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valentinak56 [21]
3 years ago
9

An ice cube at 0 degrees Celsius is heated until it melts and the water reaches 20 degrees Celsius. The amount if heat needed fo

r this is 5.000x10^3J. What is the mass of the ice cube?
Chemistry
1 answer:
Troyanec [42]3 years ago
4 0

Answer:

20 i guess

Explanation:

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Structure of tetrazine
Vika [28.1K]

Answer:

Tetrazine is a compound that consists of a six-membered aromatic ring containing four nitrogen atoms with the molecular formula C2H2N4.

(See the image)

Hope it helps!

3 0
3 years ago
Consider the reaction 2NO(g) 1 O2(g) ¡ 2NO2(g) Suppose that at a particular moment during the reaction nitric oxide (NO) is reac
GalinKa [24]
<h2>a) The rate at which NO_2 is formed is 0.066 M/s</h2><h2>b) The rate at which molecular oxygen O_2 is reacting is 0.033 M/s</h2>

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.

2NO(g)+O_2(g)\rightarrow 2NO_2(g)

The rate in terms of reactants is given as negative as the concentration of reactants is decreasing with time whereas the rate in terms of products is given as positive as the concentration of products is increasing with time.

Rate in terms of disappearance of NO = -\frac{1d[NO]}{2dt} = 0.066 M/s

Rate in terms of disappearance of O_2 = -\frac{1d[O_2]}{dt}

Rate in terms of appearance of NO_2= \frac{1d[NO_2]}{2dt}

1. The rate of formation of NO_2

-\frac{d[NO_2]}{2dt}=\frac{1d[NO]}{2dt}

\frac{1d[NO_2]}{dt}=\frac{2}{2}\times 0.066M/s=0.066M/s

2. The rate of disappearance of O_2

-\frac{1d[O_2]}{dt}=\frac{d[NO]}{2dt}

-\frac{1d[O_2]}{dt}=\frac{1}{2}\times 0.066M/s=0.033M/s

Learn more about rate law

brainly.com/question/13019661

https://brainly.in/question/1297322

7 0
3 years ago
Why do you think the soccer ball fell more slowly than the golf ball
alexgriva [62]

Answer:

kinetic energy than the potential energy it carries

Explanation:

3 0
3 years ago
At what temperature would a 1.50 m nacl solution freeze, given that the van't hoff factor for nacl is 1.9? kf for water is 1.86
IRINA_888 [86]
Win to ki ne lei ma ki mna mata ki ss kk
5 0
3 years ago
Calculate the rate constant at 200.°C for a reaction that has a rate constant of 8.30 × 10−4 s−1 at 90.°C and an activation ener
Sergeu [11.5K]

Answer:

23.0 s⁻¹ is rate constant

Explanation:

Using the Arrhenius equation:

k = A * e^(-Ea/RT)

Where k is rate constant

A is frequency factor (1.5x10¹¹s⁻¹)

Ea is activation energy = 55800J/mol

R is gas constant (8.314J/molK)

And T is absolute temperature (24°C + 273 = 297K)

Replacing:

k = 1.5x10¹¹s⁻¹ * e^(-55800J/mol/8.314J/molK*297K)

k = 1.5x10¹¹s⁻¹ * 1.53x10⁻¹⁰

k = 23.0 s⁻¹ is rate constant    i hope this helpsss

Explanation:

4 0
3 years ago
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