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Lostsunrise [7]
3 years ago
13

How do you answer this chem equation: 836.8J=14.5g*Cm*64

Chemistry
1 answer:
Alja [10]3 years ago
8 0
Is that a thing? Maybe you've said it wrong. Otherwise, I don't think you don't learn this in middle school.

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What is the empirical formula for propene C3H6
USPshnik [31]

The empirical formula is the simplest formula attainable while maintaining the ratio so it will be CH2.

Explanation:

The empirical formula of a chemical compound is the simplistic positive integer ratio of atoms being in a compound. A simple example of this thought is that the empirical formula of sulfur monoxide, or SO, would simply be SO, as is the empirical formula of disulfur dioxide, S2O2.

3 0
3 years ago
An astronaut has a mass of 60kg. Calculate their weight When they are on the surface of the moon.
IceJOKER [234]

Answer:

I think 0kg because they are flying.

7 0
3 years ago
A closed system initially containing 1×10^-3 hydrogen 2×10^-3M iodine at 448 degree Celsius and is allowed to reach equilibrium.
GaryK [48]

Answer:

Kc = 50.5

Explanation:

We determine the reaction:

H₂  +  I₂   ⇄   2HI

Initially we have 0.001 molesof H₂

and 0.002 moles of I₂

If we have produced 0.00187 moles of HI in the equilibrium we have to know, how many moles of I₂ and H₂, have reacted.

           H₂     +      I₂      ⇄   2HI

In:     0.001       0.002           -

R:       x                 x                2x

Eq:  0.001-x    0.002-x      0.00187  

x = 0.00187/2 = 9.35×10⁻⁴ moles that have reacted

So in the equilibrium we have:

0.001 - 9.35×10⁻⁴ = 6.5×10⁻⁵  moles of H₂

0.002 - 9.35×10⁻⁴ = 1.065×10⁻³ moles of I₂

Expression for Kc is =  (HI)² / (H₂) . (I₂)

0.00187 ² /  6.5×10⁻⁵ . 1.065×10⁻³ = 50.5

5 0
3 years ago
A certain reaction with an activation energy of 185 kJ/mol was run at 505 K and again at 525 K . What is the ratio of f at the h
frosja888 [35]

Answer:

The ratio of f at the higher temperature to f at the lower temperature is 5.356

Explanation:

Given;

activation energy, Ea = 185 kJ/mol = 185,000 J/mol

final temperature, T₂ = 525 K

initial temperature, T₁ = 505 k

Apply Arrhenius equation;

Log(\frac{f_2}{f_1} ) = \frac{E_a}{2.303 \times R} [\frac{1}{T_1} -\frac{1}{T_2} ]

Where;

\frac{f_2}{f_1}  is the ratio of f at the higher temperature to f at the lower temperature

R is gas constant = 8.314 J/mole.K

Log(\frac{f_2}{f_1} ) = \frac{E_a}{2.303 \times R} [\frac{1}{T_1} -\frac{1}{T_2} ]\\\\Log(\frac{f_2}{f_1} ) = \frac{185,000}{2.303 \times 8.314} [\frac{1}{505} -\frac{1}{525} ]\\\\Log(\frac{f_2}{f_1} ) = 0.7289\\\\\frac{f_2}{f_1}  = 10^{0.7289}\\\\\frac{f_2}{f_1}  = 5.356

Therefore, the ratio of f at the higher temperature to f at the lower temperature is 5.356

5 0
3 years ago
Pls somebody help me with this question! <br> The pictures have to be in order!
irina1246 [14]

Answer:

I believe the organelle should replace the question mark!

3 0
2 years ago
Read 2 more answers
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