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3 years ago

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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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

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3 years ago

If I have one mole of sulfur, how many atoms would that be?

krok68 [10]

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.

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3 years ago

3. How many molecules are in 0.500 moles of sulfur?

satela [25.4K]

Explanation:

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

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3 years ago

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

gladu [14]

Because there are so many different values of numbers, it would be impractical to use 1Ω, 2Ω, 3Ω... etc... Using colored bands helps make reading it a little easier to the trained eye. There are hundreds of thousands, if not tens of millions of different resistors would need to exist to cover every value. So you just use something called "preferred values" with their resistance values posted on them instead.

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3 years ago

A student was asked to prepare 500.0 mL of 6.0 M NaOH. The student measured 120.0 g of NaOH and placed it in a 1000 mL beaker. T

Morgarella [4.7K]

Answer:

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

Explanation:

To prepare a solution of a solid, the appropriate mass is taken and accurately weighed in a weighing balance and then made up to mark with distilled water.

From

n= CV

n = number of moles m/M( m= mass of solid, M= molar mass of compound)

C= concentration of substance

V= volume of solution

m=120g

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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3 years ago