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3 years ago

Calculate the number of moles of solute present and the number of grams

Chemistry

1 answer:

nekit [7.7K]3 years ago

6 0

Answer:

The answer is: 18 moles and 1341, 72 grams of KCl

Explanation:

The molarity is defined as the moles of solute ( in this case KCl) in 1 liter of solution:

1L solution-----3 moles of KCl

6L solution-----x= (6L solutionx 3 moles of KCl)/1 L solution= <em>18 moles of KCl</em>

We calculate the weight of 1 mol of KCl:

Weight 1 mol KCl= Weight K + Weight Cl= 39,09 g + 35, 45 g=74, 54 g/mol

1 mol KCl----- 74, 54 g

18 mol KCl----x= (18 mol KCl x 74, 54 g)/1 mol KCl=<em>1341, 72 g</em>

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What happens to the atomic radius when an electron is gained?

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A. The negative ionic radius is larger than the neutral atomic radius.

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2 years ago

How much concentrated 18M sulfuric acid is needed to prepare 250mL of a 6.0M solution?

slavikrds [6]

C₁ * V₁ = C₂ * V₂

18 * V₁ = 6.0 * 250

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hope this helps!

5 0

3 years ago

A student prepares a 1.8 M aqueous solution of 4-chlorobutanoic acid (C2H CICO,H. Calculate the fraction of 4-chlorobutanoic aci

Kruka [31]

Answer:

Percentage dissociated = 0.41%

Explanation:

The chemical equation for the reaction is:

$C_3H_6ClCO_2H_{(aq)} \to C_3H_6ClCO_2^-_{(aq)}+ H^+_{(aq)}$

The ICE table is then shown as:

$C_3H_6ClCO_2H_{(aq)} \ \ \ \ \to \ \ \ \ C_3H_6ClCO_2^-_{(aq)} \ \ + \ \ \ \ H^+_{(aq)}$

Initial (M) 1.8 0 0

Change (M) - x + x + x

Equilibrium (M) (1.8 -x) x x

$K_a = \frac{[C_3H_6ClCO^-_2][H^+]}{[C_3H_6ClCO_2H]}$

where ;

$K_a = 3.02*10^{-5}$

$3.02*10^{-5} = \frac{(x)(x)}{(1.8-x)}$

Since the value for $K_a$ is infinitesimally small; then 1.8 - x ≅ 1.8

Then;

$3.02*10^{-5} *(1.8) = {(x)(x)}$

$5.436*10^{-5}= {(x^2)$

$x = \sqrt{5.436*10^{-5}}$

$x = 0.0073729 \\ \\ x = 7.3729*10^{-3} \ M$

Dissociated form of 4-chlorobutanoic acid = $C_3H_6ClCO_2^- = x= 7.3729*10^{-3} \ M$

Percentage dissociated = $\frac{C_3H_6ClCO^-_2}{C_3H_6ClCO_2H} *100$

Percentage dissociated = $\frac{7.3729*10^{-3}}{1.8 }*100$

Percentage dissociated = 0.4096

Percentage dissociated = 0.41% (to two significant digits)

3 0

3 years ago

Read 2 more answers

There are two steps in the usual industrial preparation of acrylic acid, the immediate precursor of several useful plastics. In

KIM [24]

Answer:

ΔH = -470.4kJ

Explanation:

It is possible to sum 2 or more reactions to obtain the ΔH of the reaction you want to study (Hess's law). Using the reactions:

1. CaC2(s) + 2H2O(l) → C2H2(g) + Ca(OH)2(s)ΔH = −414kJ

2. 6C2H2(g) + 3CO2(g) + 4H2O(g) → 5CH2CHCO2H(g)ΔH = 132kJ

6 times the reaction 1.

6CaC2(s) + 12H2O(l) → 6C2H2(g) + 6Ca(OH)2(s)ΔH = −414kJ*6 = -2484kJ

This reaction + 2:

6CaC2(s) + 3CO2(g) + 16H2O(l) → + 6Ca(OH)2(s) + 5CH2CHCO2H(g) ΔH = -2484kJ + 132kJ = -2352kJ

As we want to calculate the net change enthalpy in the formation of just 1 mole of acrylic acid we need to divide this last reaction in 5:

6/5CaC2(s) + 3/5CO2(g) + 16/5H2O(l) → + 6/5Ca(OH)2(s) + CH2CHCO2H(g) ΔH = -2352kJ / 5

4 0

3 years ago

Water combines with carbon dioxide Choose one: A. only in the atmosphere. B. to produce hydrogen sulfide. C. to precipitate calc

katovenus [111]

Water combines with carbon dioxide to produce slightly acidic groundwater

that dissolves limestone and forms caves.

This is because the reaction between water and carbondioxide to form

bicarbonate ions( HCO₃⁻). The bicarbonate ions dissociate into Hydrogen

atoms thereby increasing the acidity.

The acidic environment results in the formation of acidic groundwater that

dissolves limestone and forms caves.

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8 0

3 years ago