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Sergeeva-Olga [200]
3 years ago
14

A certain mass of gàs occupied 5.0dm cube at 2 atm and 10°C .calculate d no of moles present​

Chemistry
1 answer:
zvonat [6]3 years ago
6 0

Answer: 0.43mol

Explanation:Please see attachment for explanation

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Help please I really really need it now ​
postnew [5]
1. A 2. B 5. B this is all I know hope it helps
5 0
2 years ago
Lead(II) nitrate and ammonium iodide react to form lead(II) iodide and ammonium nitrate according to the reaction Pb(NO3)2(aq)+2
Setler79 [48]

Answer:

a) volume of ammonium iodide required =349 mL

b) the moles of lead iodide formed = 0.0436 mol

Explanation:

The reaction is:

Pb(NO_{3})_{2}+2NH_{4}I -->PbI_{2}+2NH_{4}NO_{3}

It shows that one mole of lead nitrate will react with two moles of ammonium iodide to give one mole of lead iodide.

Let us calculate the moles of lead nitrate taken in the solution.

Moles=molarityX volume (L)

Moles of lead nitrate = 0.360 X 0.121 =0.0436 mol

the moles of ammonium iodide required = 2 X0.0436 = 0.0872 mol

The volume of ammonium iodide required will be:

volume=\frac{moles}{molarity}=\frac{0.0872}{0.250}=0.349L=349mL

the moles of lead iodide formed = moles of lead nitrate taken = 0.0436 mol

7 0
3 years ago
Zinc oxide adopts a face-centered cubic arrangement. Both Zinc ions and oxide ions adopt the face-centered cubic arrangement; wi
vichka [17]

Answer:

5.41 ×10⁻²²

Explanation:

We were told right from the question that both the Zinc ions and the Zinc oxide adopts a face-centered cubic arrangement.

Then, the number of ZnO molecule in one unit cell = 4

The standard molar mass of ZnO = 81.38g

Avogadro's constant = 6.023 × 10²³ mole

∴

The mass of one unit cell of zinc oxide can be calculated as:

= \frac{4*81.38}{6.023*10^{23}}

= 5.40461564×10⁻²²

≅ 5.41 ×10⁻²²

∴ The mass of one unit cell of zinc oxide = 5.41 ×10⁻²²

3 0
3 years ago
-1
Phantasy [73]
Carbonation isn’t a force that causes such
7 0
3 years ago
A metal, M, forms an oxide having the formula MO2 containing 59.93% metal by mass. Determine the atomic weight in g/mole of the
Damm [24]

Answer:

See solution.

Explanation:

Hello there!

In this case, according to the given information, it turns out possible for us to set up the formula for the calculation of the by-mass percentage of the metal:

\%  M=\frac{m_M}{m_M+2*m_O}*100 \%\\\\59.93\%  =\frac{m_M}{m_M+32.00}*100 \%

Thus, we solve for the molar mass of the metal to obtain:

59.93\% (m_M+32.00) =m_M*100 \%\\\\m_M*59.93\% +1917.76\% =m_M*100 \%\\\\m_M=47.86g/mol

For the subsequent problems, we proceed as follows:

a.

4.00gO_2*\frac{1molO_2}{32.00gO_2}=0.125molO_2

b.

0.400molH_2S*\frac{2molH}{1molH_2S}*\frac{6.022x10^{23}atomsH}{1molH}=4.82x10^{23}atomsH

c.

0.235gNH_3*\frac{1molNH_3}{17.03gNH_3} *\frac{3molH}{1molNH_3}*\frac{6.022x10^{23}atomsH}{1molH}=2.49x10^{22}atomsH

Regards!

7 0
3 years ago
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