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aniked [119]
4 years ago
13

Why is the composition of Earth's atmosphere different from the composition of the atmospheres of Venus and Mars? A. Venus, Mars

, and Earth all have similar chemical compositions. B. Earth has life while Venus and Mars do not. C. Venus and Mars have different chemical compositions from Earth. D. Venus and Mars are slightly smaller than Earth.
Chemistry
1 answer:
rewona [7]4 years ago
6 0

Answer:

they all have different chemical compositions from earth

Venus is more filled with carbon dioxide and nitrogen

earth is filled more with oxygen and nitrogen

mars is filled more with carbon dioxide and carbon monoxide

so ur answer is C

it cant be B as scientists and NASA are still in the look out for life, D has nothing to do with chemicals and atmosphere, and its not A cuz they all dont have the same similiar chemical.

~batmans wife dun dun dun....

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A chermist weighs out 400.0 g of Ca How many moles of Ca is this?
ra1l [238]

Answer:

<h3>The answer is 0.1 mol</h3>

Explanation:

To find out the number of moles we use the formula

n =  \frac{m}{M}  \\

where

n is the number of moles

M is the molar mass

m is the mass of the substance

From the question

m = 400 g

M of Ca = 40 g/mol

We have

n =  \frac{40}{400}  =  \frac{1}{10}  \\

We have the final answer as

<h3>0.1 mol</h3>

Hope this helps you

4 0
4 years ago
4. True or false: You should use Avogadro's number to convert from moles to molecules
Dmitriy789 [7]

True, to convert from moles to atoms, multiply the molar amount by Avogadro's number. To convert from atoms to moles, divide the atom amount by Avogadro's number (or multiply by its reciprocal).

8 0
4 years ago
There are 3.0 g of NO and 3.0 g of CO reacting. what is the limiting reactant ?
xxTIMURxx [149]

Answer:

NO will be the limiting reagent.

Explanation:

The balanced equation is:

2 NO + 2 CO → N₂ + 2 CO₂

By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:

  • NO: 2 moles
  • CO: 2 moles
  • N₂: 1 mole
  • CO₂: 2 moles

Being the molar mass of each compound:

  • NO: 30 g/mole
  • CO: 28 g/mole
  • N₂: 28 g/mole
  • CO₂: 44 g/mole

Then by stoichiometry the following quantities of mass participate in each reaction:

  • NO: 2 moles* 30 g/mole= 60 g
  • CO: 2 moles* 28 g/mole= 56 g
  • N₂: 1 mole* 28 g/mole= 28 g
  • CO₂: 2 moles* 44 g/mole= 88 g

The limiting reagent is one that is consumed first in its entirety, determining the amount of product in the reaction. When the limiting reagent is finished, the chemical reaction will stop.

To determine the limiting reagent, you can use a simple rule of three as follows: If 56 grams of CO react with 60 grams of NO, 3 grams of CO react with how much mass of NO?

mass of NO=\frac{3 grams of CO*60 grams of NO}{56 grams of CO}

mass of NO= 3.21 grams

But 3.21 grams of NO are not available, 3 grams are available. Since you have less moles than you need to react with 3 grams of CO, <u><em>NO will be the limiting reagent. </em></u>

<u><em></em></u>

4 0
3 years ago
Help please it's due in 12 hours<br> (20 points)
Rufina [12.5K]

Answer:

1) K and Rb

2) 20 protons, 20 neutrons, and 20 electrons.

Explanation:

1) If it's in the same column, it has the same # of valence electrons (e- on the outer shell), therefore similar chemical properties.

2)

1) atomic #, 20, is the number of protons

2) subtract the mass number from # of protons, which you get 20 neutrons

3) Number of electrons in a neutral atom = number of electrons, which is 20. You can also check yourself by looking at electron configuration on a full periodic table.

8 0
2 years ago
The activation energy of an uncatalyzed reaction is 95 kJ / mol. The addition of a catalyst lowers the activation energy to 55 k
dybincka [34]

Answer:

The rate of the catalyzed reaction will increase by a 1.8 × 10⁵ factor.

Explanation:

The rate of a reaction (r) is proportional to the rate constant (k). We can find the rate constant using the Arrhenius equation.

k=A.e^{-Ea/R.T}

where,

A: collision factor

Ea: activation energy

R: ideal gas constant

T: absolute temperature (125°C + 273 = 398 K)

For the uncatalized reaction,

kU=A.e^{-95\times 10^{3} kJ/mol /(8.314J/K.mol).398K}=3.4\times 10^{-13}A

For the catalized reaction,

kC=A.e^{-55\times 10^{3} kJ/mol /(8.314J/K.mol).398K}=6.0\times 10^{-8}A

The ratio kC to kU is 6.0 × 10⁻⁸A/3.4 × 10⁻¹³A = 1.8 × 10⁵

8 0
4 years ago
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