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vagabundo [1.1K]
3 years ago
9

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

Oxygen followed by an equal to sign followed by Carbon dioxide followed by an addition sign followed by Water. Just below Butane 29 grams is written and just below Carbon dioxide and Water is a parentheses below which 133 grams is written Which of the following is the correct estimate of the amount of oxygen used in the interaction? Sum of 133 g and 29 g Difference between 133 g and 29 g Twice the sum of 133 g and 29 g Twice the difference between133 g and 29 g
Chemistry
2 answers:
OverLord2011 [107]3 years ago
3 0
<span>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, t</span>he correct estimate of the amount of oxygen used in the interaction is the difference between 133 g and 29 g.
Yuri [45]3 years ago
3 0

Answer:

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.

Explanation:

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NikAS [45]

Answer:

After you have waived your rights, the confession must still be a voluntary confession. Whether a confession is voluntary or not is based on the “totality of the circumstances”. The Courts have held that a police officer simply lying does not result in an “involuntary confession”.

Explanation:

7 0
3 years ago
Natural gas is almost entirely methane. A container with a volume of 2.65L holds 0.120mol of methane. What will the volume be if
mrs_skeptik [129]

The final volume of the methane gas in the container is 6.67 L.

The given parameters;

  • <em>initial volume of gas in the container, V₁ = 2.65 L</em>
  • <em>initial number of moles of gas, n₁ = 0.12 mol</em>
  • <em>additional concentration, n = 0.182 mol</em>

The total number of moles of gas in the container is calculated as follows;

n_t = 0.12 + 0.182 = 0.302 \ mol

The final volume of gas in the container is calculated as follows;

PV = nRT\\\\\frac{V}{n} = \frac{RT}{P} \\\\\frac{V_1}{n_1} = \frac{V_2}{n_2} \\\\V_2 = \frac{V_1 n_2}{n_1} \\\\V_2 = \frac{2.65 \times 0.302}{0.12} \\\\V_2 = 6.67 \ L

Thus, the final volume of the methane gas in the container is 6.67 L.

Learn more here:brainly.com/question/21912477

5 0
3 years ago
Which element has properties most similar to Bromine? <br> A) F <br> B) Kr <br> C) S <br> D) Se
Stells [14]
Fluorine (F) will have properties similar to Bromine (Br) because it belongs to the same group as Bromine.
7 0
3 years ago
Read 2 more answers
The daily production of carbon dioxide from an 780.0 mw coal-fired power plant is estimated to be 3.3480 x 104 tons (not metric)
valkas [14]

The production of CO_{2} is 3.3480\times 10^{4} tons/day. Converting mass into kg,

1 ton=907.185 kg, thus,

3.3480\times 10^{4} tons=3.037\times 10^{7} kg

Thus, production of CO_{2} will be 3.037\times 10^{7} kg/ day.

The specific volume of CO_{2} is 0.0120 m^{3}/kg.

Volume of CO_{2} produced per day can be calculated as:

V=Specific volume\times mass

Putting the values,

V=0.0120 m^{3}/kg\times 3.037\times 10^{7} kg=364440 m^{3}/day

Thus, volume of CO_{2} produced per year will be:

V=\frac{365 days}{1 year}(364440 m^{3}/day)=1.33\times 10^{8}m^{3}/year

Thus, in 4 year volume of CO_{2} produced will be:

V=1.33\times 10^{8}m^{3}/year\times 4 years=5.32\times 10^{8}m^{3}

8 0
3 years ago
PLEEASE HELP TIMED WILL GIVE BRAINLIEST TO CORRECT ANSWER NLY PLEASEEEE HELPPPP
Arturiano [62]

Answer:

1) 1.15 mol

2) M=0.45

3) 22.5 mL

4) 6.25 mL

Explanation:

1)

550 mL= 0.55 L

M= mol solute/ L solution

mol solute= M * L solution

mol solute= (2.1 M * 0.55 L )        M=1.15 mol solute

2)

155 mL = 0.155 L

80 g  -> 1 mol NH4NO3

5.61 g -> x

x= (5.61 g * 1 mol NH4NO3)/80 g       x= 0.07 mol NH4NO3

M=(0.07 mol NH4NO3)/0.155 L         M=0.45

3) M1V1=M2V2

V1= M2V2/M1

V1= (0.500 M * 0.225 L)/5.00 M             V1=0.0225 L =22.5 mL

4) M1V1=M2V2

V1= M2V2/M1

V1= (0.25 M * 0.45 L)/ 18.0 M

V1=6.25 x 10^-3 L = 6.25 mL

8 0
3 years ago
Read 2 more answers
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