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ArbitrLikvidat [17]
3 years ago
12

2. How many moles of hydrogen atoms are there in 154 mL of 0.18 M H2SO4? Write your

Chemistry
1 answer:
katrin [286]3 years ago
6 0

2. 0.05544 moles of hydrogen atom are present in 154 mL 0.18 M solution of H2SO4.

3. 15.2506 heat in Joules is absorbed by 150.0 mL of pure water that is heated from 21.2°C to  45.5°C.

4. 0.75 M is the concentration of Na+ ions in 25.0 mL of 1.50 M NaOH is reacted with 25.0 mL of  1.50 M HCI

Explanation:

Number of moles of H2SO4 can be calculated by the given volume and molarity from the formula:

molarity = \frac{number of moles}{volume of teh solution}

number of moles = molarity × volume of the solution of H2SO4

    number of moles = 0.18 × 0.154 litres

                                   = 0.02772 moles of H2SO4.

Since 1 mole of H2SO4 contains 2 moles of hydrogen

so, 0.02772 moles of H2SO4 will have x moles

\frac{1}{2} = \frac{0.0272}{x}

2 × 0.02772 = x

0.05544 moles of hydrogen atom are present in 154 mL 0.18 M solution of H2SO4.

3. The heat absorbed is calculated from the formula:

ΔH = cp × m × ΔT   ( ΔT = change in temperature in Kelvin, m in kg, cp= specific heat of water)

putting the values in formula:

ΔH = 4.184 × 0.15 × ( 318.65-294.35)

     =  4.184 × 0.15× 24.3

        = 15.2506 Joules of heat is absorbed.

4. The concentration of Na+ ions

the balanced equation is

NaOH + HCl⇒ NaCl + H20

from one mole of NaCl 1 mole of NaCl i.e one 1 mole of Na+ ions is formed.

number of moles of NaOH is calculated by the formula:

Molarity = \frac{number of moles}{volume of the solution}

number of moles = 0.025L × 1.50M

                             = 0.0375 moles of NaOH

so 1 mole of NaOH produces 1 mole of Na= ions

hence, 0.0375 moles produces x moles of Na+ ions

\frac{1}{1} = \frac{0.0375}{x}

moles of Na+ is produced.  

concentration of NaOH in 50 ml solution because NaOH and HCl of 25 ml reacted.

applying the molarity formula from above

Molarity =  \frac{0.0375}{0.05}

              =  0.75 M

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Exactly 15.0 g of a substance can be dissolved in 150.0 g of water what is the solubility of the substance in grams per 100 g of
Leokris [45]
<span>(15.0 g) / (150.0 g) x (100 g) = 10.0 g/100 g H2O </span>
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3 years ago
if the mass of 191 grams NaCl reacted with 74 frams of calcium hydroxide and 80 grams of sodium hydroxide is produced, what mass
nadya68 [22]
<h3>Answer:</h3>

110.98 g/mol

<h3>Explanation:</h3>

The reaction between NaCl and Ca(OH)₂ is given by the equation;

2NaCl(aq) + Ca(OH)₂(s) → 2NaOH(aq) + CaCl₂(aq)

We are required to determine the mass of CaCl₂ produced,

We will use the following simple steps;

Step 1: Moles of NaCl and Ca(OH)₂ given

Number of moles = Mass ÷ Molar mass

Moles of NaCl

Mass of NaCl = 191 g

Molar mass NaCl = 58.44 g/mol

Number of moles = 191 g ÷ 58.44 g/mol

                             = 3.268 moles

                             = 3.27 Moles

Moles of Ca(OH)₂

Mass of Ca(OH)₂ = 74 g

Molar mass of Ca(OH)₂ = 74.093 g/mol

Number of moles = 74 g ÷ 74.093 g/mol

                             = 0.998 mole

                              = 1.0 mole

However, from the equation  2 moles of NaCl requires 1 mole of Ca(OH)₂

Therefore, from the amount of reactants available NaCl was in excess and Ca(OH)₂ is the limiting reactant .

Step 2: Moles of CaCl₂ produced

From the equation

1 mole of Ca(OH)₂ reacts with NaCl to produce 1 mole of CaCl₂

Therefore; the mole ratio of Ca(OH)₂ to CaCl₂ is 1: 1

Thus;

Moles of CaCl₂ produced is 1.0 moles

Step 3: Mass of CaCl₂ produced

Moles of CaCl₂ = 1.0 mole

Molar mass CaCl₂ = 110.98 g/mol

But; mass = number of moles × Molar mass

Therefore;

Mass of CaCl₂ = 1.0 mole × 110.98 g/mol

                       = 110.98 g CaCl₂

3 0
3 years ago
Anna is excited about conducting her lab today. She takes out her lab sheet and reads over the lab. She goes and gathers all of
trapecia [35]

Answer:

Anna didn't wear a lab coat

Explanation:

The first thing anyone who is to carry out a procedure or an experiment in a lab needs to do when he/she enters a lab is to wear a lab coat. This is a lab safety rule that protects the individual from been affected by spillage that could occur as a result of the experiment been conducted.

From the narration in the question, <u>it is not stated anywhere that Anna wore a lab coat before embarking on her lab procedure</u>.

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Give at least 3 possible Science Investigatory Project
garri49 [273]

Used cooking oil as a substitute for diesel?

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calculate the volume (in mL) of 0.100 M CaCl2 needed to produce 1.00 g of CaCO3 (s). There is an excess of Na2CO3. Volume of cal
shepuryov [24]

Answer:

100. mL

Explanation:

Step 1: Write the balanced equation for the double displacement reaction

CaCl₂ + Na₂CO₃ ⇒ 2 NaCl + CaCO₃

Step 2: Calculate the moles corresponding to 1.00 g of CaCO₃

The molar mass of CaCO₃ is 100.09 g/mol.

1.00 g × 1 mol/100.09 g = 0.0100 mol

Step 3: Calculate the moles of CaCl₂ required to produce 0.0100 moles of CaCO₃

The molar ratio of CaCl₂ to CaCO₃ is 1:1. The moles of CaCl₂ required are 1/1 × 0.0100 mol = 0.0100 mol.

Step 4: Calculate the volume of 0.100 M CaCl₂ that contains 0.0100 mol

0.0100 mol × 1 L/0.100 mol × 1000 mL/1 L = 100. mL

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