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snow_lady [41]
3 years ago
12

Please help with questions 14-17 ! and thanks - will mark you brainliest

Chemistry
1 answer:
KiRa [710]3 years ago
8 0

Answer: 14 :12 water molecules,  

Explanation: 17: 6 molecules

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1. What are nonrenewable resources?
arlik [135]
They are natural resources that are in limited supply or that take a long time to create
8 0
3 years ago
In Europe gasoline is often sold in liters. If a typical fill-up requires 11 gal of gas, how many liters are necessary?
egoroff_w [7]

Answer:

41,64 liters

Explanation:

In order to change the units we need to know the equivalence of said units.

1 gal of gas equals to 3,78541 liters.

We can use cross-multiplication in order to convert it into liters:

\frac{1 gal}{3.78541 L} = \frac{11 gal}{X}  \\\\X= \frac{3.78541 L x 11 gal}{1 gal}

A typical fill-up requires 41,64 liters.

3 0
4 years ago
How to know how many valence electrons are in an element
Alenkinab [10]

Answer:

Hope it helped

Explanation:

For neutral atoms, the number of valence electrons is equal to the atom's main group number. The main group number for an element can be found from its column on the periodic table. For example, carbon is in group 4 and has 4 valence electrons. Oxygen is in group 6 and has 6 valence electrons.

7 0
3 years ago
At 95°C, the pressure of a sample of chlorine is 1.15atm. What will the pressure be at 105°C, assuming constant volume?
Jobisdone [24]

Answer:

95+273=368k

105+273=378k

(1 .15/368)×378

=1.18

4 0
3 years ago
At 35°C, Kc = 1.6 multiplied by10-5 for the following reaction
Brums [2.3K]

Answer : The equilibrium concentrations of all species NO,Cl_2\text{ and }NOCl are, 0.05 M, 0.043 M and 0.975 M respectively.

Explanation : Given,

Moles of  NO = 2 mole

Moles of  Cl_2 = 1 mole

Volume of solution = 1 L

Initial concentration of NO = 2 M

Initial concentration of Cl_2 = 1 M

The given balanced equilibrium reaction is,

                            2NO(g)+Cl_2(g)\rightleftharpoons 2NOCl(g)

Initial conc.          2 M            1 M            0

At eqm. conc.    (2-2x) M   (1-x) M         (2x) M

The expression for equilibrium constant for this reaction will be,

K_c=\frac{[NOCl]^2}{[NO]^2[Cl_2]}

The K_c for reverse reaction = \frac{1}{1.6\times 10^{-5}}

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

\frac{1}{1.6\times 10^{-5}}=\frac{(2x)^2}{(2-2x)^2\times (1-x)}

By solving the term 'x', we get :

x = 0.975

Thus, the concentrations of NO,Cl_2\text{ and }NOCl at equilibrium are :

Concentration of NO = (2-2x) M  = (2 - 2 × 0.975) M = 0.05 M

Concentration of Cl_2 = (1-x) M = 1 - 0.975 = 0.043 M

Concentration of NOCl = x M = 0.975 M

Therefore, the equilibrium concentrations of all species NO,Cl_2\text{ and }NOCl are, 0.05 M, 0.043 M and 0.975 M respectively.

6 0
3 years ago
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