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IrinaVladis [17]

2 years ago

8

a 3.81 g sample of NaHCO3 was completely decomposed. After decomposition, Na2CO3 had mass of 2.86g. Determine mass of H2CO3 prod

uced.

1 answer:

qaws [65]2 years ago

8 0

Answer:

1.67g H2CO3 are produced

Explanation:

Based on the reaction:

2NaHCO3 → Na2CO3 + H2CO3

<em>2 moles of NaHCO3 produce 1 mole of Na2CO3 and 1 mole of H2CO3</em>

To solve this question we need to find the moles of Na2CO3 = Moles of H2CO3. With their moles we can find the mass of H2CO3 as follows:

<em>Moles Na2CO3 -Molar mass: 105.99g/mol-</em>

2.86g Na2CO3 * (1mol/105.99g) = 0.02698 moles Na2CO3 = Moles H2CO3

<em>Mass H2CO3 -Molar mass: 62.03g/mol-</em>

0.02698 moles * (62.03g/mol) =

<h3>1.67g H2CO3 are produced</h3>

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20cm of 0.09M solution of H2SO4. requires 30cm of NaOH for complete neutralization. Calculate the

kirill115 [55]

Answer:

Choice A: approximately $0.12\; \rm M$ .

Explanation:

Note that the unit of concentration, $\rm M$ , typically refers to moles per liter (that is: $1\; \rm M = 1\; \rm mol\cdot L^{-1}$ .)

On the other hand, the volume of the two solutions in this question are apparently given in $\rm cm^3$ , which is the same as $\rm mL$ (that is: $1\; \rm cm^{3} = 1\; \rm mL$ .) Convert the unit of volume to liters:

$V(\mathrm{H_2SO_4}) = 20\; \rm cm^{3} = 20 \times 10^{-3}\; \rm L = 0.02\; \rm L$ .
$V(\mathrm{NaOH}) = 30\; \rm cm^{3} = 30 \times 10^{-3}\; \rm L = 0.03\; \rm L$ .

Calculate the number of moles of $\rm H_2SO_4$ formula units in that $0.02\; \rm L$ of the $0.09\; \rm M$ solution:

$\begin{aligned}n(\mathrm{H_2SO_4}) &= c(\mathrm{H_2SO_4}) \cdot V(\mathrm{H_2SO_4})\\ &= 0.02 \; \rm L \times 0.09 \; \rm mol\cdot L^{-1} = 0.0018\; \rm mol \end{aligned}$ .

Note that $\rm H_2SO_4$ (sulfuric acid) is a diprotic acid. When one mole of $\rm H_2SO_4$ completely dissolves in water, two moles of $\rm H^{+}$ ions will be released.

On the other hand, $\rm NaOH$ (sodium hydroxide) is a monoprotic base. When one mole of $\rm NaOH$ formula units completely dissolve in water, only one mole of $\rm OH^{-}$ ions will be released.

$\rm H^{+}$ ions and $\rm OH^{-}$ ions neutralize each other at a one-to-one ratio. Therefore, when one mole of the diprotic acid $\rm H_2SO_4$ dissolves in water completely, it will take two moles of $\rm OH^{-}$ to neutralize that two moles of $\rm H^{+}$ produced. On the other hand, two moles formula units of the monoprotic base $\rm NaOH$ will be required to produce that two moles of $\rm OH^{-}$ . Therefore, $\rm NaOH$ and $\rm H_2SO_4$ formula units would neutralize each other at a two-to-one ratio.

$\rm H_2SO_4 + 2\; NaOH \to Na_2SO_4 + 2\; H_2O$ .

$\displaystyle \frac{n(\mathrm{NaOH})}{n(\mathrm{H_2SO_4})} = \frac{2}{1} = 2$ .

Previous calculations show that $0.0018\; \rm mol$ of $\rm H_2SO_4$ was produced. Calculate the number of moles of $\rm NaOH$ formula units required to neutralize that

$\begin{aligned}n(\mathrm{NaOH}) &= \frac{n(\mathrm{NaOH})}{n(\mathrm{H_2SO_4})}\cdot n(\mathrm{H_2SO_4}) \\&= 2 \times 0.0018\; \rm mol = 0.0036\; \rm mol\end{aligned}$ .

Calculate the concentration of a $0.03\; \rm L$ solution that contains exactly $0.0036\; \rm mol$ of $\rm NaOH$ formula units:

$\begin{aligned}c(\mathrm{NaOH}) &= \frac{n(\mathrm{NaOH})}{V(\mathrm{NaOH})} = \frac{0.0036\; \rm mol}{0.03\; \rm L} = 0.12\; \rm mol \cdot L^{-1}\end{aligned}$ .

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MakcuM [25]

Answer:

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Explanation:

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What temperature does ice freeze

Gwar [14]

Answer:

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Explanation:

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3 years ago

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How many atoms are in 73.9g of potassium oxide

denis-greek [22]

Answer: $4.69(10)^{23} atoms$

Explanation:

Firstly, we have to find the Molecular mass of potassium oxide ( $K_{2}0$ ):

$K$ atomic mass: 39 u

$O$ atomic mass: 16 u

$K_{2}O$ molecular mass: $39(2) g/mol+16g/mol=94 g/mol$

This means that in 1 mole of $K_{2}O$ there are $94 g$ and we need to find how many moles there are in $73.9 g$ $K_{2}O$ :

1 mole of $K_{2}O$ ----- $94 g$ of $K_{2}O$

$X$ ----- $73.9 g$ of $K_{2}O$

$X=\frac{(73.9 g)(1 mole)}{94 g}$

$X=0.78 mole$ This is the quantity of moles in 73.9 g of potassium oxide

Now we can calculate the number of atoms in 73.9 g of potassium oxide by the following relation:

$N_{atoms}=(X)(N_{A})$

Where:

$N_{atoms}$ is the number of atoms in 73.9g of potassium oxide

$N_{A}=6.0221(10)^{23}/mol$ is the Avogadro's number, which is determined by the number of particles (or atoms) in a mole.

Then:

$N_{atoms}=(0.78 mole)(6.0221(10)^{23}/mol)$

$N_{atoms}=4.69(10)^{23} atoms$ This is the quantity of atoms in 73.9g of potassium oxide

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3 years ago

What kind of elements do not give, receive or share electrons?

galina1969 [7]

The group of elements on the periodic table that does not participate in chemical bonding at all would be the noble gas elements. These either do not give, receive, and or share electrons.

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