All categories

3 years ago

Please help asap! This is the last question for my chemistry test!

Chemistry

1 answer:

stepan [7]3 years ago

4 0

Answer:

i took the test but failed and just got that question right so b

Explanation:

You might be interested in

All substances taking part in a certain interaction are shown below: Butane is written followed by an addition sign followed by

OverLord2011 [107]

The Law of Conservation of Mass simply states that the total amount of mass should not change in a chemical reaction that is isolated (no other objects can enter the reaction). The total mass of the reactants must be equal to the total mass of the products. Thus, the correct estimate of the amount of oxygen used in the interaction is the difference between 133 g and 29 g.

3 0

3 years ago

Read 2 more answers

Convert 99 moles of CO2 to atoms

lorasvet [3.4K]

Answer:

1.78 × 10²⁶ Atoms

Explanation:

Relation between number of molecules and moles is,

No. of Molecules = Moles × 6.022 × 10²³ Molecules/mol

No. of Molecules = 99 mol × 6.022 × 10²³ Molecules/mol

No. of Molecules = 5.96 × 10²⁵ Molecules

Also, In CO₂ Molecule there are 3 atoms.

So,

No. of atoms = 5.96 × 10²⁵ Molecules × 3

No. of atoms = 1.78 × 10²⁶ Atoms

3 0

3 years ago

Three mixtures were prepared from three very narrow molar mass distribution polystyrene samples with molar masses of 10,000, 30,

8_murik_8 [283]

Answer:

(a). 46,666.7 g/mol; 78,571.4 g/mol

(b). 86950g/mol; 46,666.7 g/mol.

(c). 86950g/mol; 43,333.33 g/mol

Explanation:

So, we are given the molar masses for the three samples as: 10,000, 30,000 and 100,000 g mol−1.

Thus, the equal number of molecule in each sample = ( 10,000 + 30,000 + 100,000 ) / 3 = 46,666.7 g/mol.

The average molar mass = [ ( 10,000)^2 + (30,000)^2 + 100,000)^2] ÷ 10,000 + 30,000 + 100,000 = 78,571. 4 g/mol.

(b). The equal masses of each sample = 3/[ ( 1/ 10,000) + (1/30,000 ) + (1/100,000) ] = 20930.23 g/mol.

Average molar mass = ( 10,000 + 30,000 + 100,000 ) / 3 = 46,666.7 g/mol.

(c). Equal masses of the two samples = (0.145 × 10,000) + (0.855 × 100,000)/ 0.145 + 0.855 = 86950g/mol.

The weight average molar mass = 1.7 + 10,000 + 100,000/ 1.7 + 1 = 43,333.33 g/mol.

6 0

3 years ago

I will mark brainlist if correct !! :) PLS HELPPP

Soloha48 [4]

<h2>Answer: Option B</h2>

Option A is incorrect. Heated water cannot freeze.

Option B is correct. Heated water will evaporate.

Option C is incorrect. Heated water cannot create condensation.

Option D is incorrect. Water can't melt when heated unless it's been frozen beforehand.

6 0

3 years ago

Butane (C4 H10(g), mc031-1.jpgHf = –125.6 kJ/mol) reacts with oxygen to produce carbon dioxide (CO2 , mc031-2.jpgHf = –393.5 kJ/

ankoles [38]

The balanced chemical equation for the combustion of butane is:

$2C_{4}H_{10}(g) +13 O_{2}(g)-->8CO_{2}(g)+10H_{2}O(g)$

Δ $H_{reaction}^{0}$ = Σ $n_{products}$ Δ $H_{f}^{0}_{(products)}$ -Σ $n_{reactants}$ Δ $H_{f}^{0}_{(reactants)}$

= $[{8*(-393.5kJ/mol)}+{10*(-241.82kJ/mol)}]-[{2*(-125.6kJ/mol)}+13*(0 kJ/mol)}]$ =[-3148kJ/mol+(-2418.2kJ/mol)]-[(-251.2kJ/mol)+0]

= -5315 kJ/mol

Calculating the enthalpy of combustion per mole of butane:

$1mol C_{4}H_{10}*( \frac{-5315kJ}{2mol C_{4}H_{10} })=-2657.5 \frac{kJ}{molC_{4}H_{10}}$

Therefore the heat of combustion per one mole butane is -2657.5 kJ/mol

Correct answer: -2657.5 kJ/mol

6 0

3 years ago