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agasfer [191]
2 years ago
7

Which sample contains the largest number of oxygen atoms? Select one: a. 8.0 g of carbon dioxide b. 8.0 g of potassium chlorate

c. 8.0 g of calcium perchlorate d. 8.0 g of sodium hydroxide
Chemistry
1 answer:
adell [148]2 years ago
5 0

The sample with the largest number of oxygen atoms will be calcium perchlorate.

<h3>Number of atoms in a compound</h3>

Since we are not looking at the number of moles, the mass of the compounds has no bearing on the number of atoms of oxygen.

  • The chemical formula for carbon dioxide is CO_2. Thus, it has 2 atoms of oxygen.
  • The chemical formula for potassium chlorate is KClO_3. Thus, it has 3 oxygen atoms.
  • The chemical formula for calcium perchlorate is Ca(ClO_4)_2. Thus, it has 8 atoms of oxygen.
  • The chemical formula for sodium hydroxide is NaOH. Thus, it has 1 atom of oxygen.

Therefore, the compound with the largest number of oxygen atoms is calcium perchlorate.

More on the number of atoms in compounds can be found here: brainly.com/question/1686912

#SPJ1

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The concentration of CI ion in a sample of H,0 is 15.0 ppm. What mass of CI ion is present in 240.0 mL of H,0, which has a densi
Doss [256]

Answer:

Mass of solute = 0.0036 g

Explanation:

Given data:

Concentration of Cl⁻ = 15.0 ppm

Volume of water = 240 mL

Mass of Cl⁻ present = ?

Solution:

1 mL = 1 g

240 mL = 240 g

Formula:

ppm = mass of solute / mass of sample ×1,000,000

by putting values,

15.0 ppm = (mass of solute / 240 g) ×1,000,000

Mass of solute = 15.0 ppm ×  240 g / 1,000,000

Mass of solute = 0.0036 g

8 0
3 years ago
Calculate the grams of sulfur dioxide, SO2, produced when a mixture of 35.0 g of carbon disulfide and 30.0 g of oxygen reacts. W
alekssr [168]

Answer:

58.9g of SO2 is produced

8g of oxygen remains unconsumed

Explanation:

The burning of Carbon disulfide (CS2) in oxygen. gives the reaction:

CS2 (g) + 3O2 (g) → CO2 (g) + 2SO2 (g)

Molar mass of CS2 = 76.139 g/mol

Molar mass of O2 = 15.99 g/mol

Molar mass of SO2 = 64.066 g/mol

Number of moles of CS2 = 35g/ 76.139 g/mol =0.46 moles

Number of moles of O2 = 30g/15.999 g/mol =1.88 moles

From the chemical reaction

1 mole of CS2 react with 3 moles of O2 to give 2 moles of SO2

Thus 0.46 moles of CS2 reacts to form 2× 0.46 = 0.92 moles of SO2

Mass of SO2 produced = 0.92×64.07 = 58.9g of SO2 is produced

thus 0.46 moles of CS2 reacts with 3 × 0.46 moles of O2 which is =1.38 moles of O2

Thus oxygen is the limiting reactant with 1.88 - 1.38 = 0.496~~0.5 mole remaining

Or 8g of oxygen

58.9g of SO2 is produced

oxygen is the limiting

4 0
3 years ago
At a certain temperature this reaction follows first-order kinetics with a rate constant of 2.01 s^−1
ddd [48]

Answer:

0.80 seconds (2 significant figures)

Explanation:

The equation of the reaction is given as;

CICH2CH2Cl (g) --> CH2CHCI (g) + HCl(g)

Rate constant (k) = 2.01 s^-1

From the units of the rate constant, this is a first order reaction.

Initial Concentration = 1.34 M

t = ?

Final concentration = 20% of 1.34 = 0.268 M

The integrated rate law for a first order reaction is given as;

ln[A] = ln[A]o - kt

ln(0.268) = ln(1.34) - 2.01(t)

-2.01(t) = - 1.6094

t = 0.8007 ≈ 0.80 seconds (2 significant figures)

6 0
3 years ago
For the following reaction, 22.0 grams of nitrogen monoxide are allowed to react with 5.80 grams of hydrogen gas . nitrogen mono
mash [69]

NO is the limiting reagent and 4.34 g is the amount of the excess reagent that remains after the reaction is complete

<h3>What is a limiting reagent?</h3>

The reactant that is entirely used up in a reaction is called as limiting reagent.

The reaction:

2NO(g) +2H_2(g) → N_2 +2H_2O

Moles of nitrogen monoxide

Molecular weight: M_(_N_O_)=30g/mol

n_(_N_O_) =\frac{mass}{molar \;mass}

n_(_N_O_) =\frac{22.0}{30g/mol}

n_(_N_O_) = 0.73 mol

Moles of hydrogen

Molecular weight: M_(_H_2_)=30g/mol

n_(_H_2_) =\frac{mass}{molar \;mass}

n_(_H_2_) =\frac{5.80g}{2g/mol}

n_(_H_2_) = 2.9 mol

Hydrogen gas is in excess.

NO is the limiting reagent.

The amount of the excess reagent remains after the reaction is complete.

n_(_N_2_) = (2.9 mol- 0.73 mol NO x \frac{1 \;mol \;of \;H_2}{2 \;mole \;of \;NO}) x \frac{2g \;of \;H_2}{mole \;of \;H_2}

n_(_N_2_) =4.34 g

Learn more about limiting reagents here:

brainly.com/question/26905271

#SPJ1

5 0
1 year ago
PLEASE HELP SOMEONE PLEASEEEE
crimeas [40]

Answer:

the answer is b

hope this helps you!!

Have a lovely day!

-millie :D

4 0
3 years ago
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