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3 years ago

C12H26O + SO3+NaOH ----> C12H25NaSO4+ H2O

Chemistry

1 answer:

Vlad [161]3 years ago

4 0

Answer:

We can solve this by the method of which i solved your one question earlier

so again here molar mass of C12H25NaSO4 is 288.372 and number of moles for 11900 gm of C12H25NaSO4 will be = 11900/288.372

which is almost = 41.26 moles

so to get one mole of C12H25NaSO4 we need one mole of C12H26O

so for 41.26 moles of C12H25NaSO4 it will require 41 26 moles of C12H26O

so the mass of C12H26O = 41.26× its molar mass

C12H26O = 41.26×186.34

= 7688.38 gm!!

so the conclusion is If you need 11900 g of C12H25NaSO4 (Sodium Lauryl Sulfate) you need C12H26O 7688.38 gm !!

Again i d k wether it's right or wrong but i tried my best hope it helped you!!

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How many grams are there in 7.4 x 1023 molecules of H2SO4?

bazaltina [42]

7.4x10^23 = molecules of silver nitrate sample

6.022x10^23 number of molecules per mole (Avogadro's number)

Divide molecules of AgNO3 by # of molecules per mol

7.4/6.022 = 1.229 mols AgNO3 (Sig Figs would put this at 1.3)

(I leave off the x10^23 because they both will divide out)

Use your periodic table to find the molar weight of silver nitrate.

107.868(Ag) + 14(N) + 3(16[O]) = 169.868g/mol AgNO3

Now multiply your moles of AgNO3 with your molar weight of AgNO3

1.229mol x 169.868g/mol = 208.767g AgNO3

4 0

3 years ago

Copper (II) sulfate pentahydrate may be heated to drive off the water of hydration. If 5 g of water are produced, what was the o

____ [38]

Answer:

13.9g of CuSO4•5H2O

Explanation:

We'll begin by writing the balanced equation for the reaction. This is given below:

CuSO4•5H2O → CuSO4 + 5H2O

Next, we shall determine the mass of CuSO4•5H2O heated and the mass of H2O produced from the balanced equation.

This is illustrated below

Molar mass of CuSO4• 5H2O = 63.5 + 32 + (16x4) + 5(2x1 + 16)

= 63.5 + 32 + 64 + 5(18) = 249.5g/mol

Mass of CuSO4•5H2O from the balanced equation = 1 x 249.5 = 249.5g

Molar mass of H2O = (2x1) + 16 = 18g/mol

Mass of H2O from the balanced equation = 5 x 18 = 90g

From the balanced equation above,

249.5g of CuSO4•5H2O produced 90g of H2O.

Now, we can determine the mass of CuSO4•5H2O needed to produce 5g of H2O. This can be achieved as shown below:

From the balanced equation above,

249.5g of CuSO4•5H2O produced 90g of H2O.

Therefore, Xg of CuSO4•5H2O will produce 5g of H2O i.e

Xg of CuSO4•5H2O = (249.5 x 5)/90

Xg of CuSO4•5H2O = 13.9g

Therefore, 13.9g of CuSO4•5H2O is needed to produce 5g of H2O.

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3 years ago

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2 years ago

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Element X had a mass of 300g and a 1/2 life of 10 years. how many grams will remain after 40 years?

kiruha [24]

<h3>Answer:</h3>

18.75 grams

<h3>Explanation:</h3>

Half-life refers to the time taken by a radioactive material to decay by half of the original mass.
In this case, the half-life of element X is 10 years, which means it takes 10 years for a given mass of the element to decay by half of its original mass.
To calculate the amount that remained after decay we use;

Remaining mass = Original mass × (1/2)^n, where n is the number of half-lives

Number of half-lives = Time for the decay ÷ Half-life

= 40 years ÷ 10 years

= 4

Therefore;

Remaining mass = 300 g × (1/2)⁴

= 300 g × 1/16

= 18.75 g

Hence, a mass of 300 g of an element X decays to 18.75 g after 40 years.

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ikadub [295]

Answer:

I think the answer is gravity

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