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sergij07 [2.7K]

2 years ago

6

What is the total mass (in grams) of 1.000 l of an aqueous 1.500 m nacl solution with a density of 1.058 g/ml? (do not include t

he unit in your answer)?

1 answer:

lawyer [7]2 years ago

7 0

The density is an unique property of a compound in solid or liquid phase. The density is the mass of the compound in its unit volume. Now due to the addition of an particular amount of solid like in this case salt (NaCl) the density of the water will change and the density given will be the density of the salted water.

The volume of the salted water is 1L or 1000mL. The density of the salted water is 1.058g/mL. Thus the 1 mL of the solution have the mass 1.058 g. Thus 1000 mL of the solution will have mass 1058.

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A piece of wire contains 1.52x10^22 atoms of copper. Calculate the miles of copper in the wire

melamori03 [73]

Moles of Copper =

$1.52$ × $10^{22}atoms$ × $\frac{1 mol }{6.022 * 10^{23}atoms}$

= 0.0252 mol Cu

6 0

2 years ago

If 36.9 mL of B2H6 reacted with excess oxygen gas, determine the actual yield of B2O3 if the percent yield of B2O3 was 75.7%. (T

zhuklara [117]

Answer: The actual yield of $B_2O_3$ is 60.0 g

Explanation:-

The balanced chemical reaction :

$B_2H_6(l)+3O_2(g)\rightarrow B_2O_3(s)+3H_2O(l)$

Mass of $B_2H_6$ = $Density\times Volume=1.131g/ml\times 36.9ml=41.7g$

$\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}$

$\text{Moles of} B_2H_6=\frac{41.7g}{27.668g/mol}=1.51moles$

According to stoichiometry:

1 mole of $B_2H_6$ gives = 1 mole of $B_2O_3$

1.51 moles of $B_2H_6$ gives = $\frac{1}{1}\times 1.51=1.51$ moles of $B_2O_3$

Theoretical yield of $B_2O_3=moles\times {Molar mass}}=1.14mol\times 69.62g/mol=79.3g$

Percent yield of $B_2O_3$ = $75.7\%$

$\%\text{ yield}=\frac{\text{Actual yield}}{\text{Theoretical yield}}\times 100$

$75.7\%=\frac{\text{Actual yield}}{79.3}\times 100$

${\text{Actual yield}}=60.0g$

Thus the actual yield of $B_2O_3$ is 60.0 g

7 0

2 years ago

Which of the following contains maximum number of atoms? a) 2.0g hydrogen b) 2.0g oxygen c) 2.0g nitrogen d) 2.0g methane

Fynjy0 [20]

A) 2.0g hydrogen is your answer

7 0

2 years ago

A 50.0 mL sample of 0.600 M calcium hydroxide is mixed with 50.0 mL sample of 0.600 M hydrobromic acid in a Styrofoam cup. The t

TEA [102]

Explanation:

The reaction equation will be as follows.

$Ca(OH)_{2}(aq) + 2HBr(aq) \rightarrow CaBr_{2}(aq) + 2H_{2}O(l)$

So, according to this equation, 1 mole $Ca(OH)_{2}$ = 2 mol HBr = 1 mol $CaBr_{2}$

Therefore, calculate the number of moles of calcium hydroxide as follows.

No. of moles of $Ca(OH)_{2}$ = $V \times Molarity$

= $50 \times 0.6$

= 30 mmol

Similarly, calculate the number of moles of HBr as follows.

No. of moles of HBr = $M \times V$

= $50 \times 0.6$

= 30 mmol

This means that the limiting reactant is HBr.

So, no. of moles of $CaBr_{2}$ = $30 \times \frac{1}{2}$

= 15 mmol

Hence, calculate the amount of heat released as follows.

Heat released in the reaction(q) = $m \times s \times \Delta T$

as, m = mass of solution

and, Density = $\frac{mass}{volume}$

or, mass = Density × Volume

= 1.08 g/ml \times (50 + 50) ml

= 108 g

where, s = specific heat of solution = 4.18 j/g.k

and, change in temperature $\Delta T$ = $(26 - 23)^{o}C$

= $3
^{o}C$

Hence, the heat released will be as follows.

q = $m \times s \times \Delta T$

q = $108 \times 4.18 \times 3^{o}C$

= 1354.32 joule

or, = 1.354 kJ (as 1 kJ = 1000 J)

Also, $\Delta H_{rxn}$ = $\frac{-q}{n}$

= $\frac{-1.354}{15 \times 10^{-3}}$

= -90.267 kJ/mol

Thus, we can conclude that the enthalpy change for the given reaction is -90.267 kJ/mol.

6 0

2 years ago

A 0.20 mol sample of MgCl2(s) and a 0.10 mol sample of KCl(s) are dissolved in water and diluted to 500 mL. What is the concentr

igor_vitrenko [27]

Answer:

1 M

Explanation:

Magnesium chloride will furnish chloride ions as:

$MgCl_2\rightarrow Mg^{2+}+2Cl^-$

Given :

Moles of magnesium chloride = 0.20 mol

Thus, moles of chlorine furnished by magnesium chloride is twice the moles of magnesium chloride as shown below:

$Moles =2\times 0.20\ moles$

Moles of chloride ions by magnesium chloride = 0.40 moles

Potassium chloride will furnish chloride ions as:

$KCl\rightarrow K^{+}+Cl^-$

Given :

Moles of potassium chloride = 0.10 moles

Thus, moles of chlorine furnished by potassium chloride is same as the moles of potassium chloride as shown below:

Moles of chloride ions by potassium chloride = 0.10 moles

Total moles = 0.40 + 0.10 moles = 0.50 moles

Given, Volume = 500 mL = 0.5 L (1 mL = 10⁻³ L)

Concentration of chloride ions is:

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

$Molarity_{Cl^-}=\frac{0.50}{0.5}$

<u> The final concentration of chloride anion = 1 M</u>

8 0

2 years ago