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3 years ago

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Q8. The titration of 15.00 mL of HBr solution of unknown concentration requires 18.44 mL of a 0.100 M KOH solution to reach the

equivalence point. What is the concentration of the unknown HBr solution in M? Tro, Nivaldo J.. Chemistry (p. 201). Pearson Education. Kindle Edition.

1 answer:

lara [203]3 years ago

8 0

Answer: 0.123 M

Explanation:

According to the neutralization law:

$n_1M_1V_1=n_2M_2V_2$

where,

$M_1$ = molarity of $HBr$ solution = ?

$V_1$ = volume of $HBr$ solution = 15.00 ml

$M_2$ = molarity of $KOH$ solution = 0.100 M

$V_2$ = volume of $KOH$ solution = 18.44 ml

$n_1$ = valency of $HBr$ = 1

$n_2$ = valency of $KOH$ = 1

$1\times M_1\times 15.00=1\times 0.100\times 18.44$

$M_1=0.123$

Therefore, the concentration of the unknown HBr solution is 0.123 M

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How many atoms are in 25.00 g of B?

klio [65]

Answer:

There are 1.393 x 10²⁴ atoms in 25.00 g of B.

Explanation:

Hey there!

We are given a value, in grams, that we need to convert to a number of atoms.

We can convert grams to atoms by using Avogadro's Number ( $N_A$ ). This number is equivalent to $6.022 \times 10^{23}$ .

This number can be used to convert any values to:

atoms
molecules
formula units
moles

In order to do this problem, we will need to use dimensional analysis (DA). This process allows us to convert from grams to atoms.

We need to set up our ratios in order to work this out. We can use a periodic table to help us through this next part of the problem.

<u>1. Locating the number of moles of B in the sample</u>

We first need to find the amount of moles of boron (B) there are in the sample.

Checking a periodic table, the atomic mass in atomic mass units (amu) is 10.81 amu.

Atomic mass units can easily be converted to grams and these units can be used interchangeably.

Therefore, for each atom of boron, it weighs 10.81 grams to us. This is equivalent to the mass of one mole of boron.

To find the number of moles, we have two possible ratios we can use:

$\displaystyle \frac{1 \ mole \ B}{10.81 \ grams \ B}$

$\displaystyle \frac{10.81 \ grams \ B}{1 \ mole \ B}$

These ratios mean the same thing, but we need to convert our final unit to moles.

We are given a sample in grams, and when dividing our units, we need to keep moles.

Since the first portion of our expression is in grams, we need to have grams in the bottom of our expression.

$\displaystyle 25.00 \ \text{grams B} \ \times \frac{1 \text{mole B}}{10.81 \ \text{grams B}}$

We can now simplify the expression. Our <u>grams B</u> unit will cancel out, so we are therefore left with <u>moles B</u> remaining.

<u>2. Locating the number of atoms in the sample</u>

Now with our equation, we can convert our number of moles that would be solved if we stopped with the above. However, we need to convert to atoms.

We use Avogadro's number and create a ratio with that of moles.

$\displaystyle \frac{6.022 \times 10^{23}\text{atoms}}{1 \text{mole B}}$

$\displaystyle \frac{1 \text{mole B}}{6.022 \times 10^{23} \text{atoms}}$

We need to cancel out our moles and end with atoms, so we must have moles in the denominator. Therefore, we use the first ratio.

Using our previous expression, we multiply by this new ratio and solve the expression.

$\displaystyle 25.00 \ \text{grams B} \ \times \frac{1 \text{mole B}}{10.81 \ \text{grams B}} \ \times \frac{6.022 \times 10^{23}\text{atoms}}{1 \text{mole B}}$

This expression can now be operated. You will need a calculator to perform this calculation.

<u>Our numerator is:</u>

$[(25.00 \times 1 \times (6.022 \times 10^{23})]$

Plugging this into a calculator, we get:

$1.5055 \times 10^{25}$

<u>Our denominator is:</u>

$(1 \times 10.81 \times 1)$

This simplifies to:

$10.81$

<u>Dividing our numerator and denominator:</u>

<u /> $\displaystyle \frac{1.5055 \times 10^{25}}{10.81}$ <u />

Plugging this into a calculator, we get:

$1.392691952 \times 10^{24}$

<u>3. Simplifying with significant figures</u>

Now, we need to take into account that we have significant figures. We are given this original value:

$25.00$

This value has four significant figures, which means we need to round our value we received above to four significant figures.

$\approx 1.393$

Our units are added as well as our scientific notation:

$1.393 \times 10^{24} \ \text{atoms of B}$

Therefore, our final answer is choice A.

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All minerals on the mohs hardness scale are hard enough to scratch a piece of glass is it true or false

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The thermal stability of group 2 hydroxides increases down the group ? give reasons

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ANSWERS:

Group 2 metal carbonates, nitrates and hydroxides decompose to heat to give the corresponding metal oxide and release CO2, NO2 and O2, and H2O respectively. The thermal stability increases down the group as theionic character of the compounds increases down the group.

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What is the resistance of a 150 W lightbulb connected to a 24 V voltage source?

Luda [366]

Answer:

3.84 Ω

Explanation:

From the question given above, the following data were obtained:

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V = IR

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I = V/R

P = V/R × V

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Where:

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I => Current

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With the above formula (i.e P = V²/R), we can calculate resistance as illustrated below:

Electrical power (P) = 150 W

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Resistance (R) =?

P = V²/R

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Cross multiply

150 × R = 576

Divide both side by 150

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What term is best described as the rate of a chemical reaction at any instant in time?

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