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AleksandrR [38]
3 years ago
12

An

Chemistry
1 answer:
Basile [38]3 years ago
4 0

Answer:

The answer of this question is molecule

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A car traveling with constant speed travels 150 km in 7200 s. What is the speed of the car?
Advocard [28]
Speed=distance/time  

D=150km

S=7200s 

Divide the numbers. 

150/7200=0.02km/s <---final answer
5 0
3 years ago
What is the molality of a solution of water and kcl if the freezing point of the solution is –3mc030-1.jpgc?
Natasha_Volkova [10]
We will use the expression for freezing point depression ∆Tf
     ∆Tf = i Kf m
Since we know that the freezing point of water is 0 degree Celsius, temperature change ∆Tf is 
     ∆Tf = 0C - (-3°C) = 3°C 
and the van't Hoff Factor i is approximately equal to 2 since one molecule of KCl in aqueous solution will produce one K+ ion and  one Cl- ion:
     KCl → K+ + Cl- 
Therefore, the molality m of the solution can be calculated as 
     3 = 2 * 1.86 * m
     m = 3 / (2 * 1.86)
     m = 0.80 molal
4 0
3 years ago
Help quick
Mila [183]

Answer: it is in the correct order

Explanation:

5 0
2 years ago
12. Which atoms or group of atoms make up the functional group of ethanol (CH3CH2OH)?
Montano1993 [528]

Answer:

Option A. the hydroxyl group (-OH)

Explanation:

Ethanol, CH₃CH₂OH belongs to the class of organic compound called alkanol.

They have general formula as R–OH

Where

R => is an alkyl group

OH => is the hydroxyl group

The hydroxyl group (OH) is the functional group of the alkanol (alcohol)

8 0
2 years ago
At 2000°C, the equilibrium constant for the reaction below is Kc = 4.10 ´ 10–4 . If 0.600 moles of NO is placed in a 1.0-L react
erastova [34]

Answer:

At equilibrium, the concentration of N_{2 (g)} is going to be 0.30M

Explanation:

We first need the reaction.

With the information given we can assume that is:

N_{2 (g)} + O_{2 (g)} ⇄ 2NO_{(g)}

If there is placed 0.600 moles of NO in a 1.0-L vessel, we have a initial concentration of 0.60 M NO; and no N_{2 (g)} nor  O_{2 (g)} present. Immediately, N_{2 (g)} andO_{2 (g)} are going to be produced until equilibrium is reached.

By the ICE (initial, change, equilibrium) analysis:

I: [N_{2 (g)}]=0   ;     [O_{2 (g)} ]= 0    ; [NO_{(g)}]=0.60M

C: [N_{2 (g)}]=+x   ;     [O_{2 (g)} ]= +x    ; [NO_{(g)}]=-2x

E: [N_{2 (g)}]=0+x   ;     [O_{2 (g)} ]= 0+x   ; [NO_{(g)}]=0.60-2x

Now we can use the constant information:

K_{c}=\frac{[products]^{stoichiometric coefficient} }{[reactants]^{stoichiometric coefficient} }

4.10* 10^{-4} =\frac{(0.60-2x)^{2}}{(x)*(x)}

4.10* 10^{-4}= \frac{(0.60-2x)^{2}}{x^{2} }

4.10* 10^{-4} * x^{2}= (0.60-2x)^{2}}

\sqrt{4.10* 10^{-4} * x^{2}}= \sqrt{(0.60-2x)^{2}}}

0.0202 x =0.60 - 2x

2x+0.0202x=0.60

x=\frac{0.60}{2.0202}= 0.30

At equilibrium, the concentration of N_{2 (g)} is going to be 0.30M

3 0
2 years ago
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