moles of Cu and H₂O is produced.

8 0
2 years ago
A combustion analysis of 5.214 g of a compound yields 5.34 g co 2 ​ , 1.09 g h 2 ​ o, and 1.70 g n 2 ​ . if the molar mass of th
nekit [7.7K]
Answer is: C₃H₃N₃O₃.
Chemical reaction: CₓHₓNₓOₓ + O₂ → aCO₂ + x/2H₂ + x/2N₂.
m(CₐHₓNₓ) = 5,214 g.
m(CO₂) = 5,34 g.
m(H₂) = 1,09 g.
m(N₂) = 1,70 g.
n(CO₂) = n(C) =  5,34 g ÷ 44 g/mol = 0,121 mol.
n(H₂O) = 1,09 g ÷18 g/mol = 0,06 mol.
n(H) = 2 · 0,0605 mol = 0,121 mol.
n(N₂) = 1,7 g ÷ 28 g/mol = 0,0607 mol.
n(N) = 0,0607 mol · 2 = 0,121 mol.
n(C) : n(H) : n(N) = 0,121 mol : 0,121 mol : 0,121 mol /: 0,121
n(C) : n(H) : n(N) = 1 : 1 : 1.
M(CHN) = 27 g/mol.
m(O₂) = 8,13 g - 5,214 g = 2,914 g.
n(O₂) = 2,914 g ÷ 32 g/mol = 0,09 mol.
n(CₓHₓNₓOₓ) = 5,214 g ÷ 129,1 g/mol = 0,0404 mol.
n(CₓHₓNₓOₓ) : n(CO₂) = 1 : 3.


3 0
4 years ago
Read 2 more answers
Calculate the atomic mass of gallium, symbol Ga. Gallium has two isotopes: Ga-69 and Ga-71. The relative abundance of Ga-69 is 6
amm1812

Answer:

The atomic mass of gallium (Ga) = <u>69.723 g/mol</u>

Explanation:

Given: Two isotopes of Gallium (Ga) are Gallium-69 (⁶⁹Ga) and Gallium-71 (⁷¹Ga)

<u>For ⁶⁹Ga: </u>

Relative abundance = 60.12% = 60.12 ÷ 100 = 0.6012; Atomic mass = 68.9257 g/mol

<u>For ⁷¹Ga:</u>

Relative abundance = 39.88% = 39.88 ÷ 100 = 0.3988; Atomic mass = 70.9249 g/mol

∴ The atomic mass of Ga = (Relative abundance of ⁶⁹Ga × Atomic mass of ⁶⁹Ga) + (Relative abundance of ⁷¹Ga × Atomic mass of ⁷¹Ga)

⇒ Atomic mass of Ga = (0.6012 × 68.9257 g/mol) + (0.3988 × 70.9249 g/mol) = <u>69.723 g/mol</u>

<u>Therefore, the atomic mass of gallium (Ga) = 69.723 g/mol</u>

8 0
4 years ago
Which pair of dispersed phases and dispersing media can never form a colloid?
lana66690 [7]

Answer:

Option B Liquid and Gas

7 0
3 years ago
If 1/4 cups are in 1 serving jow many servings are in 4 cups​
mote1985 [20]

Answer:

One serving is 1/4 cup therefore 1 cup would be 4 servings. Since we have 9 cups we would then multiply 9 x 4 and the answer would be 36 servings. You have a ratio of cups to serving, so 1/4 cup : 1 serving, or .

Explanation:

7 0
3 years ago
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