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ASHA 777 [7]
3 years ago
5

According to our textbook, write the formula for the anhydrous compound that was part of the mixture called natron that was used

by the Egyptians. What did they use this compound for and what was the name of the resulting hydrate that formed?
Chemistry
1 answer:
bearhunter [10]3 years ago
8 0

Answer:

The formula for the anhydrous compound that was part of the mixture called natron that was used by the Egyptians is Na2(CO3)10(H2O).

They use this compound for medicine, cookery, agriculture, in glass-making and to dehydrate egyptian mummies.

Compound of sodium carbonate and sodium bicarbonate was the name of the resulting hydrate that formed.

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When benzene (c6h6) reacts with bromine (br2) bromobenzene(c6h5br) is obtained: c6h6 + br2 → c6h5br + hbr what is the theoretica
saveliy_v [14]

Answer A) : We have to calculate the number of moles of Benzene involved in the reaction,

30 g / 78 moles of benzene = 0.384 moles


For bromine it will be the same process,

65 g / 159.8 moles = 0.406 moles


By observing the reaction given above we can say that the reaction ratio of bromine and benzene is 1 : 1


We need to find the mass of bromobenzene,

which should be, 6(12) + 5 (1)+ 79.90 = 156.9 g/mol


So, the mass of bromobenzene will be 156.9 g/mol X 0.3846 mol = 60.343 g


Hence the theoretical yield will be 60.34 g


Answer B) : To calculate the actual yield we have to divide it with theoretical yield.


(56.7g / 60.343 g ) X100% = 93.96 %


Here, we can say that we got 93.96 % of actual yield.


As we know it is impossible to get 100% yield in any reaction.

3 0
3 years ago
What is the relationship between the number of each atom used to form compound and oxidation number?
deff fn [24]

Answer:Explanation:

In compounds, all other atoms are assigned an oxidation number so that the sum of the oxidation numbers on all the atoms in the species equals the charge on the species.

6 0
3 years ago
Calculate the energy, in joules, required to ionize a hydrogen atom when its electron is initially in the n =2 energy level. The
qaws [65]

Answer:

E_{ionization}=5.45\times 10^{-19}\ J

Explanation:

E_n=-2.18\times 10^{-18}\times \frac{1}{n^2}\ Joules

For transitions:

Energy\ Difference,\ \Delta E= E_f-E_i =-2.18\times 10^{-18}(\frac{1}{n_f^2}-\frac{1}{n_i^2})\ J=2.18\times 10^{-18}(\frac{1}{n_i^2} - \dfrac{1}{n_f^2})\ J

\Delta E=2.18\times 10^{-18}(\frac{1}{n_i^2} - \dfrac{1}{n_f^2})\ J

So, n_i=2 and n_f=\infty (As the hydrogen has to ionize)

Thus,

\Delta E=2.18\times 10^{-18}(\frac{1}{2^2} - \dfrac{1}{{\infty}^2})\ J

\Delta E=2.18\times 10^{-18}(\frac{1}{2^2})\ J

E_{ionization}=5.45\times 10^{-19}\ J

4 0
3 years ago
Thermochemistry
max2010maxim [7]
THe Answer is 105 J

hope this helps and please brainliest, I was first :)
4 0
3 years ago
A gas effuses 1.55 times faster than propane (C3H8)at the
stepladder [879]

Answer:  The mass of the gas is 18.3 g/mol.

Explanation:

To calculate the rate of diffusion of gas, we use Graham's Law.

This law states that the rate of effusion or diffusion of gas is inversely proportional to the square root of the molar mass of the gas. The equation given by this law follows:

\text{Rate of diffusion}\propto \frac{1}{\sqrt{\text{Molar mass of the gas}}}

\frac{Rate_{X}}{Rate_{C_3H_8}}=1.55

\frac{Rate_{X}}{Rate_{C_3H_8}}=\sqrt{\frac{M_{C_3H_8}}{M_{X}}}

1.55=\sqrt{\frac{44}{M_{X}}

Squaring both sides and solving for M_{X}

M_{X}=18.3g/mol

Hence, the molar mas of unknown gas is 18.3 g/mol.

5 0
3 years ago
Read 2 more answers
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