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same number of each kind of atom appears in the reactants and in the products

The number of oxygen atoms in 19.3 g of sodium sulfate (Na₂SO₄) is 3.27×10²³ atoms

We'll begin by calculating the number of mole in 19.3 g of sodium sulfate (Na₂SO₄).

Mass of Na₂SO₄ = 19.3 g

Molar mass of Na₂SO₄ = (23×2) + 32 +(16×4)

= 46 + 32 + 64

= 142 g/mol

Mole of Na₂SO₄ =?

Mole = mass / molar mass

Mole of Na₂SO₄ = 19.3 / 142

Mole of Na₂SO₄ = 0.136 mole

Recall:

1 mole of Na₂SO₄ contains 4 moles of O.

Therefore,

0.136 mole of Na₂SO₄ will contain = 0.136 × 4 = 0.544 mole of O

Finally, we shall determine the number of atoms in 0.544 mole of O.

From Avogadro's hypothesis,

1 mole of O = 6.02×10²³ atoms

Therefore,

0.544 mole of O = 0.544 × 6.02×10²³

0.544 mole of O = 3.27×10²³ atoms

Thus, 19.3 g of sodium sulfate (Na₂SO₄) contains 3.27×10²³ atoms of oxygen.

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Answer:

3.27·10²³ atoms of O

Explanation:

To figure out the amount of oxygen atoms in this sample, we must first evaluate the sample.

The chemical formula for sodium sulfate is Na₂SO₄, and its molar mass is approximately 142.05.

We will use stoichiometry to convert from our mass of Na₂SO₄ to moles of Na₂SO₄, and then from moles of Na₂SO₄ to moles of O using the mole ratio; then finally, we will convert from moles of O to atoms of O using Avogadro's constant.

19.3g Na₂SO₄ · · ·

After doing the math for this dimensional analysis, you should get a quantity of approximately 3.27·10²³ atoms of O.

Answer:

Explanation:

Hello.

In this case, when ribose (C₅H₁₀O₅) yields carbon dioxide (CO₂) we write:

Which needs to be balanced by adding water and hydrogen ions:

You can also see that there are 20 transferred electrons, since the carbon atoms in the ribose have 0 as their oxidation state and the carbon atoms in the carbon dioxide have +4 as the oxidation state, thus, each carbon transfers  4 electrons, a five carbon atoms transfer 20 electrons overall.

In such a way, since the carbon is increasing its oxidation state, such half reaction is an oxidation half reaction.

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Answer:

The specific heat of the copper is 0.771 cal/ grams °C

Explanation:

Step 1: Data given

Mass of the piece of copper = 15.0 grams

The temperature of the wire changes from 12.0 °C to 79.0 °C

The amount of heat absorbed is 775 cal

Step 2: Calculate the specific heat of copper

Q = m*c*ΔT

⇒with Q = the heat absorbed = 775 cal

⇒with m = the mass of the copper = 15.0 grams

⇒with c = the specific heat of copper = TO BE DETERMINED

⇒with ΔT = The change in temperature = 79.0 °C - 12.0 °C = 67.0 °C

775 cal = 15.0 grams * c * 67.0 °C

c = 0.771 cal/gm °C

The specific heat of the copper is 0.771 cal/ grams °C

The reform reaction between steam and gaseous methane (CH4) produces "synthesis gas," a mixture of carbon monoxide gas and dihydrogen gas. Synthesis gas is one of the most widely used industrial chemicals, and is the major industrial source of hydrogen.

Suppose a chemical engineer studying a new catalyst for the reform reaction finds that 924 liters per second of methane are consumed when the reaction is run at 261°C and 0.96atm. Calculate the rate at which dihydrogen is being produced. Give your answer in kilograms per second. Round your answer to 2 significant digits.

Answer: The rate at which dihydrogen is being produced is 0.12 kg/sec

Explanation:

The balanced chemical equation is ;

According to ideal gas equation:

P = pressure of gas = 0.96 atm

V = Volume of gas = 924 L

n = number of moles

R = gas constant =

T =temperature =

According to stoichiometry:

1 mole of methane produces = 3 moles of hydrogen

Thus 20.2 moles of methane produces = moles of hydrogen

Mass of hydrogen =

Thus the rate at which dihydrogen is being produced is 0.12 kg/sec

Half-life of a radioactive element

The Half-Life of a radioactive element osbtime taken for half the nucleus of the atom of the element to decay

Calculating the original amount of U-238

• The number of moles present in each element is first determined:
• Number of moles = mass/molar mass

For Pb-206:

mass = 0.37 mg = 0.37 * 10⁻³ g

molar mass = 206 g/mol

number of moles = 0.37 * 10⁻³ g/206 g/mol

number of moles of Pb-206 = 1.79 * 10⁻⁶ moles

For U-238:

mass = 0.95 mg = 0.95 * 10⁻³ g

molar mass = 238 g/mol

number of moles = 0.95 * 10⁻³ g/238 g/mol

number of moles = 3.99 * 10⁻⁶ moles

Assuming that all the Pb-206 were formed from U-238

• Initial moles of U-238 = Moles of present U-238 + molesw of present Pb-208

Initial moles of U-238 = 3.99 * 10⁻⁶ moles + 1.79 * 10⁻⁶ moles

Initial moles of U-238 = 5.78 * 10⁻⁶ moles

One mole of U-238 contains = 6.02 * 10²³ atoms

5.78 * 10⁻⁶ moles of U-238 will contain 6.02 * 10²³ * 5.78 * 10⁻⁶ atoms

Number of atoms of U-238 initially present = 3.48 * 10¹⁸ atoms

Therefore, the number of atoms of U-238 initially present is 3.48 * 10¹⁸ atoms

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