In a titration of hydrobromic acid (HBr) against sodium hydroxide (NaOH), 22.20 mL of 0.150 M NaOH is required to neutralize 36.60 mL of HBr. Calculate the molarity of the HBr. M1V1=M2V2

Answers

Answer 1

Answer:

[HBr] = 0.0910 M

Explanation:

NaOH → Strong base

HBr → Strong acid.

This is a titration that involves a strong acid and a strong base, so the recaction is about acid base equilibrium.

Formula for titration in the equivalence point is:

mmoles of acid = mmoles of base

M . Volume of acid = M . volume of base

We replace data:

M . 36.60 mL = 0.150 M . 22.20 mL

M = (0.150 M . 22.20 mL) / 36.60 mL

M = 0.0910 mol/L


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Answer:

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Answer:

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Answer:

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gases have maximum density and rigidity true or false​

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Answer:

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Answer:

they become water vapor and after that, they become a type of precipitation

The evaporation of a drop of water in an air stream is determined by the rate of transfer of heat and water vapour between the drop surface and the ambient air. If these transfer processes are governed by molecular diffusion the appropriate equations take a well known form. I hope this can help

What are changed after fushing the cover of the jar? ​

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Answer:

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Answer:

When you first set the water balloon on top of the jar, the air pressure outside the jar and the air pressure inside the jar are the same. ... So now there is less air, less stuff, in the jar

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A catalyst is a substance that increases the rate of a chemical reaction by lowering the activation energy without being used up in the reaction. ... Because it is neither a reactant nor a product, a catalyst is shown in a chemical equation by being written above as the yield arrow

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kinetic, mechanical


B)

mechanical, heat


C)

electric, mechanical


D)

mechanical, potential

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the answer to your question is option A.

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Calcium 3 and chlorine 6
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Calculate the number of atoms in 13.2 mole copper

Answers

ABCDEFGHIJKLM NOP QRSTUVWXYZ

In the given question, [tex]7.949 \times 10^{24}[/tex] atoms are present in 13.2 mole of copper.

A mole is defined as the amount of a substance that contains exactly [tex]6.022\times 10^{23}[/tex] 'elementary entities' of the given substance.

The number of atoms in 13.2 moles of copper (Cu) can be calculated using Avogadro's number.

To calculate the number of atoms in 13.2 moles of copper, we can multiply the number of moles by Avogadro's number:

Number of atoms = [tex]\rm 13.2\ moles\ \times[/tex] [tex]6.022\times 10^{23}[/tex] atoms/mole

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Therefore, in 13.2 moles of copper, there are approximately [tex]7.949 \times 10^{24}[/tex] atoms.

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Answer:

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Just did the question

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Answer:

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Explanation:

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Answers

Answer:

See explanation

Explanation:

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Answer:

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CO2 AND H2O are the products of the following reaction

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Answer:

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If 18.5 grams of CuCl2 react with 22.8 grams of NaNO3, what mass of NaCl can be formed?

CuCl2 + NaNO3 → Cu(NO3)2 + NaCl

12.4 g NaCl
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Answers

Answer:

From above calculations, gives the limited amount of NaCl. So, 15.7 g of NaCl would form from given amounts of reactants.

Explanation:

thank me later

What is the limiting reactant when 19.9 g CuO react with 2.02 g H2?

Answers

Answer:

Explanation:

use the equation

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=19.9/79.5

=0.250moles of CuO

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Answers

Answer:

C₂H₆

Explanation:

The formula for ethane is C₂H₆

The second one is the answer

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Answers

D - they are isotopes of the same atom

Question 27
How many molecules are in 28 g of CO2?
O 3.5 x 10-21
O 7.42 x 1026
3.83 x 1023
O 3.8 x 1044

Answers

Answer:

3.83x10^23

Explanation:

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How many grams of Potassium are in 10 moles of potassium?

Answers

Answer:

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Explanation:

Help pleaseeee

you wish to make a potassium carbonate solution with a molar concentration of 0.25mol/dm³,but you only have 11.7g of potassium carbonate. what is the maximum volume of solution you can make?​

Answers

Answer:

0.337 dm³

Explanation:

M(K2CO3) = 138.2 g/mol

1) 11.7 g * (1 mol/138.2 g) =0.08466 mol

2) 0.08466 mol * (1 dm³/0.25 mol) = 0.3386 dm³ ≈ 0.337 dm³

What is a gas’s temperature in Celsius when it has a volume of 25 L, 203 mol, 143.5 atm?

Answers

To Find :

Gas’s temperature in Celsius when it has a volume of 25 L, 203 mol, 143.5 atm .

Solution :

We know, by ideal gas equation :

[tex]PV =nRT\\\\T = \dfrac{PV}{nR}[/tex]

Here, R is universal gas constant and [tex]R = 0.082 \ L \ atm \ mol^{-1} \ K^{1}[/tex]

Putting all given values and R in above equation, we get :

[tex]T = \dfrac{PV}{nR}\\\\T = \dfrac{143.5\times 25}{203 \times 0.082}\ K\\\\T = 215.52\ K[/tex]

Hence, this is the required solution.

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