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gladu [14]
3 years ago
11

Lana balanced an equation so that the result was 2C2H3Br + 5O2 → 4CO2 + 2H2O + 2HBr. Which most likely represents the starting e

quation?
2C4H3Br + 5O2 → 4CO2 + 2H2O + 2HBr
C2H3Br + 5O2 → 4CO2 + H2O + 2HBr
C4H3Br + O2 → CO2 + H2O + HBr
C2H3Br + O2 → CO2 + H2O + HBr
Chemistry
2 answers:
Elina [12.6K]3 years ago
5 0

Answer:

C2H3Br + O2 → CO2 + H2O + HBr

Explanation:

So it should be D :)

ratelena [41]3 years ago
4 0

Explanation:

C2H3Br + O2 → CO2 + H2O + HBr

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At what temperature, would the volume of a gas
PtichkaEL [24]

Explanation:

P1V1 = nRT1

P2V2 = nRT2

Divide one by the other:

P1V1/P2V2 = nRT1/nRT2

From which:

P1V1/P2V2 = T1/T2

(Or P1V1 = P2V2 under isothermal conditions)

Inverting and isolating T2 (final temp)

(P2V2/P1V1)T1 = T2 (Temp in K).

Now P1/P2 = 1

V1/V2 = 1/2

T1 = 273 K, the initial temp.

Therefore, inserting these values into above:

2 x 273 K = T2 = 546 K, or 273 C.

Thus, increasing the temperature to 273 C from 0C doubles its volume, assuming ideal gas behaviour. This result could have been inferred from the fact that the the volume vs temperature line above the boiling temperature of the gas would theoretically have passed through the origin (0 K) which means that a doubling of temperature at any temperature above the bp of the gas, doubles the volume.

From the ideal gas equation:

V = nRT/P or at constant pressure:

V = kT where the constant k = nR/P. Therefore, theoretically, at 0 K the volume is zero. Of course, in practice that would not happen since a very small percentage of the volume would be taken up by the solidified gas.

8 0
3 years ago
To draw the Lewis structure of the polyatomic ion,CIO3 you would have to _____ those in the structures of Cl, O, O, and O. Selec
NikAS [45]
D. Add one electron to, 
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4 years ago
How much heat is lost when 575 grams molten iron at 1825 k becomes solid iron at 293 k? The melting point or iron is 1811 k.​
Galina-37 [17]

Answer:

146 kJ  

Explanation:

There are two heat flows in this question.  

Heat lost on cooling + heat lost on solidifying = 0  

                 q₁              +                 q₂                   = 0  

              mCΔT          +             nΔHsol              = 0  

Data:  

       m = 575 g  

       C = 0.449 J·K⁻¹g⁻¹  

    T_i = 1825 K  

    T_f = 1811 K  

ΔHsol = -13.8 kJ·mol⁻¹  

Calculations:  

(a) Heat lost on cooling  

ΔT = T_f - T_i = 1811 K - 1825 K = -14 K  

q₁ = mCΔT = 575 g × 0.449 J·K⁻¹g⁻¹ × (-14 K) = -361 J = -3.61 kJ  

(b) Heat lost on solidifying  

n = \text{575 g} \times \dfrac{\text{1 mol}}{\text{55.84 g}} = \text{10.30 mol}\\\\q_{2} = n\Delta_{\text{sol}}H = \text{10.30 mol} \times \dfrac{\text{-13.8 kJ}}{\text{1 mol}}= \text{-142.1 kJ}

(c) Total heat lost  

q = q₁ + q₂ = -3.61 kJ - 142.1 kJ = -146 kJ  

The heat lost was 146 kJ.

 

5 0
3 years ago
What is the reaction quotient?
ikadub [295]

Answer:

A. It measures the [products] / [reactants].

Explanation:

  • The reaction quotient aids in figuring out which direction a reaction is likely to proceed, given either the pressures or the concentrations of the reactants and the products.
  • It is calculated from the concentration of the products and reactants at mixing up.
  • Reaction quotient (Q) =  [products] / [reactants]

So, the right choice answer:

A. It measures the [products] / [reactants].

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Answer:

this really works https://chem.libretexts.org/Courses/Saint_Marys_College_Notre_Dame_IN/CHEM_122-02_(Under_Construction)/1%3A_Review_from_CHEM_121/1.05%3A_Units%2C_Measurement_Uncertainty%2C_and_Significant_Figures_(Worksheet)

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