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lara31 [8.8K]
2 years ago
6

Ricky and Shemane, who are stoked to be learning stoich, were working diligently on a chemistry assignment in class. Ricky reali

zed that in most chemical reactions, there’s an excess of one reactant. How would Shemane correctly explain to Ricky why this occurs?
the reaction will continue until all of the limiting reactant is consumed.

reactions do not take place unless there are unequal amounts of reactants.


the reaction will continue until all of the excess reactant is consumed.


reactions occur too quickly if equal amounts of reactants are used.
Chemistry
1 answer:
kondaur [170]2 years ago
6 0
38373737267383838282828228)
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Based on the name, which substance is a covalent compound?
V125BC [204]
QUICK ANSWER

The name of the covalent compound N2O5 is dinitrogen pentoxide, more commonly known as nitrogen pentoxide. This covalent compound is part of a bigger group of compounds, nitrogen oxides, created purely from nitrogen and oxygen

6 0
2 years ago
Read 2 more answers
If 5.0 liters H2 (g) at STP is heated to a temperature of 985, pressure remaining constant, the new volume of the gas will be?
vlada-n [284]

Answer:

V_2=18 \ L \ H_2

General Formulas and Concepts:

<u>Chemistry - Gas Laws</u>

  • STP (Standard Conditions for Temperature and Pressure) = 22.4 L per mole at 1 atm, 273 K
  • Charles' Law: \frac{V_1}{T_1} =\frac{V_2}{T_2}

Explanation:

<u>Step 1: Define</u>

Initial Volume: 5.0 L H₂ gas

Initial Temp: 273 K

Final Temp: 985 K

Final Volume: ?

<u>Step 2: Solve for new volume</u>

  1. Substitute:                    \frac{5 \ L \ H_2}{273 \ K} =\frac{x \ L \ H_2}{985 \ K}
  2. Cross-multiply:             (5 \ L \ H_2)(985 \ K) = (x \ L \ H_2)(273 \ K)
  3. Multiply:                        4925 \ L \ H_2 \cdot K = 273x \ L \ H_2 \cdot K
  4. Isolate <em>x</em>:                       18.0403 \ L \ H_2 = x
  5. Rewrite:                         x=18.0403 \ L \ H_2

<u>Step 3: Check</u>

<em>We are given 2 sig figs as the smallest. Follow sig fig rules and round.</em>

<em />18.0403 \ L \ H_2 \approx 18 \ L \ H_2<em />

5 0
3 years ago
For the overall reaction below, which of the following is the correctly written rate law? Overall reaction: O3(g)+2NO2(g)→N2O5(g
zavuch27 [327]

Answer: Rate=k[O_3][NO_2]^2

Explanation:

Rate law says that rate of a reaction is directly proportional to the concentration of the reactants each raised to a stoichiometric coefficient determined experimentally called as order.

O_3(g)+NO_2(g)\rightarrow NO_3(g)+O_2(g)   slow

NO_3(g)+NO_2(g)\rightarrow N_2O_5(g)   fast

To determine the net chemical equation, we will simply add the above two equations, we get:

O_3(g)+2NO_2(g)\rightarrow N_2O_5(g)+O_2(g)

Rate=k[O_3][NO_2]^2

Order with respect to O_3 is 1 and Order with respect to NO_2 is 2.

Thus the rate law will be:  Rate=k[O_3][NO_2]^2

6 0
3 years ago
What can be said about a reaction with H = 62.4 kJ/mol and S = 0.145 kJ/(mol·K)?
Zigmanuir [339]
Answer:

At 430.34 K the reaction will be at equilibrium, at  T > 430.34 the reaction will be spontaneous, and at T < 430.4K the reaction will not occur spontaneously.

Explanation:

1) Variables:

G = Gibbs energy
H = enthalpy
S = entropy

2) Formula (definition)

G = H + TS

=> ΔG = ΔH - TΔS

3) conditions

ΔG < 0 => spontaneous reaction
ΔG = 0 => equilibrium
ΔG > 0 non espontaneous reaction

4) Assuming the data given correspond to ΔH and ΔS

ΔG = ΔH - T ΔS = 62.4 kJ/mol + T 0.145 kJ / mol * K

=>  T = [ΔH - ΔG] / ΔS

ΔG = 0 =>  T = [ 62.4 kJ/mol - 0 ] / 0.145 kJ/mol*K = 430.34K

This is, at 430.34 K the reaction will be at equilibrium, at  T > 430.34 the reaction will be spontaneous, and at T < 430.4K the reaction will not occur spontaneously.

3 0
3 years ago
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The energy diagram illustrated in the model corresponds to what student B
KatRina [158]

Answer: Exothermic reaction

Explanation:

Exothermic reactions lose energy, or ΔH, when they react.

4 0
3 years ago
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