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3 years ago

Answer for 1A and 1B

Chemistry

1 answer:

SCORPION-xisa [38]3 years ago

8 0

Happy birthday to you and enjoy your day even if it is without a wonderful time in the Lord

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The volume of 7.91 M HCl needed to make 196.1 mL of 2.13 M HCl is ____.

kumpel [21]

Answer:

a) 52.8

Explanation:

M1V1 = M2V2

(7.91 M)(x ml) = (2.13 M) (196.1 ml)

(7.91M) (xml) = 417.693 M.ml

x ml = 417.693/ 7.91

x = 52.8

6 0

3 years ago

Read 2 more answers

How many grams of Lioh are in 1.7 moles please help

natali 33 [55]

Answer:

23.94834 gram

Explanation:

4 0

3 years ago

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A 100.0-mL sample of 1.00 M NaOH is mixed with 50.0 mL of 1.00 M H2SO4 in a large Styrofoam coffee cup calorimeter fitted with a

worty [1.4K]

Answer:

THE ENTHALPY CHANGE IN KJ/MOLE IS +114 KJ/MOLE.

Explanation:

Heat = mass * specific heat capacity * temperature rise

Total volume = 100 + 50 = 150 mL

Total mass = density * volume

Total mass = 1 * 150 mL = 150 g

So therefore, the heat evolved during the reaction is:

Heat = 150 * 4.18 * ( 31.4 - 22.3)

Heat = 150 * 4.18 * 9.1

Heat = 5705.7 J

Equation for the reaction:

2 NaOH(aq) + H2SO4(aq) → Na2SO4(aq) + 2 H2O(l)

From the equation, 2 moles of NaOH reacts with 1 mole of H2SO4 to produce 1 mole of Na2SO4 and 2 moles of water

50 mL of 1 M of H2SO4 contains

50 * 1 / 1000 mole of acid

= 0.05 mole of acid

The production of 1 mole of water evolved 5705.7 J of heat and hence the enthalpy changein kJ per mole will be:

0.05 mole of H2SO4 produces 5705.7 J of heat

1 mole of H2SO4 will produce 5705.7 / 0.05 J

= 114,114 J / mole

In kj/mole = 114 kJ/mole.

Hence, the enthalpy change of the reaction in kJ /mole is +114 kJ/mole.

5 0

3 years ago

Solids are usually more dense than:

MrMuchimi

Solids are usually more dense than liquids and gases.

6 0

3 years ago

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To a 25.00 mL volumetric flask, a lab technician adds a 0.150 g sample of a weak monoprotic acid, HA , and dilutes to the mark w

Elis [28]

<u>Answer:</u> The number of moles of weak acid is $4.24\times 10^{-3}$ moles.

<u>Explanation:</u>

To calculate the moles of KOH, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}}\text{Volume of solution (in L)}}$

We are given:

Volume of solution = 43.81 mL = 0.04381 L (Conversion factor: 1L = 1000 mL)

Molarity of the solution = 0.0969 moles/ L

Putting values in above equation, we get:

$0.0969mol/L=\frac{\text{Moles of KOH}}{0.04381}\\\\\text{Moles of KOH}=4.24\times 10^{-3}mol$

The chemical reaction of weak monoprotic acid and KOH follows the equation:

$HA+KOH\rightarrow KA+H_2O$

By Stoichiometry of the reaction:

1 mole of KOH reacts with 1 mole of weak monoprotic acid.

So, $4.24\times 10^{-3}mol$ of KOH will react with = $\frac{1}{1}\times 4.24\times 10^{-3}=4.24\times 10^{-3}mol$ of weak monoprotic acid.

Hence, the number of moles of weak acid is $4.24\times 10^{-3}$ moles.

6 0

3 years ago