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masha68 [24]
3 years ago
6

The brightness of a star depends on its _____ a. color b. composition of atmosphere c. distance from Earth, and stars that are c

loser look _____ a. brighter b. dimmer c. white
Chemistry
1 answer:
Yanka [14]3 years ago
5 0

Answer:

Either B or C. Composition or the Distance from the Earth.

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Carbon tetrachloride can be produced by the following reaction: Suppose 1.20 mol of and 3.60 mol of were placed in a 1.00-L flas
hjlf

The given question is incomplete. The complete question is :

Carbon tetrachloride can be produced by the following reaction:

CS_2(g)+3Cl_2(g)\rightleftharpoons S_2Cl_2(g)+CCl_4(g)

Suppose 1.20 mol CS_2(g) of and 3.60 mol of Cl_2(g)  were placed in a 1.00-L flask at an unknown temperature. After equilibrium has been achieved, the mixture contains 0.72 mol  of CCl_4. Calculate equilibrium constant at the unknown temperature.

Answer: The equilibrium constant at unknown temperature is 0.36

Explanation:

Moles of  CS_2 = 1.20 mole

Moles of  Cl_2 = 3.60 mole

Volume of solution = 1.00  L

Initial concentration of CS_2 = \frac{moles}{volume}=\frac{1.20mol}{1L}=1.20M

Initial concentration of Cl_2 = \frac{moles}{volume}=\frac{3.60mol}{1L}=3.60M

The given balanced equilibrium reaction is,

                 CS_2(g)+3Cl_2(g)\rightleftharpoons S_2Cl_2(g)+CCl_4(g)

Initial conc.         1.20 M        3.60 M                  0                  0

At eqm. conc.     (1.20-x) M   (3.60-3x) M   (x) M        (x) M

The expression for equilibrium constant for this reaction will be,

K_c=\frac{[S_2Cl_2]\times [CCl_4]}{[Cl_2]^3[CS_2]}

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

K_c=\frac{(x)\times (x)}{(3.60-3x)^3\times (1.20-x)}

Given :Equilibrium concentration of CCl_4 , x = \frac{moles}{volume}=\frac{0.72mol}{1L}=0.72M

K_c=\frac{(0.72)\times (0.72)}{(3.60-3\times 0.72)^3\times (1.20-0.72)}

K_c=0.36

Thus equilibrium constant at unknown temperature is 0.36

4 0
2 years ago
The soda was cold before it was placed in the koozie. the koozie...
umka21 [38]

Answer:

a

Explanation:

3 0
2 years ago
Read 2 more answers
Determine the approximate amount of potassium hydrogen phthalate, KHP, that you will need to neutralize 6.00 ml of 0.100 M NaOH.
Sveta_85 [38]

Answer:

potassium hydrogen phthalate KHP MOLAR MASS = 204.233 glmol

to get 1000 ml

Molar concentration = Mass concentration/Molar Mass

mass concentration = molar concentration x molar mass

mass concentration=0.1 M,

molar mass= 204.233 g/mol

so to get 1L

mass conc = 204.233 x 0.1

= 20.4233g  for 1L or 1000 ml

to get 6.00 ml

if 20.4233g is for 1000ml

then to 6.00 ml

= 20.4233 x 6 / 1000

= 0.123g for 6.00 ml

according to the equation below

NaOH(aq) + KHC8H4O4(aq) --> KNaC8H4O4(aq) + H2O(l)

number of moles of NaOH is equal to that of KHP

so the same amount will be needed too, which is

= 0.123g

6 0
2 years ago
The process that makes atmospheric nitrogen available to plants by mutualistic and free-living bacteria is called
artcher [175]
Nitrogen fixation is the process that makes atmospheric nitrogen available to plants by mutualistic and free-living bacteria. The process is undertaken by the rhizobium bacteria that live in root roots of plants such as legumes. The mutualistic relationship is that the plant supplies the bacteria with a habitat in which to live, water, and nutrients, and the bacteria supply nitrogen for making plant proteins. 
3 0
3 years ago
Which is the strongest acid listed in the table?
andrew11 [14]

Answer:

Hydrofluoric acid.

Explanation:

To know which of the acid is the strongest, let us determine the pka of each acid. This is illustrated below:

1. Acetic acid

Ka = 1.8x10^-5

pKa =..?

pKa = –logKa

pKa = –Log 1.8x10^-5

pKa = 4.74

2. Benzoic acid

Ka = 6.5x10^-5

pKa =..?

pKa = –logKa

pKa = –Log 6.5x10^-5

pKa = 4.18

3. Hydrofluoric acid.

Ka = 6.8x10^-4

pKa =..?

pKa = –logKa

pKa = –Log 6.8x10^-4

pKa = 3.17

4. Hypochlorous acid

Ka = 3.0x10^-8

pKa =..?

pKa = –logKa

pKa = –Log 3.0x10^-8

pKa = 7.52

Note: the smaller the pKa value, the stronger the acid.

The pka of the various acids as calculated above is given below:

Acid >>>>>>>>>>>>>>>>>> pKa

1. Acetic acid >>>>>>>>>> 4.74

2. Benzoic acid >>>>>>>> 4.18

3. Hydrofluoric acid >>>> 3.17

4. Hypochlorous acid >> 7.52

From the above illustration, we can see that hydrofluoric acid has the lowest pKa value. Therefore, hydrofluoric acid is the strongest among them.

6 0
3 years ago
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