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3 years ago

Which statement is true about HCI?

Chemistry

1 answer:

katen-ka-za [31]3 years ago

3 0

Answer:

d- it is an acid because it increases the concentration of hydronium ions

Explanation:

HCl; hydrochloric acid is an acid because it increases the concentration of hydronium ions when in solutions.

An acid is a substance that interacts with water to produce excess hydroxonium ions H₃O⁺ in an aqueous solution.

For example:

HCl + H₂O → H₃O⁺ + Cl⁻

HCl is classified as a strong acid due to its complete ionization when in solutions.

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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.

1. Locating the number of moles of B in the sample

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 grams B unit will cancel out, so we are therefore left with moles B remaining.

2. Locating the number of atoms in the sample

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.

Our numerator is:

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

Plugging this into a calculator, we get:

$1.5055 \times 10^{25}$

Our denominator is:

$(1 \times 10.81 \times 1)$

This simplifies to:

$10.81$

Dividing our numerator and denominator:

$\displaystyle \frac{1.5055 \times 10^{25}}{10.81}$

Plugging this into a calculator, we get:

$1.392691952 \times 10^{24}$

3. Simplifying with significant figures

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.

8 0

2 years ago

A piece of metal with a mass of 159 g is placed in a 50 mL graduated cylinder. The water level rises from 20. mL to 41 mL. What

almond37 [142]

Volume of water displaced is equal to the volume of the metal. So 41 - 20 = 21 ml. Density = mass / volume so density = 159 / 21 = 7.57 g/ml

8 0

3 years ago

What state of matter is at 35 Celsius

Otrada [13]

Different forms of matter have different melting/boiling points. For example, at 100 degrees Celsius, H2O (water) will turn from lliquid to gas. But NaOH (table salt) doesn't even go from solid to liquid until some 800 degrees Celsius. So, in order to figure out which state matter is at 35 Celsius, you'd have to be more specific about what kind of matter...

7 0

3 years ago

A 25.0-mL sample of 0.100M Ba(OH)2(aq) is titrated with 0.125 M HCl(aq).

morpeh [17]

Answer:

20.0

Explanation:

NaOH = (25.0) (0.100m) \ 0.125M = 20.0mL

8 0

3 years ago

A chemistry student adds a quantity of an unknown solid compound X to 5.00 L of distilled water at 15.° C . After 10 minutes of

Yuliya22 [10]

Answer:

34 g/100 mL

Explanation:

The solubility of a compound can be expressed in g/100mL, for this we must divide the mass of the compound that dissolves in the solute by the volume of the solvent.

The solvent, in this case, is water, and that mass of the solute X that dissolved is the mass that was recovered after the solvent was drained and evaporated. So the solubility of X (S) is:

S = 0.17 kg/5L

S = 170g/5000mL

S = 170g/(5*1000)mL

S = 34 g/100 mL

8 0

2 years ago