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klasskru [66]
3 years ago
6

One mole of nitrogen and one mole of neon are combined in a closed container at STP.How big is the container?V = __________ L

Chemistry
1 answer:
cestrela7 [59]3 years ago
5 0

Answer : The volume of container will be 44.8 L

Explanation :

At STP condition,  

The temperature and pressure are 273 K and 1 atm respectively.

As we know that at STP, 1 mole of substance occupies 22.4 L volume of gas.

As per question,

1 mole of nitrogen gas occupies 22.4 L volume

and,

1 mole of neon gas also occupies 22.4 L volume

Thus, total volume will be:

Total volume of container = 22.4 + 22.4 = 44.8 L

Hence, the volume of container will be 44.8 L

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Lyrx [107]

Answer:

C

Explanation:

The airplanes speed would decrease

7 0
3 years ago
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Analyze: What do the elements in each column of the periodic table have in common?
ludmilkaskok [199]

Answer:

Each column is called a group. The elements in each group have the same number of electrons in the outer orbital. Those outer electrons are also called valence electrons. They are the electrons involved in chemical bonds with other elements.

Explanation:

Each column is called a group. The elements in each group have the same number of electrons in the outer orbital. Those outer electrons are also called valence electrons. They are the electrons involved in chemical bonds with other elements.

7 0
3 years ago
Vanadium has an atomic mass of 50.9415 amu. It has two common isotopes. One isotope has a mass of 50.9440 amu and a relative abu
grandymaker [24]

Answer:

Average atomic mass of the  vanadium = 50.9415 amu

Isotope (I) of vanadium' s abundance = 99.75 %= 0.9975

Atomic mass of Isotope (I) of vanadium ,m= 50.9440 amu

Isotope (II) of vanadium' s abundance =(100%- 99.75 %) = 0.25 % = 0.0025

Atomic mass of Isotope (II) of vanadium ,m' = ?

Average atomic mass of vanadium =

m × abundance of isotope(I) + m' × abundance of isotope (II)

50.9415 amu =50.9440 amu× 0.9975 + m' × 0.0025

m'= 49.944 amu

Explanation:

6 0
2 years ago
A mixture of 0.10 mol of NO, 0.050 mol of H2, and 0.10 mol of H2O is placed in a 1.0-L vessel at 300 K. The following equilibriu
tatyana61 [14]

Answer:

[H2] =    0.012 M

[N2] =    0.019 M

[H2O] =  0.057 M

Explanation:

The strategy here is to account for the species at equilibrium given that the concentration of [NO]=0.062M at equilibrium is known and the quantities initially present and its stoichiometry.

                  2NO(g)         +    2H2(g)    ⇒        N2(g)      +         2H2O(g)

i  mol            0.10                   0.050                                             0.10

c mol            -0.038                -0.038                +0019                +0.038                                                

e mol            0.062                 0.012                  00.019               0.057

Since the volume of the vessel is 1.0 L, the concentrations in molarity are:

[NO] =   0.062 M

[H2] =    0.012 M

[N2] =    0.019 M

[H2O] =  0.057 M

5 0
2 years ago
Which of the following is the ultimate byproduct of weathering?
NikAS [45]
Soil
the answer is soil

6 0
2 years ago
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