Answer:
Option A. 2, 3, 2
Explanation:
We'll begin by balancing the equation. This can be achieved by doing the following:
Fe + Cl2 —> FeCl3
There are 2 atoms of Cl on the left side and 3 atoms on the right side. It can be balance by putting 3 in front of Cl2 and 2 in front of FeCl3 as shown below:
Fe + 3Cl2 —> 2FeCl3
There are 2 atoms of Fe on the right side and 1 atom on the left side. It can be balance by putting 2 in front of Fe as shown below:
2Fe + 3Cl2 —> 2FeCl3
Now the equation is balanced.
The coefficients are : 2, 3, 2
The preparation of lead (ii) sulphate from lead (ii) carbonate occurs in two steps:
- insoluble lead carbonate is converted to soluble lead (ii) nitrate
- soluble lead (ii) nitrate is reacted with sulphuric acid to produce lead (ii) sulphate.
<h3>How can a solid sample of lead (ii) sulphate be prepared from lead (ii) carbonate?</h3>
Lead (ii) carbonate and lead (ii) sulphate are both insoluble salts of lead.
In order to prepare lead (ii) sulphate, a two step process is performed.
In the first step, Lead (ii) carbonate is reacted with dilute trioxonitrate (v) acid to produce lead (ii) nitrate.
- PbCO₃ + 2HNO₃ → Pb(NO₃)₂ + CO₂ + H₂O
In the second step, dilute sulfuric acid is reacted with the lead (ii) nitrate to produce insoluble lead (ii) sulphate which is filtered and dried.
- Pb(NO₃)₂ + H₂SO₄ → PbSO₄ + 2HNO₃
In conclusion, lead (ii) sulphate is prepared in two steps.
Learn more about lead (ii) sulphate at: brainly.com/question/188055
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Yes because she is holding the weight of the box.
Answer:
HCl
Explanation:
<em>Choices:</em>
<em>CO: 28.01g/mol</em>
<em>NO₂: 46g/mol</em>
<em>CH₄: 16.04g/mol</em>
<em>HCl: 36.4g/mol</em>
<em>CO₂: 44.01g/mol</em>
<em />
It is possible to identify a substance finding its molar mass (That is, the ratio between its mass in grams and its moles). It is possible to find the moles of the gas using general ideal gas law:
PV = nRT
<em>Where P is pressure of gas 0.764atm; V its volume, 0.279L; n moles; R gas constant: 0.082atmL/molK and T its absolute temperature, 295.85K (22.7°C + 273.15).</em>
Replacing:
PV = nRT
PV / RT = n
0.764atm*0.279L / 0.082atmL/molKₓ295.85K = n
<em>8.786x10⁻³ = moles of the gas</em>
<em />
As the mass of the gas is 0.320g; its molar mass is:
0.320g / 8.786x10⁻³moles = 36.4 g/mol
Based in the group of answer choices, the identity of the gas is:
<h3>HCl</h3>
<em />
Answer:
n = 4 there are 16 orbitals