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KonstantinChe [14]

3 years ago

9

Sound generally travels faster through solids than through gases because the particles (atoms or molecules) of a solid are _____

than the particles of a gas.

1 answer:

Tju [1.3M]3 years ago

4 0

Either closer or denser

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Consider the balanced equation. 2HCl + Mg MgCl2 + H2 If 40.0 g of HCl react with an excess of magnesium metal, what is the theor

Serga [27]

Answer:

Theoretical yield of hydrogen is 1.11 g

Explanation:

Balanced equation, $Mg+2HCl\rightarrow MgCl_{2}+H_{2}$

As Mg remain present in excess therefore HCl is the limiting reagent.

According to balanced equation, 2 moles of HCl produce 1 mol of $H_{2}$ .

Molar mass of HCl = 36.46 g/mol

So, 40.0 g of HCl = $\frac{40.0}{36.46}moles$ of HCl = 1.10 moles of HCl

Hence, theoretically, number of moles of $H_{2}$ are produced from 1.10 moles of HCl = $(\frac{1}{2}\times 1.10)moles=0.550moles$

Molar mass of $H_{2}$ = 2.016 g/mol

So, theoretical yield of $H_{2}$ = $(0.550\times 2.016)g=1.11g$

7 0

3 years ago

Read 2 more answers

Starting with lead (ii) carbonate describe how a solid sample of lead (ii) sulphate can be prepared

tatuchka [14]

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

#SPJ1

4 0

2 years ago

Calculate the molarity of H3PO4 when you added 57.3 g into 3,820 mL of water

motikmotik

Answer:

0.153M

Explanation:

57.3/97.994 (molar mass)=0.585 moles of H3PO4

.0585/3.820L=0.153M

5 0

3 years ago

_______ law states the volume of a definite quantity of dry gas is inversely proportional to the pressure, provided the temperat

nasty-shy [4]

Boyles law --i hope this helps

5 0

3 years ago

Calculate the mass of ammonia (NH3) that contains a billion (1.00 * 10^12) hydrogen atoms.

Lynna [10]

Answer:

The mass of ammonia (NH3) that contains $1.00 \times 10^12$ hydrogen atoms is $9.4\times 10^{-12}$ g.

Explanation:

As $6.022\times 10^{23}$ atoms of hydrogen = 1 mole of the hydrogen atom

Therefore, $10^{12}$ atoms of hydrogen $= \frac{1}{6.022 \times 10^{23}}\times 10^{12}=1.66\times 10^{-12}$ moles of the hydrogen atom.

Now, there are 3 moles of hydrogen atoms in 1 mole of ammonia $(NH_3)$ .

As the mass of 1 mole of ammonia is 17g, so

when there are 3 moles of hydrogen atoms, then the mass of ammonia = 17 g

Therefore, when there are $1.66\times 10^{-12}$ moles of hydrogen atoms, then the mass of ammonia $= \frac{17}{3}\times 1.66\times 10^{-12}=9.4\times 10^{-12$ g.

Hence, the mass of ammonia $(NH_3)$ that contains $1.00 \times 10^12$ hydrogen atoms is $9.4\times 10^{-12}$ g.

3 0

3 years ago