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Ede4ka [16]
2 years ago
7

Imagine you had HCl with a concentration of exactly 0.10 mol/dm3. If 0.023 dm3 of a sodium hydroxide solution, NaOH (aq), could

exactly neutralize 0.040 dm3 of the HCl solution, what is the concentration of the NaOH (aq)
Chemistry
1 answer:
polet [3.4K]2 years ago
8 0

Answer:

Explanation:

Step 1: Calculate the amount of sodium hydroxide in moles

Volume of sodium hydroxide solution = 25.0 ÷ 1,000 = 0.0250 dm3

Rearrange:

Concentration in mol/dm3 =

Amount of solutein mol = concentration in mol/dm3 × volume in dm3

Amount of sodium hydroxide = 0.100 × 0.0250

= 0.00250 mol

Step 2: Find the amount of hydrochloric acid in moles

The balanced equation is: NaOH(aq) + HCl(aq) → NaCl(aq) + H2O(l)

So the mole ratio NaOH:HCl is 1:1

Therefore 0.00250 mol of NaOH reacts with 0.00250 mol of HCl

Step 3: Calculate the concentration of hydrochloric acid in mol/dm3

Volume of hydrochloric acid = 20.00 ÷ 1000 = 0.0200 dm3

Concentration in mol/dm3 =

Concentration in mol/dm3 =

= 0.125 mol/dm3

Step 4: Calculate the concentration of hydrochloric acid in g/dm3

Relative formula mass of HCl = 1 + 35.5 = 36.5

Mass = relative formula mass × amount

Mass of HCl = 36.5 × 0.125

= 4.56 g

So concentration = 4.56 g/dm3

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torisob [31]

Answer:

6) 225.8 g

7) 1.01 g

8) 21.624 g

Explanation:

6) Molar Mass of Lithium Bromine: 86.845

86.845 * 2.6 = 225.8 g

7) Molar Mass of Neon: 20.180

20.180 * 0.05 = 1.01 g

8) Molar Mass of Water Vapor: 18.02

18.02 * 1.2 = 21.624 g

5 0
2 years ago
If 801 J of heat is available, what is the mass in grams of iron (specific heat = 0.45 J/g・°C) that can be heated from 22.5°C to
inysia [295]

Answer:

The correct answer will be "18.25 g".

Explanation:

The given values are:

Specific heat,

C = 0.45 J/g・°C

Heat involved,

q =  801 J

Temperature,

ΔT = 120.0°C-22.5°C

     = 97.5°C

As we know,

⇒  C = \frac{q}{m \Delta T}

On substituting the given values, we get

⇒  0.45=\frac{801}{m(97.5)}

⇒  m = 18.25 \ g

3 0
3 years ago
Balance the following Equation:
lisabon 2012 [21]

Answer:

HCl is the limiting reactant. It will completely be consumed (1.37 moles)

Option D is correct

Explanation:

Step 1: Data given

Mass of Zinc (Zn) = 50.0 grams

Mass of Hydrogen chloride (HCl) = 50.0 grams

atomic mass Zn = 65.38 g/mol

Molar mass HCl = 36.46 g/mol

Step 2: The balanced equation

Zn + 2HCl → ZnCl2 + H2

Step 3: Calculate moles

Moles = mass / molar mass

Moles Zn = 50.0 grams / 65.38 g/mol

Moles Zn = 0.764 moles

Moles HCl = 50.0 grams / 36.46 g/mol

Moles HCl = 1.37 moles

Step 4: Calculate limiting reactant

For 1 mol Zn we need 2 moles HCl to produce 1 mol ZnCl2 and 1 mol H2

HCl is the limiting reactant. It will completely be consumed (1.37 moles)

Zn is in excess. There will react 1.37/2 = 0.685 moles

There will remain 0.764 -0.685 = 0.079 moles

7 0
3 years ago
In this reaction: Mg (s) + I₂ (s) → MgI₂ (s), if 10.0 g of Mg reacts with 60.0 g of I₂, and 53.88 g of MgI₂ form, what is the pe
ad-work [718]

We know the law of conservation of mass

  • It states that mass is neither formed nor destroyed in any chemical reaction.
  • Mass of reactants=Mass of products.

Here

  • Mg and I_2 are reactants
  • MgI_2 is product with some yield.
  • Mass of reactants=10+60.0=70.0g
  • Mass of MgI_2=53.88g
  • Mass of yield=Product-MgI_2=70-53.88=16.12g

Lets find the percentage

\\ \tt\hookrightarrow \dfrac{Mass\:of\:yield}{Total\:mass}\times 100

\\ \tt\hookrightarrow \dfrac{16.12}{70}\times 100

\\ \tt\hookrightarrow 0.23028(010)=23.028\%

6 0
2 years ago
Read 2 more answers
What is the similarity between simple diffusion and facilitated diffusion?
MakcuM [25]

they are both types of passive transport which means they require no energy. They both work with the concentration gradient which means they go from a high concentration area to a low concentration area. The differences are simple diffusion just goes though the membrane of a cell while facilitated diffusion uses a protein channel

Simple diffusion: it is the process where molecules move from a area of high concentration to an are of lower concentration. There is no energy needed in simple diffusion. For example when sodium is highly concentration in a cell, it moves outside of the cell where sodium is less concentration. it takes no energy as simple diffusion is random and molecules move according to their concentration.

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4 0
3 years ago
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