This is dimensional analysis-

A gas occupies a volume of 0.20 L at 25 kPa. What volume will the gas occupy at 2.0 kPa?

ycow [4]

Answer:

2 L

Explanation:

5 0

3 years ago

N2H4 + N2O4 --> N2 + H2O

ahrayia [7]

Answer:

The limiting reagent is N2O4
14,09g

Explanation:

First, we adjust the reaction.

$2N_{2} H_{4}$ + $N_{2} O_{4}$ ⇄ $6N_{2} + 4H_{2}O$

Second, we assume that the participating moles are equal to the stoichiometric ratios because we do not know the amounts of the reagents.

We can determinate what is the limiting reagent comparing of product amounts which can be formed from each reactant.

Using $N_{2} H_{4}$ to form $H_{2}O$

$molH_{2} O = 1mol N_{2} H_{4} } . \frac{4 mol H_{2} O}{2mol N_{2} H_{4} }. \frac{18\frac{g}{mol} H_{2} O}{1mol H_{2} O_} } . \frac{ 1 mol N_{2}H_{4} }{32,04\frac{g}{mol} N_{2} H_{4} }$

$molH_{2} O = 1, 125 mol$

Using $N_{2} O_{4}$ to form $H_{2} O$

$molH_{2} O = 1mol N_{2} O_{4} } . \frac{4 mol H_{2} O}{1mol N_{2} O_{4} }. \frac{18\frac{g}{mol} H_{2} O}{1mol H_{2} O_} } . \frac{ 1 mol N_{2}O_{4} }{92\frac{g}{mol} N_{2} O_{4} }$

$molH_{2} O = 0,783 mol$

The limiting reagent is N2O4, because can produce only 0, 783 mol of H2O.

This is the minimum measure can be formed of each product.

∴ $MassOfH_{2}O = 0,783mol . 18\frac{g}{mol}$

$MassOfH_{2}O = 14,09g$

5 0

4 years ago

Andi performs the calculation that is shown below.

Sergio039 [100]

Answer: -

A) 3.59

Explanation: -

The number of significant figures in 2.06 = 3

The number of significant figures in 1.743 = 4

The number of significant figures in 1.00 = 3

During multiplication of significant figures, the number of significant figures in the answer would be the smallest value that is 3 in this case.

2.06 x 1.743 x 1.00 = 3.5908.

The answer to 3 significant figures is

2.06 x 1.743 x 1.00 = 3.59

Thus, Andi when performs the calculation that is shown below.

(2.06)(1.743)(1.00) the answer be reported using the correct number of significant figures as A) 3.59

8 0

3 years ago

Read 2 more answers

A solution of NaCl ( aq ) is added slowly to a solution of lead nitrate, Pb ( NO 3 ) 2 ( aq ) , until no further precipitation o

3241004551 [841]

Answer:

The molarity of the $Pb(NO_3)_2$ solution = 0.1895 M

Explanation:

The mass of the $PbCl_2$ obtained = 10.55 g

Molar mass of $PbCl_2$ = 278.1 g/mol

The formula for the calculation of moles is shown below:

$moles = \frac{Mass\ taken}{Molar\ mass}$

Thus,

$Moles= \frac{10.55\ g}{278.1\ g/mol}$

$Moles_{PbCl_2}= 0.0379\ mol$

From the reaction:-

$Pb(NO_3)_2+2NaCl\rightarrow PbCl_2+2NaNO_3$

1 mole of $PbCl_2$ is produced from 1 mole of $Pb(NO_3)_2$

Also,

0.0379 mole of $PbCl_2$ is produced from 0.0379 mole of $Pb(NO_3)_2$

Mole of $Pb(NO_3)_2$ = 0.0379 mol

Volume = 200.0 mL = 0.2 L ( 1 mL = 0.001 L )

Molarity is defined as the number of moles present in one liter of the solution. It is basically the ratio of the moles of the solute to the liters of the solution.

The expression for the molarity, according to its definition is shown below as:

$Molarity=\frac{Moles\ of\ solute}{Volume\ of\ the\ solution}$

Thus,

$Molarity=\frac{0.0379}{0.2}\ M = 0.1895\ M$

<u>The molarity of the $Pb(NO_3)_2$ solution = 0.1895 M</u>

4 0

3 years ago

how would it affect your results if you used a wet Erlenmeyer flask instead of a dry one when transferring your acid solution fr

nikklg [1K]

It would dilute the acid?

4 0

3 years ago