Chemistry, 28.08.2019 01:30 jfw0504

In the laboratory a student combines 28.7 ml of a 0.240 m barium chloride solution with 17.5 ml of a 0.331 m chromium(ii) chloride solution. what is the final concentration of chloride anion ?

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Answer from: robert7248

Answer : The final concentration of chloride ion is, 0.549 M

Explanation :

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

$Moles =Molarity \times {Volume\ of\ the\ solution}$

Barium chloride will furnish chloride ions as:

$BaCl_2\rightarrow Ba^{2+}+2Cl^-$

Given :

For barium chloride :

Molarity = 0.240 M

Volume = 28.7 mL

The conversion of mL to L is shown below:

1 mL = 10⁻³ L

Thus, volume = 28.7×10⁻³ L

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

$Moles =2\times 0.240 \times {28.7\times 10^{-3}}\ moles$

Moles of chloride ions by barium chloride = 0.013776 moles

Chromium(II) chloride will furnish chloride ions as:

$CrCl_2\rightarrow Cr^{2+}+2Cl^-$

Given :

For chromium(II) chloride :

Molarity = 0.331 M

Volume = 17.5 mL

The conversion of mL to L is shown below:

1 mL = 10⁻³ L

Thus, volume = 17.5×10⁻³ L

Thus, moles of chlorine furnished by chromium(II) chloride is twice the moles of chromium(II) chloride as shown below:

$Moles =2\times 0.331 \times {17.5\times 10^{-3}}\ moles$

Moles of chloride ions by chromium(II) chloride = 0.011585 moles

Total moles = 0.013776 moles + 0.011585 moles = 0.025361 moles

Total volume = (28.7×10⁻³ L) + (17.5×10⁻³ L) = 46.2×10⁻³ L

Concentration of chloride ions is:

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

$\text{Molarity of chloride anion}=\frac{0.025361}{46.2\times 10^{-3}}=0.549M$

Therefore, the final concentration of chloride anion is 0.549 M

Answer from: Quest

when 25.0 ml of 0.700 mol/l naoh was mixed in a calorimeter with 25.0 ml of 0.700 mol/l hcl, both initially at 20.0 °c, the temperature increased to 22.1 °c. the heat capacity of the calorimeter is 279 j/°c.

the equation for the reaction is:

naoh + hcl → nacl + h₂o

moles of hcl = 0.0250 l hcl ×

0.700

mol hcl

1 l hcl = 0.0175 mol hcl

volume of solution = (25.0 + 25.0) ml = 50.0 ml

mass of solution = 50.0 ml soln ×

1.00

ml soln

= 50.0 g soln

δt=t2-t1

= (22.1 – 20.0) °c = 2.1 °c

the heats involved are

heat from neutralization + heat to warm solution + heat to warm calorimeter = 0

q1+q2+q3=0

nδh+mcδt+cδt=0

0.0175 mol ×

δh

+ 50.0 g × 4.184 j·g⁻¹°c⁻¹ × 2.1 °c + 279 j°c⁻¹ × 2.1 °c = 0

0.0175 mol ×

δh

+ 439.32 j + 585.9 j = 0

0.0175 mol ×

δh

= -1025.22 j

δh=1025.22j0.0175 mol= -58 600 j/mol = -58.6 kj/mol

Answer from: Quest

The answer is c) pyrite

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In the laboratory a student combines 28.7 ml of a 0.240 m barium chloride solution with 17.5 ml of a...