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adelina 88 [10]
3 years ago
10

Find the density of a cube on Earth that weighs 1.5 kg and has a side-length of 10 cm.

Chemistry
1 answer:
tamaranim1 [39]3 years ago
3 0

Answer:

1500kg/m^3

Explanation:

Formula:

d=m/v

Given:

m=1.5kg

v=1000cm^3

(The side length of a cube is always equal to the others)

Required:

d=?

Solution:

d=m/v

d=1.5kg/1000cm^3

d=1.5kg/0.001m^3

d=1500kg/m^3

Hope this helps ;) ❤❤❤

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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
3 years ago
If I have one mole of sulfur, how many atoms would that be?
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Answer:

Atoms of sulfur = 9.60⋅g32.06⋅g⋅mol−1×6.022×1023⋅mol−1

Explanation:

because the units all cancel out, the answer is clearly a number, ≅2×1023 as required.

7 0
3 years ago
3. How many molecules are in 0.500 moles of sulfur?
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Explanation:

No of molecules=0.500×6.023×10²³=3.011×10²³ molecules

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Why are fixed resistors’ values indicated by color bands rather than printing the numeric value on their exterior?
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Answer:

The concentration of the solution will be much lower than 6M

Explanation:

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From

n= CV

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V= volume of solution

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M= 40gmol-1

V=500ml

120/40= C×500/1000

C= 120/40× 1000/500

C=6M

This solution will not be exactly 6M if the student follows the procedure outlined in the question. The actual concentration will be much less than 6M.

This is because, solutions are prepared in a standard volumetric flask. Using a 1000ml beaker, the student must have added more water than the required 500ml thereby making the actual concentration of the solution less than the expected 6M.

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