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astra-53 [7]
2 years ago
14

What is the empirical formula of a compound that is 41.4% Strontium, 13.24%

Chemistry
1 answer:
Pepsi [2]2 years ago
6 0

Answer:

Empirical Formula N2O6Sr Strontium Nitrate

Explanation:

N=13.2% O=45.4% Sr=41.4%

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In each pair, choose the larger of the indicated quantities or state that the samples are equal:
romanna [79]

<u> There are more total ions in 2.2 moles of </u>MgCl_2

<h3>What are ions?</h3>

Ion, any atom or group of atoms that bears one or more positive or negative electrical charges. Positively charged ions are called cations; negatively charged ions, anions.

Ions are formed by the addition of electrons to, or the removal of electrons from, neutral atoms or molecules or other ions; by combination of ions with other particles; or by rupture of a covalent bond between two atoms in such a way that both of the electrons of the bond are left in association with one of the formerly bonded atoms.

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8 0
1 year ago
Can someone help me and fast
jasenka [17]

Answer:

It is likely B.

Explanation:

8 0
3 years ago
A certain reaction is exothermic in the forward direction. The reaction has more moles of gas on the product side. Which of the
alexgriva [62]

Answer:

Decreasing the pressure

6 0
4 years ago
Read the following chemical equation. Mg(s) + H+ Cl− → Mg2+ Cl− + H2 What most likely happens during this reaction?
spin [16.1K]

Answer:

Magnesium loses two electrons.

Explanation:

  • As clear from the reaction Mg converted from <em>Mg(s) to Mg²⁺</em>, so Mg converted from the oxidation state (0) to (2+).

<em>∴ Mg losses two electrons.</em>

  • Cl⁻ remains as it is, so it is considered as a catalyst and neither loss nor gain any electrons.

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6 0
3 years ago
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A compound composed of only carbon and chlorine is 85.5% chlorine by mass. propose a lewis structure for the lightest of the pos
aivan3 [116]

Answer is in picture below.

Use 100 grams of the compound:

ω(Cl) = 85.5% ÷ 100%.

ω(Cl) = 0.855; mass percentage of the chlorine in the compound.

m(Cl) = 0.855 · 100 g.

m(Cl) = 85.5 g; mass of chlorine.

m(C) = 100 g - 85.5 g.

m(C) = 14.5 g; mass of carbon.

n(Cl) = m(Cl) ÷ M(Cl).

n(Cl) = 85.5 g ÷ 35.45 g/mol.

n(Cl) = 2.41 mol; amount of chlorine.

n(C) = 14.5 g ÷ 12 g/mol.

n(C) = 1.21 mol; amount of carbon.

n(Cl) : n(C) = 2.41 mol : 1.21 mol = 2 : 1.

This compound is dichlorocarbene CCl₂.

4 0
3 years ago
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