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yarga [219]
2 years ago
14

What is the total number of atoms in ammonium hydroxide? I will give Brainliest! :)

Chemistry
1 answer:
Karo-lina-s [1.5K]2 years ago
6 0

Answer:Ammonia is a compound that contains one nitrogen atom and three hydrogen atoms. Ordinarily having a characteristic of a pungent and gaseous compound.

Elemet: Hydrogen, Symbol: H, # of Atoms:5

Elemet:Nitrogen, Symbol: N, # of Atoms:1

Elemet:Oxygen, Symbol: O, #of Atoms:1

Explanation:

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Determine the number of atoms in 73.0 grams of calcium, Ca. (The
Elden [556K]
# of atoms per mol = Avogadro’s # (6.022 x 10^23)

Number of mols = mass of substance / molar mass

73 g / 40.08 g = 1.8 mols of Ca in 73 grams

1.8 mols x avagadro’s # = 1.1 x 10^24 atoms in 73 grams of Ca
7 0
2 years ago
Convert 2.39 x 1022 molecules of water into moles of water
torisob [31]

Answer:

0.0397\ \text{mol}

Explanation:

Number of molecules of water = 2.39\times 10^{22}

N_A = Avogadro's number = 6.022\times 10^{23}\ \text{mol}^{-1}

Number of moles is given by

\dfrac{2.39\times 10^{22}}{6.022\times 10^{23}}

=0.0397\ \text{mol}

The number of moles of water is 0.0397\ \text{mol}.

6 0
3 years ago
The correct mathematical expression for finding the molar solubility (s) of calcium phosphate is
GalinKa [24]

Answer is: The molar solubility of calcium phosphate is 108s⁵ = Ksp.

<span> Balanced chemical reaction: Ca</span>₃(PO₄)₂(s) → 3Ca²⁺(aq) + 2PO₄³⁻(aq).<span>
[Ca²</span>⁺] = 3s(Ca₃(PO₄)₂) = 3s.<span>
[PO</span>₄³⁻] = 2s.<span>
Ksp = [Ca²</span>⁺]³ · [PO₄³⁻]².<span>
Ksp = (3s)³ · (2s)².
Ksp = 108s</span>⁵.

s = ⁵√(Ksp ÷ 108).

6 0
3 years ago
The molar mass of aniline is 93 g/mol. what is its molecular formula
Mashutka [201]

Answer:

Molecular formula of aniline is C6H5NH2.

Explanation:

As we know, molecular mass can be calculated as

Molar mass =  C6H5NH2

Molar mass = (6*12)+(1*7)+(28)

Molar mass  = 93 g/mol

4 0
3 years ago
If you begin with 2.7 g Al and 4.05 g Cl2, what mass of AlCl3 can be produced?
Tresset [83]
<span>atomic weights: Al = 26.98, Cl = 35.45 In this reaction; 2Al = 53.96 and 3Cl2 = 212.7 Ratio of Al:Cl = 53.96/212.7 = 0.2537 that is approximately four times the mass Cl is needed. Step 2: (a) Ratio of Al:Cl = 2.70/4.05 = 0.6667 since the ratio is greater than 0.2537 the divisor which is Cl is not big enough to give a smaller ratio equal to 0.2537. so Cl is limiting (b)since Cl is the limiting reactant 4.05g will be used to determine the mass of AlCl3 that can be produced. From Step 1: 212.7g of Cl will produce 266.66g AlCl3 212.7g = 266.66g 4.05g = x x = 5.08g of AlCl3 can be produced (c) Al:Cl = 0.2537 Al:Cl = Al:4.05 = 0.2537 mass of Al used in reaction = 4.05 x 0.2537 = 1.027g Excess reactant = 2.70 - 1.027 = 1.67g King Leo · 9 years ago</span>
8 0
3 years ago
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