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2 years ago

Solution A is yellow when alizarin yellow is added and blue when thymol

Chemistry

1 answer:

alexira [117]2 years ago

3 0

Answer:

D. 6.3 x 10-5 mol/L NaOH

Explanation:

Alizarin yellow is an indicator that is yellow when pH < 10.1. In the same way, thymol blue is blue when pH > 9

That means the pH of the solution is between 9 - 10.1

Any acid as HCl could have a pH of these.

The solution of 3.2x10⁻⁴M NaOH has a pH of:

pOH = -log[OH-] = 3.49

pH = 14-pOH = 10.51. The pH of the solution is not 10.5

Now, the solution of 6.3x10⁻⁵M NaOH has a pH:

pOH = -log[OH-] = 4.2

pH = 14-pOH = 9.8

The pH of the solution could be 9.8. Right option is:

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The type of change that occurs when a solid becomes a liquid is _____.

Brilliant_brown [7]

This is going to be a <em>phase </em>change!

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3 years ago

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The equilibrium constant Kp for the reaction (CH3),CCI (g) = (CH3),C=CH, (g) + HCl (g) is 3.45 at 500. K. (5.00 x 10K) Calculate

Karolina [17]

<u>Answer:</u> The value of $K_p$ for the reaction is 6.32 and concentrations of $(CH_3)_2C=CH,HCl\text{ and }(CH_3)_3CCl$ is 0.094 M, 0.094 M and 0.106 M respectively.

<u>Explanation:</u>

Relation of $K_p$ with $K_c$ is given by the formula:

$K_p=K_c(RT)^{\Delta ng}$

where,

$K_p$ = equilibrium constant in terms of partial pressure = 3.45

$K_c$ = equilibrium constant in terms of concentration = ?

R = Gas constant = $0.0821\text{ L atm }mol^{-1}K^{-1}$

T = temperature = 500 K

$\Delta n_g$ = change in number of moles of gas particles = $n_{products}-n_{reactants}=2-1=1$

Putting values in above equation, we get:

$3.45=K_c\times (0.0821\times 500)^{1}\\\\K_c=\frac{3.45}{0.0821\times 500}=0.084$

The equation used to calculate concentration of a solution is:

$\text{Molarity}=\frac{\text{Moles}}{\text{Volume (in L)}}$

Initial moles of $(CH_3)_3CCl(g)$ = 1.00 mol

Volume of the flask = 5.00 L

So, $\text{Concentration of }(CH_3)_3CCl=\frac{1.00mol}{5.00L}=0.2M$

For the given chemical reaction:

$(CH_3)_3CCl(g)\rightarrow (CH_3)_2C=CH(g)+HCl(g)$

Initial: 0.2 - -

At Eqllm: 0.2 - x x x

The expression of $K_c$ for above reaction follows:

$K_c=\frac{[(CH_3)_2C=CH]\times [HCl]}{[(CH_3)_3CCl]}$

Putting values in above equation, we get:

$0.084=\frac{x\times x}{0.2-x}\\\\x^2+0.084x-0.0168=0\\\\x=0.094,-0.178$

Negative value of 'x' is neglected because initial concentration cannot be more than the given concentration

Calculating the concentration of reactants and products:

$[(CH_3)_2C=CH]=x=0.094M$

$[HCl]=x=0.094M$

$[(CH_3)_3CCl]=(0.2-x)=(0.2-0.094)=0.106M$

Hence, the value of $K_p$ for the reaction is 6.32 and concentrations of $(CH_3)_2C=CH,HCl\text{ and }(CH_3)_3CCl$ is 0.094 M, 0.094 M and 0.106 M respectively.

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3 years ago

Name the type of chemical reaction : CaO(s) + H2O(l) -----------Ca(OH)2(aq)

Anna [14]

It is called Synthesis reaction

5 0

3 years ago

The molar mass is determined by measuring the freezing point depression of an aqueous solution. A freezing point of -5.20°C is r

Dima020 [189]

Answer:

The empirical formula is C2H4O3

The molecular formula is C4H8O6

The molar mass is 152 g/mol

Explanation:

The complete question is: An unknown compound contains only carbon, hydrogen, and oxygen. Combustion analysis of the compound gives mass percents of 31.57% C and 5.30% H. The molar mass is determined by measuring the freezing-point depression of an aqueous solution. A freezing point of -5.20°C is recorded for a solution made by dissolving 10.56 g of the compound in 25.0 g water. Determine the empirical formula, molar mass, and molecular formula of the compound. Assume that the compound is a nonelectrolyte.

Step 1: Data given

Mass % of Carbon = 31.57 %

Mass % of H = 5.30 %

Freezing point = -5.20 °C

10.56 grams of the compound dissolved in 25.0 grams of water

Kf water = 1.86 °C kg/mol

Step 2: Calculate moles of Carbon

Suppose 31.57% = 31.57 grams

moles C = mass C / Molar mass C

moles C = 31.57 grams / 12.0 g/mol = 2.63 moles

Step 3: Calculate moles of Hydrogen:

Moles H = 5.30 grams / 1.01 g/mol

moles H = 5.25 moles

Step 4: Calculate moles of Oxygen

Moles O = ( 100 - 31.57 - 5.30) / 16 g/mol

Moles O = 3.95 moles

Step 5: We divide by the smallest number of moles

C: 2.63 / 2.63 = 1 → 2

H: 5.25/2.63 = 2 → 4

O: 3.95/ 2.63 = 1.5 → 3

The empirical formula is C2H4O3

The molar mass of the empirical formula = 76 g/mol

Step 6: Calculate moles solute

Freezing point depression = 5.20 °C = m * 1.86

m = 5.20 / 1.86

m = 2.80 molal = 2.80 moles / kg

2.80 molal * 0.025 kg = 0.07 moles

Step 7: Calculate molar mass

Molar mass = mass / moles

Molar mass = 10.56 grams / 0.07 moles

Molar mass = 151 g/mol

Step 8: Calculate molecular formula

151 / 76 ≈ 2

We have to multiply the empirical formula by 2

2*(C2H4O3) = C4H8O6

The molecular formula is C4H8O6

The molar mass is 152 g/mol

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3 years ago

Which is the electron configuration for zinc?

alisha [4.7K]

[AR] 3d10 4s2 is the configuration for zinc

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3 years ago

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