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3 years ago

What is the name of this Acid? H2CO3*

Chemistry

2 answers:

zzz [600]3 years ago

7 0

ANSWER: carbonic acid.

Dimas [21]3 years ago

6 0

H2CO3 is carbonic acid

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What is the name of the rock? a. limestone b. breccia c. sandstone d. conglomerate

Virty [35]

A is the best answer
Limestone

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3 years ago

Read 2 more answers

Determine the empirical formula of a compound containing 48.38 grams of carbon, 6.74 grams of hydrogen, and 53.5 grams of oxygen

Darya [45]

Answer: The empirical formula of the compound becomes $CH_2O$

Explanation:

The empirical formula is the chemical formula of the simplest ratio of the number of atoms of each element present in a compound.

We are given:

Mass of C = 48.38 g

Mass of H = 6.74 g

Mass of O = 53.5 g

The number of moles is defined as the ratio of the mass of a substance to its molar mass. The equation used is:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ ......(1)

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

Step 1: Converting the given masses into moles.

Molar mass of C = 12 g/mol

Molar mass of H = 1 g/mol

Molar mass of O = 16 g/mol

Putting values in equation 1, we get:

$\text{Moles of C}=\frac{48.38g}{12g/mol}=3.023 mol$

$\text{Moles of H}=\frac{6.74g}{1g/mol}=6.74 mol$

$\text{Moles of O}=\frac{53.5g}{1g/mol}=3.34 mol$

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

Calculating the mole fraction of each element by dividing the calculated moles by the least calculated number of moles that is 3.023 moles

$\text{Mole fraction of C}=\frac{3.023}{3.023}=1$

$\text{Mole fraction of H}=\frac{6.74}{3.023}=2.23\approx 2$

$\text{Mole fraction of O}=\frac{3.34}{3.023}=1.105\approx 1$

Step 3: Taking the mole ratio as their subscripts.

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

Hence, the empirical formula of the compound becomes $CH_2O$

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3 years ago

How are oxidation-reduction reactions related to how you use energy?

Alexxandr [17]

Answer:

In oxidation reduction reactions, one species gets reduced by taking on electron(s) and another species gets oxidized by losing electrons. The movement of electrons can be used to do work. ... The electron flow can be run through a wire and these electrons can be used to do work (like run a battery). Hope this helps.

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2 years ago

An acetate buffer solution is prepared by combining 50. mL of 0.20 M acetic acid,

natima [27]

Answer:

The answer is "Option B".

Explanation:

$\to CH_3COOH + NaOH \longleftrightarrow CH_3COONa + H_2O\\\\\to CH_3COONa + NaOH\longleftrightarrow CH3COONa\\\\\therefore \ mol\ NaOH = (5 \ E-3\ L)\times(0.10 \ \frac{mol}{L}) = 5 \ E-4\ mol\\\\$

$\to mol\ CH_3COOH = (0.05 \ L)\times(0.20 \frac{mol}{L}) = 0.01 \ mol\\\\\to C \ CH_3COOH = \frac{(0.01 \ mol - 5 \ E-4\ mol) }{(0.105 \ L)}\\\\\to C \ CH_3COOH = 0.0905 \ M\\\\\therefore \ mol \ CH_3COONa = (0.05\ L )\times (0.20 \ \frac{mol}{L}) = 0.01 \ mol\\\\$

$\to C \ CH_3COONa = \frac{(0.01\ mol + 5 \ E-4\ mol)}{(0.105\ L )}\\\\\to C \ CH_3COONa = 0.1 \ M\\\\\therefore Ka = ([H_3O^{+}]\times \frac{(0.1 + [H_3O^+]))}{(0.0905 - [H_3O^+])} = 1.75\ E-5\\\\\to 0.1[H_3O^+] + [H_3O^+]^2 = (1.75 E-5)\times (0.0905 - [H_3O^+])\\\\$

$\to [H_3O^+]^2 \ 0.1[H_3O^+] = 1.584\ E-6 - 1.75\ E-5[H_3O^+]\\\\\to [H_3O^+]^2 + 0.1000175[H_3O^+] - 1.584 \ E-6 = 0\\\\\to [H_3O^+] = 1.5835\ E-5 \ M\\\\\therefore pH = - \log [H_3O^+]\\\\\to pH = - \log (1.5835 \ E-5)\\\\ \to pH = 4.8004 > 4.7$

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3 years ago

The force most responsible for holding the nucleus of an atom together is the:

DedPeter [7]

Strong nuclear force. I think that's the one one that holds the nucleons together.

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3 years ago