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erma4kov [3.2K]
2 years ago
15

Which of the following shows 13,400 as having 4 significant figures?

Chemistry
1 answer:
Gnom [1K]2 years ago
5 0
A would be your answer
You might be interested in
Look at pic and help
natita [175]

Answer:

Answer

Explanation:

1) Other gasses / or argon (based on what u took in ur school )

2) oxygen

3? Nitrogen

3 0
2 years ago
Read 2 more answers
Which is the best example for genetic diversity?
DanielleElmas [232]

Answer:

Genetic Diversity Examples

  • Different breeds of dogs. ...
  • Different varieties of rose flower, wheat, etc.
  • There are more than 50,000 varieties of rice and more than a thousand varieties of mangoes found in India.

Genetic diversity is the total number of genetic characteristics in the genetic makeup of a species, it ranges widely from the number of species to differences within species and can be attributed to the span of survival for a species.

Explanation:

i hope this helps u.

8 0
3 years ago
4. Calculate the relative molecular masses of the following substances [RAM: H=1, O=16, S=32, C=12, N=14] a. CH2(NH2)COOH b. H2S
8090 [49]

Answer:

a. CH2(NH2)COOH

molecular \: mass = (12 \times 2) + (5 \times 1) + (14 \times 1) + (16 \times 2)  \\  = 24 + 5 + 14 + 32 \\  = 75 \: grams

b. H2SO4

molecular \: mass = (2 \times 1) + 32 + (16 \times 4) \\  = 2 + 32 + 64\\  = 98 \: grams

4 0
2 years ago
The reaction between nitrogen dioxide and carbon monoxide is NO2(g)+CO(g)→NO(g)+CO2(g)NO2(g)+CO(g)→NO(g)+CO2(g) The rate constan
Alex17521 [72]

Answer : The rate constant at 525 K is, 0.0606M^{-1}s^{-1}

Explanation :

According to the Arrhenius equation,

K=A\times e^{\frac{-Ea}{RT}}

or,

\log (\frac{K_2}{K_1})=\frac{Ea}{2.303\times R}[\frac{1}{T_1}-\frac{1}{T_2}]

where,

K_1 = rate constant at 701K = 2.57M^{-1}s^{-1}

K_2 = rate constant at 525K = ?

Ea = activation energy for the reaction = 1.5\times 10^2kJ/mol=1.5\times 10^5J/mol

R = gas constant = 8.314 J/mole.K

T_1 = initial temperature = 701 K

T_2 = final temperature = 525 K

Now put all the given values in this formula, we get:

\log (\frac{K_2}{2.57M^{-1}s^{-1}})=\frac{1.5\times 10^5J/mol}{2.303\times 8.314J/mole.K}[\frac{1}{701K}-\frac{1}{525K}]

K_2=0.0606M^{-1}s^{-1}

Therefore, the rate constant at 525 K is, 0.0606M^{-1}s^{-1}

8 0
3 years ago
How small do you think an individual atom is
Rudiy27
Tiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiny. atoms are smaller than anything else
4 0
3 years ago
Read 2 more answers
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