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3 years ago

How many equivalents are in 0.60 mole of Mg2+

1 answer:

5 0

Equivalents express the combining power of an element for example what mass of an element can combine with or displace one gram of hydrogen or eight grams of oxygen.

Since Mg is divalent, one gram of Mg2+ can combine with 2 grams of hydrogen

Therefore, 0.6 mole of Mg2+ can combine with 0.6 x 2 = 1.2 moles of hydrogen

Hence there are 1.2 equivalents in 0.60 mole of Mg2+

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In a chromatography experiment, chlorophyll pigments are separated using paper. What is the stationary phase in this experiment?

Bond [772]

Answer:
The stationary phase in chromatography experiment is paper.

Explanation:
In chromatography experiment, the stationary phase is defined as the fixed substance that is necessary to start chromatography. In our case, this fixed substance is paper, so that makes paper our stationary phase.

Hope this helps :)

4 0

3 years ago

Read 2 more answers

Calculate the no. of each atoms presents in 10 grams of calcium carbonate.

raketka [301]

Molar Mass of Calcium carbonate:-

$\\ \sf\longmapsto CaCO_3$

$\\ \sf\longmapsto 40u+12u+3(16u)$

$\\ \sf\longmapsto 52u+48u$

$\\ \sf\longmapsto 100u$

$\\ \sf\longmapsto 100g/mol$

Given Mass=10g

$\boxed{\sf No\:of\;moles=\dfrac{Given\:Mass}{Molar\:Mass}}$

$\\ \sf\longmapsto No\:of\;moles=\dfrac{10}{100}$

$\\ \sf\longmapsto No\:of\:moles=0.1mol$

Now

$\boxed{\sf No\:of\:molecules=No\;of\:moles\times Avagadro\: No}$

$\\ \sf\longmapsto 0.1\times 6.023\times 10^{23}$

$\\ \sf\longmapsto 6.023\times 10^{22}molecules$

8 0

3 years ago

Hydrogen gas (a potential future fuel) can be formed by the reaction of methane with water according to the following equation:

rusak2 [61]

Answer:

60.42% is the percent yield of the reaction.

Explanation:

Moles of methane gas at 734 Torr and a temperature of 25 °C.

Volume of methane gas = V = 26.0 L

Pressure of the methane gas = P = 734 Torr = 0.9542 atm

Temperature of the methane gas = T = 25 °C = 298.15 K

Moles of methane gas = n

$PV=nRT$

$n=\frac{PV}{RT}=\frac{0.9542 atm\times 26.0L}{0.0821 atm L/mol K\times 298.15 K}=1.0135 mol$

Moles of water vapors at 700 Torr and a temperature of 125 °C.

Volume of water vapor = V' = 23.0 L

Pressure of water vapor = P' = 700 Torr = 0.9100 atm

Temperature of water vapor = T' = 125 °C = 398.15 K

Moles of water vapor gas = n'

$P'V'=n'RT'$

$n'=\frac{PV}{RT}=\frac{0.9100 atm\times 23.0L}{0.0821 atm L/mol K\times 398.15 K}=0.6402 mol$

$CH_4(g)+H_2O(g)\rightarrow CO(g)+3H_2(g)$

According to reaction , 1 mol of methane reacts with 1 mol of water vapor. As we can see that moles of water vapors are in lessor amount which means it is a limiting reagent and formation of hydrogen gas will depend upon moles of water vapors.

According to reaction 1 mol of water vapor gives 3 moles of hydrogen gas.

Then 0.6402 moles of water vapor will give:

$\frac{3}{1}\times 0.6402 mol=1.9208 mol$ of hydrogen gas