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aksik [14]
3 years ago
9

Help How many atoms of hydrogen(gas) form if 10 grams of aluminum react? 2Al + 6HCl = 2AlCl3 + 3H2

Chemistry
1 answer:
arlik [135]3 years ago
4 0

Answer: 3.35x10²³atoms H2

Explanation: solution attached:

Convert mass of Al to moles

Do the mole to mole ratio between Al and H2

Convert moles of H2 to atoms using Avogadro's number.

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Electrons and protons because they are essentially always the same
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How do you determine how many neutrons an element has?
Alexandra [31]

Answer:

To find the number of neutrons, subtract the number of protons from the mass number. number of neutrons

Explanation:

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2 years ago
What mass of iron is produced from adding 80.0 g of iron(II) oxide (71.85 g/mol) to 20.0 g of magnesium metal? FeO (l) + Mg (l)
nadya68 [22]

Answer:

46.0g of Iron are produced

Explanation:

Based on the chemical reaction:

FeO(l) + Mg(l) → Fe(l) + MgO(s)

<em>1 mole of Iron (II) oxide reacts per mole of Mg to produce 1 mole of iron</em>

<em />

To solve this question we need to convert each mass of reactant to moles using its respectives molar masses in order to find limitng reactant. Moles of limiting reactant = Moles of iron produced:

<em>Moles FeO (Molar mass: 71.85g/mol):</em>

80.0g * (1mol / 71.85g) = 1.11moles FeO

<em>Moles Mg (Molar mass: 24.305g/mol)</em>

20.0g * (1mol / 24.305g) = 0.823 moles Mg

As moles of Mg < Moles FeO, Mg is limiting reactant and the moles of Fe are 0.823 moles.

The mass of Iron produced is:

0.823 moles Fe * (55.845g/mol) =

46.0g of Iron are produced

6 0
2 years ago
Facts about diane abbott
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Diane Julie Abbott is a British Labour Part Politician. She has been the Member of Parliament for Stoke Newington since 1987. She became the first black woman to be elected to the House of Commons. Born in September 27th 1953 Married to David Thompson. Education: Newnham College, University of Cambridge
7 0
3 years ago
How many atoms are contained in 12.5 grams of silver? A. 6.97 × 1022 atoms B. 7.52 × 1024 atoms C. 0.116 atoms D. 1.92 × 10–25 a
satela [25.4K]
M(Ag)=12.5 g
Nₐ=6.022*10²³ mol⁻¹

n(Ag)=m(Ag)/M(Ag)

N=n(Ag)*Nₐ

N=Nₐm(Ag)/M(Ag)

N=6.022*10²³mol⁻¹*12.5g/107.868g/mol=6.97*10²²


A. 6.97 × 10²² atoms

7 0
3 years ago
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