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3 years ago

Substance c releases H+ ions into the solution it is a _______

Chemistry

2 answers:

Leya [2.2K]3 years ago

4 0

An acid is a substance or compound that releases hydrogen ions (H+) when in solution

harkovskaia [24]3 years ago

3 0

Substance C is a acid.......

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How many moles are in 1.84x10^24 formula units of CaCl2, calcium chloride?

Lady_Fox [76]

1 mole = 6.022 * 10^23 formula units.

1.84*10^24 formula units CaCl₂ * (1 mole CaCl₂/6.022*10^23 formula units CaCl₂) = 3.06 moles of CaCl₂.

There are 3.06 moles of CaCl₂.

7 0

3 years ago

Indicate the oxidation state for each of the following:

Nimfa-mama [501]

P2H2 Oxidation State - Diphosphene

P2H2 has an oxidation number of +1. In P2H2, the oxidation

number of H is -1

H2C2O4 Oxidation State - Oxalic Acid

C has an oxidation number of +3 in H2C2O4. O has an oxidation

number of -2 in H2C2O4. H2C2O4 has an oxidation number of +1.

CrO2 Oxidation State - Chromium(IV) Oxide

CrO2 has an oxidation number of +4.

<h3>Explain oxidation?</h3>

The oxidation state, also known as the oxidation number, is an atom's hypothetical charge if all of its bonds to other atoms were fully ionic.

The sum of the oxidation states of all the atoms in an ion equals the ion's charge.

A substance's more electronegative element is given a negative oxidation state. The one that is less electronegative is given a positive oxidation state.

To learn more about oxidation refer to :

brainly.com/question/27239694

#SPJ13

6 0

1 year ago

An ice-skating rink has tubes under its floor to freeze the water. salt water is cooled well below the freezing point of water a

alex41 [277]

Correct Answer: Option g: adding salt to water lowers its freezing point

Reason:
Freezing point is a colligative property. When a non-volatile solution is present in solution, it's freezing point decreases. This is referred as depression in freezing point (ΔTf). Extent of lowering in freezing point is dependent on number of particles present in system. Mathematically it is expressed as:

ΔTf = Kf X m

where, m = molality of solution
Kf = cryoscopic constant.

Hence, adding salt to water lowers the freezing point of solution.

6 0

3 years ago

So I am lost and I don't know how what any of those conversion factor things are and I'm wondering how to do #2,3, and 4. (I did

dalvyx [7]

Explanation:

#2.

A centigram is 1/100 of a gram, so that means a gram equals 100 centigrams.

Therefore you multiply 72.4 grams by 100/1 (or just 100), and get 7240 cg.

You did that one right but put the wrong unit in the answer. It is is cg ( centigrams).

#3.

1 liter is equal to 1000 milliliters, and I kiloliter is equal to 1000 liters. So one kiloliter is 1000*1000 milliliters or 1,000,000 milliliters.

The conversion factor would be

1/1000000

#4.

1 gigabyte is equal to 10^9 bytes.

I byte is equal to 10^9 bytes.

So 1 gigabyte is 10^9 * 10^9 nanobytes, or 10^18.

The conversion factor would be (1*10^18)/1.

6 0

4 years ago

compound consists of carbon, hydrogen and fluorine. In one experiment, combustion of 2.50 g of the compound produced 3.926 g of

arlik [135]

Answer: The empirical and molecular formula of the compound is $CH_2F$ and $C_{14}H_{28}F_{14}$ respectively

Explanation:

We are given:

Mass of $CO_2=3.926g$

For calculating the mass of carbon:

In 44 g of carbon dioxide, 12 g of carbon is contained.

So, in 3.926 g of carbon dioxide, $\frac{12}{44}\times 3.926=1.071g$ of carbon will be contained.

To calculate the percentage composition of element in sample, we use the equation:

$\%\text{ composition of element}=\frac{\text{Mass of element}}{\text{Mass of sample}}\times 100$ ......(1)

For Carbon:

Mass of carbon = 1.071 g

Mass of sample = 2.50 g

Putting values in equation 1, we get:

$\%\text{ composition of carbon}=\frac{1.071g}{2.50g}\times 100=42.84\%$

For Fluorine:

Mass of fluorine = 2.54 g

Mass of sample = 5.00 g

Putting values in equation 1, we get:

$\%\text{ composition of fluorine}=\frac{2.54g}{5.00g}\times 100=50.8\%$

Percent composition of hydrogen = [100 - 42.84 - 50.8] % = 6.36 %

We are given:

Percentage of C = 42.84 %

Percentage of F = 50.8 %

Percentage of H = 6.36 %

Let the mass of compound be 100 g. So, percentages given are taken as mass.

Mass of C = 42.84 g

Mass of F = 50.8 g

Mass of H = 6.36 g

To formulate the empirical formula, we need to follow some steps:

Step 1: Converting the given masses into moles.

Moles of Carbon = $\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{42.84g}{12g/mole}=3.57moles$

Moles of Hydrogen = $\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{6.36g}{1g/mole}=6.36moles$

Moles of Fluorine = $\frac{\text{Given mass of Fluorine}}{\text{Molar mass of Fluorine}}=\frac{50.8g}{19g/mole}=2.67moles$

Step 2: Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 2.67 moles.

For Carbon = $\frac{0.072}{2.67}=1.34\approx 1$

For Hydrogen = $\frac{6.36}{2.67}=2.38\approx 2$

For Fluorine = $\frac{2.67}{2.67}=1$

Step 3: Taking the mole ratio as their subscripts.

The ratio of C : H : F = 1 : 2 : 1

The empirical formula for the given compound is $CH_2F$

For determining the molecular formula, we need to determine the valency which is multiplied by each element to get the molecular formula.

The equation used to calculate the valency is:

$n=\frac{\text{Molecular mass}}{\text{Empirical mass}}$

We are given:

Mass of molecular formula = 448.4 g/mol

Mass of empirical formula = $12+(2\times 1)+19]=33g/mol$

Putting values in above equation, we get:

$n=\frac{448.4g/mol}{33g/mol}=13.6\approx 14$

Multiplying this valency by the subscript of every element of empirical formula, we get:

$C_{(14\times 1)}H_{(14\times 2)}F_{(14\times 1)}=C_{14}H_{28}F_{14}$

Hence, the empirical and molecular formula of the compound is $CH_2F$ and $C_{14}H_{28}F_{14}$ respectively

3 0

3 years ago