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2 years ago

5

When chemical W is added to a liquid, the chemical breaks down into substances X, Y, and Z. It is not possible to break down sub

stances X, Y, and Z into simpler substances. What conclusion is supported by this evidence?

2 answers:

Salsk061 [2.6K]2 years ago

6 0

Answer:

223

Explanation:

12345 [234]2 years ago

5 0

X, Y, and Z are elements.

An element is the smallest part of a substance that can exist on its own and can not be broken down into smaller unit.

From the question, W was broken down to give X, Y, Z. All three substances can not be broken down by chemical means.

The fact that X, Y and Z can not be broken down by chemical means implies that they are elements.

Learn more: brainly.com/question/13516179

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Calculate the percent ionization of propionic acid (C2H5COOH) in solutions of each of the following concentrations (Ka is given

tekilochka [14]

Answer:

a) = 0.704%

b) = 1.30%

c) = 2.60%

Explanation:

Given that:

$K_a$ = $1.34*10^{-5$

For Part A; where Concentration of A = 0.270 M

Percentage Ionization(∝) $\alpha = \sqrt{\frac{K_a}{C} }$

$\alpha = \sqrt{\frac{1.34*10^{-5}}{0.270} }$

$\alpha = \sqrt{4.9629*10^{-5}}$

$\alpha = 7.044*10^{-3$

percentage% (∝) = $7.044*10^{-3}*100$

= 0.704%

For Part B; where Concentration of B = $7.84*10^{-2$ M

$\alpha = \sqrt{\frac{1.34*10^{-5}}{7.84*10^{-2}} }$

$\alpha = \sqrt{1.709*10^{-4} }$

$\alpha = 0.0130\\$

percentage% (∝) = 0.0130 × 100%

= 1.30%

For Part C; where Concentration of C= $1.92*10^{-2} M$

$\alpha = \sqrt{\frac{1.34*10^{-5}}{1.97*10^{-2}} }$

$\alpha = \sqrt{6.802*10^{-4}}$

$\alpha =0.02608$

percentage% (∝) = 0.02608 × 100%

= 2.60%

7 0

3 years ago

. The freezing point of an aqueous solution containing a nonelectrolyte solute is – 2.79 °C. What is the boiling point of this s

Semmy [17]

Answer:

Boiling point of the solution is 100.78°C

Explanation:

This is about colligative properties.

First of all, we need to calculate molality from the freezing point depression.

ΔT = Kf . m . i

As the solute is nonelectrolyte, i = 1

0°C - (-2.79°C) = 1.86 °C/m . m . 1

2.79°C / 1.86 m/°C = 1.5 m

Now, we go to the boiling point elevation

ΔT = Kb . m . i

Final T° - 100°C = 0.52 °C/m . 1.5m . 1

Final T° = 0.52 °C/m . 1.5m . 1 + 100°C → 100.78°C

4 0

3 years ago

A 200.0 mL solution of 0.40 M ammonium chloride was titrated with 0.80 M sodium hydroxide. What was the pH of the solution after

choli [55]

Answer:

9.25

Explanation:

Let first find the moles of $NH_4Cl$ and $NaOH$

number of moles of $NH_4Cl$ = 0.40 mol/L × 200 × 10⁻³L

= 0.08 mole

number of moles of $NaOH$ = 0.80 mol/L × 50 × 10⁻³L

= 0.04 mole

The equation for the reaction is expressed as:

$NH^+_{4(aq)} \ + OH^-_{(aq)} ------> NH_{3(g)} \ + H_2O_{(l)}$

The ICE Table is shown below as follows:

$NH^+_{4(aq)} \ + OH^-_{(aq)} ------> NH_{3(g)} \ + H_2O_{(l)}$

Initial (M) 0.08 0.04 0

Change (M) - 0.04 -0.04 + 0.04

Equilibrium (M) 0.04 0 0.04

$K_a*K_b = 10^{-14} \ at \ 25^0C$

$K_a = \frac{10^{-14}}{K_b}$

$K_a = \frac{10^{-14}}{1.76*10^{-5}}$

$K_a= 5.68*10^{10}$

$pK_a = - log \ (K_a)$

$pK_a = - log \ (5.68*10^{-10})$

$pK_a = 9.25$

$pH = pKa + \ log (\frac{HB}{HA} )$ for buffer solutions

$pH = pKa + \ log (\frac{moles \ of \ base }{ moles\ of \ acid} )$ since they are in the same solution

$pH = 9.25 + \ log (\frac{0.04 }{ 0.04} )$

$pH = 9.25$

8 0

2 years ago

What is mb for CH3NH2….

drek231 [11]

Answer:

A.) $K_b = \frac{[CH_3NH_3^+][OH^-]}{[CH_3NH_2]}$

Explanation:

The general Kb expression is:

$K_b = \frac{[HA][OH^-]}{[A^-]}$

In this equation

-----> Kb = equilibrium constant

-----> [HA] = acid

-----> [A⁻] = base

Since liquids are not included in equilibrium expressions, H₂O should not be present. The products are in the numerator while the reactant are in the denominator. In this reaction, CH₃NH₂ is acting as a base and CH₃NH₃⁺ is acting as an acid.

As such, the expression is:

$K_b = \frac{[CH_3NH_3^+][OH^-]}{[CH_3NH_2]}$

7 0

2 years ago

How many mole of ZnCl2 will be produced from 55.0 g of Zn, assuming HCL is available in excess

Ilia_Sergeevich [38]

Answer:

0.84 mol

Explanation:

Given data:

Moles of ZnCl₂ produced = ?

Mass of Zn = 55.0 g

Solution:

Chemical equation:

2HCl + Zn → ZnCl₂ + H₂

Number of moles of Zn:

Number of moles = mass / molar mass

Number of moles = 55.0 g/ 65.38 g/mol

Number of moles = 0.84 mol

Now we will compare the moles of Zn with ZnCl₂ from balance chemical equation.

Zn : ZnCl₂

1 : 1

0.84 : 0.84

So from 55 g of Zn 0.84 moles of zinc chloride will be produced.

8 0

3 years ago