How much CH4 remains after 10.0 g of CH4 reacts completely with 15.0 g of O2 according to the following equation? CH4(g) + 2 O2(g) → CO2(g) + 2 H2O(g) (A) 0.00 g (B) 2.50 g (C) 5.00 g (D) 6.25 g 8 Part 2 (0.5 point) Feedback O See Hint Write a balanced nuclear equation to represent the image above that depicts nuclear fission. Assume the nuclelshown represent 2330 splitting into ? Kr and 141Ba. 8 141 92

Answers

Answer 1

(A) 0.00 g of CH4 remains after the reaction.

To determine how much CH4 remains after the reaction, we need to use the balanced chemical equation and the concept of stoichiometry.

The balanced chemical equation for the reaction is:

CH4(g) + 2 O2(g) → CO2(g) + 2 H2O(g)

From the balanced equation, we know that 1 molecule of CH4 reacts with 2 molecules of O2.

We are given that 10.0 g of CH4 and 15.0 g of O2 react. We can use this information to determine the number of moles of CH4 and O2 that react.

1 mole of CH4 weighs 16.04 g and 1 mole of O2 weighs 32.00 g.

10.0 g of CH4 is equal to:

10.0 g / 16.04 g/mol = 0.623 mol CH4

15.0 g of O2 is equal to:

15.0 g / 32.00 g/mol = 0.46875 mol O2

Since the reaction is 2:1 ratio of O2: CH4, so 0.46875 mol O2 is enough for 0.234375 mol of CH4

So, no CH4 remains after the reaction.

Therefore, the answer is (A) 0.00 g of CH4 remains after the reaction.

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Related Questions

does any solid cu(oh)2 form when 0.075 g of koh is dissolved in 1.0 l of 1.0 × 10−3 m cu(no3)2?

Answers

Yes, the precipitation of Cu(OH)₂ will form when 0.075 g of KOH is dissolved in 1.0 L of 1.0 X 10⁻³ M Cu(NO₃)₂.

Chemical reactions begin with reactants as well as products of a reaction are the substances that were produced. The common chemical formula can be used to represent a chemical reaction: Reaction products. During the chemical reactions, bonds shatter and reform. Also, the opposite outcome of the reaction will happen.

[OH⁻]₀ = 0.075g/56.105g/mol(1.01) = 1.34×10⁻³

Cu(OH)₂ (s) ⇄ Cu²⁺ (aq) + 2OH⁻(aq)

? 0.0010 1.34×10⁻³

Q = [Cu²⁺][OH⁻]² = 0.0010 ×(1.34×10⁻³)² = 1.8×10⁻⁹ > Ksp 2.2×10⁻²⁰

This means that equilibrium will shift to the left, i.e., to reactants in which in turn implies that solid Cu(OH)₂ will be formed.

moles of Cu²⁺= 1.L x 1 x 10⁻³ M= 10⁻³

moles of KOH = 0.075 g/ 56.107 g/mol=0.00133

[Cu²⁺]= 10⁻³ / 1. L= 10⁻³ M

[OH⁻]=0.00133 / 1.L= 0.00133 M

Qsp= 10⁻³ (0.00133)² =1.786 x 10⁻⁹ >> Ksp ( = 2.2 x 10⁻²⁰)

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write a nuclear equation for the fusion of h−3 with h−1 to form he−4.

Answers

rgashkjaehahnfpv kagjmpahjmsnklsmnn

This reaction releases a large amount of energy, which is what makes it useful in nuclear fusion reactors.

What is nuclear fusion?

Nuclear fusion is a process in which two or more atomic nuclei join together to form a single, heavier nucleus.

The nuclear equation for the fusion of h-3 with h-1 to form he-4 is:
3H-1 + 1H-3 -> 4He-2 + 1n0
This equation describes the fusion of two hydrogen atoms to form a helium atom and a neutron. The first hydrogen atom (H-1) has a mass of 1, while the second hydrogen atom (H-3) has a mass of 3. When the two hydrogen atoms fuse together, they form a helium atom (He-2) with a mass of 4 and a neutron (n0) with a mass of 1. This reaction releases a large amount of energy, which is what makes it useful in nuclear fusion reactors.

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0.25 moles of a gas at 760 mmHg
and 298 K are contained in a 6.1 L
bottle. What is the pressure of the
system if the amount of gas in the
bottle is reduced to 0.13 mole and
the temperature is reduced to 100 K?

Answers

Answer: 2.3*10^3 Pa.

Explanation: We can use the ideal gas law to determine the pressure of the system: PV = nRT

where P is pressure, V is volume, n is the number of moles, R is the ideal gas constant, and T is the temperature in Kelvin.

The ideal gas constant R = 8.314 J/mol*K

Given:

n1 = 0.25 moles

V1 = 6.1 L

T1 = 298 K

P1 = 760 mmHg

n2 = 0.13 moles

V2 = 6.1 L

T2 = 100 K

We can use the equation PV = nRT to find the new pressure P2, using the information provided:

P2 = (n2RT2)/V2

We know that the volume of the bottle remains the same and we can convert the pressure from mmHg to Pascals

P2 = (0.138.314100)/6.1 = 2.3*10^3 Pa

Therefore, the pressure of the system when the amount of gas in the bottle is reduced to 0.13 mole and the temperature is reduced to 100 K is 2.3*10^3 Pa.

Answer:

132.62

Explanation:

i got the answer correct in the program

what is the name and formula of the chemical reagent used to separate the cations of group 1 from ions in other groups.

Answers

When water K2CrO4 is added, a yellow precipitate of PbCrO4 forms, confirming the presence of Pb2+ in the aqueous solution.

What does grouping cations into compounds mean?

1. Grouping the cations into various categories. The anions supplied by the group reagents are used to precipitate the cations of each succeeding group as compounds. The precipitate from one group's cations is separated (usually by centrifugation followed by decantation).

In Group 1 Group 2 and transition metals, how are cations formed?

More electropositive elements, such as those from groups 1, 2, 13, d-block (transition metals), and actinides, typically form cations. The oxidation states of D-block elements are positive, and they can form cations.

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a chemist has 70 ml of 50% methane solution. how much of an 80% siolution must she add yo get a 60% solution?

Answers

a chemist has 70ml of 50% methane solution.8.75 ml of 80% methane solution to the 70 ml of 50% methane solution to get a 60% solution.

To solve this problem, you can use the formula for mixing two solutions of different concentrations:

C1V1 + C2V2 = C3V3

Where:

C1 = concentration of the first solution

V1 = volume of the first solution

C2 = concentration of the second solution

V2 = volume of the second solution

C3 = final concentration of the mixture

V3 = final volume of the mixture

We know that:

C1 = 50% = 0.50 (as a decimal)

V1 = 70 ml

C3 = 60% = 0.60 (as a decimal)

So we can rearrange the formula to solve for V2:

V2 = (C3V3 - C1V1) / C2

Since we don't know the final volume V3, we can assume it to be V1+V2

V2 = (0.60 * (V1 + V2) - 0.50 * V1) / 0.80

Solving this equation we have,

V2 = (0.60 * (70 + V2) - 0.50 * 70) / 0.80

V2 = (42 - 35) / 0.80

V2 = 7/0.8

V2 = 8.75 ml

So the chemist needs to add 8.75 ml of 80% methane solution to the 70 ml of 50% methane solution to get a 60% solution.

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