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2 years ago

What is the only nonmetal with four valence electrons.

Chemistry

1 answer:

Alja [10]2 years ago

3 0

Carbon is a non metal and also has four valence electrons.

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A 30.7 g sample of Strontium nitrate, Sr(NO3)2•nH2O, is heated to a constant mass of 22.9 g. Calculate the hydration number.

Elodia [21]

Answer:

Hydration number: 4

Explanation:

1) Mass of water in the hydrated compound

Mass of water = Mass of the hydrated sample - mass of the dehydrated compound

Mass of water = 30.7 g - 22.9 g = 7.8 g

2) Number of moles of water

Number of moles = mass in grams / molar mass

molar mass of H₂O = 2×1.008 g/mol + 15.999 g*mol = 18.015 g/mol

Number of moles of H₂O = 7.9 g / 18.015 g/mol = 0.439 mol

3) Number of moles of Strontium nitrate dehydrated, Sr (NO₃)₂

The mass of strontium nitrate dehydrated is the constant mass obtained after heating = 22.9 g

Molar mass of Sr (NO₃)₂ : 211.63 g/mol (you can obtain it from a internet or calculate using the atomic masses of each element from a periodic table).

Number of moles of Sr (NO₃)₂ = 22.9 g / 211.63 g/mol = 0.108 mol

4) Ratio

0.439 mol H₂O / 0.108 mol Sr(NO₃)₂ ≈ 4 mol H₂O : 1 mol Sr (NO₃)₂

Which means that the hydration number is 4.

4 0

3 years ago

Read 2 more answers

g Ammonia has been studied as an alternative "clean" fuel for internal combustion engines, since its reaction with oxygen produc

Gala2k [10]

$\text{Ammonia has been studied as an alternative "clean" fuel for internal combustion}$

$\text{engines, since its reaction with oxygen produces only nitrogen and water vapor,}$

$\text{and in the liquid form it is easily transported. An industrial chemist studying this}$

$\text{reaction fills a} \ \mathbf{100 \ L }\ \text{tank with} \ \mathbf{8.6 \ mol} \ \text{of ammonia gas and} \ \mathbf{28 \ mol} \ \ \text{of oxygen gas, }$

$\text{to be} \ \mathbf{2.6\ mol} \ .\ \text{Calculate the concentration equilibrium constant for the combustion of}$

$\text{ammonia at the final temperature of the mixture. Round your answer to 2 significant digits.}$

Answer:

Explanation:

From the correct question above:

The reaction can be represented as:

$\mathbf{4 NH_3_{(g)}+ 3O_{2(g)} \iff 2N_{2(g)}+ 6H_2O_{(g)} }$

From the above reaction; the ICE table can be represented as:

$\mathbf{4 NH_3_{(g)}+ 3O_{2(g)} \iff 2N_{2(g)}+ 6H_2O_{(g)} }$

I (mol/L) 0.086 0.28 0 0

C -4x -3x +2x +6x

E 0.086 - 4x 0.28 - 3x +2x +6x

At equilibrium;

The water vapor = $\dfrac{2.6 \ mol}{100 \ L} = 6x$

$x = \dfrac{2.6}{100} \times \dfrac{1}{6}$

$x = 0.00433$

$\text{equilibrium constant} ({k_c}) = \dfrac{ [N_2]^2 [H_2O]^6 }{ [[NH_3]^4] [O_2]^3 }$

$\implies \dfrac{(2x)^2 (6x)^6}{(0.086-4x)^4\times (0.28-3x)^3} \\ \\$

Replacing the value of x, we have:

$K_c = \dfrac{4 \times 46,656 \times x^8}{(0.086-4x)^4\times (0.28 -3x)^3} \\ \\ K_c = \dfrac{4 \times 46656 \times (0.00433)^8}{(0.06868)^4(0.26701)^3} \\ \\ K_c = \mathbf{5.4446 \times 10^{-8}}$

$K_c = \mathbf{5.5 \times 10^{-8} \ to \ 2 \ significant \ figures}$

5 0

3 years ago

Someone please help! I will give Brainliest to who answers the best! Write an explanation of how mass determines (or affects) th

xxTIMURxx [149]

Gravity is the force of motion pulling down objects to the ground. If there was no gravity, everyone would walking as if they were on the moon.

Mass is what gravity needs. If an object has a little amount of mass, gravity will be able to easily bring it to the ground.

If an object has a very huge amount of mass, gravity will still be able to bring it to the ground but it will be hard.

For example: An airplane has a HUGE amount of mass. Gravity pulls it down but the airplane needs to be steering up in order for it to be straight. Gravity is applied on the airplane when it is landing.

BUT..... if a table is in the way of an object it depends if it will fall down to the ground or stay on the table.

If an object has little mass and a table is in the way of gravity pulling it down to the ground, the object will stay on the table. Like a plate of food on a table.

If an object has a very big amount of mass and a table is in the way of gravity pulling it to the ground, the object will break the object and make it's make to the ground. That is mostly why most of the time people have very strong tables/ anything to hold a heavy object.

Another example is if you're lifting weights and you have little amount of mass, you're most likely to get the little sized weight. It depend on you mass.

Here are some pictures I included here as well of Mass and gravity.

Glad to help! :) :D

8 0

3 years ago

The concentrated sulfuric acid we use in the laboratory is 98.0% sulfuric acid by weight. Calculate the molality and molarity of

timama [110]

Answer : The molarity and molality of the solution is, 18.29 mole/L and 499.59 mole/Kg respectively.

Solution : Given,

Density of solution = $1.83g/cm^3=1.83g/ml$

Molar mass of sulfuric acid (solute) = 98.079 g/mole

98.0 % sulfuric acid by mass means that 98.0 gram of sulfuric acid is present in 100 g of solution.

Mass of sulfuric acid (solute) = 98.0 g

Mass of solution = 100 g

Mass of solvent = Mass of solution - Mass of solute = 100 - 98.0 = 2 g

First we have to calculate the volume of solution.

$\text{Volume of solution}=\frac{\text{Mass of solution}}{\text{Density of solution}}=\frac{100g}{1.83g/ml}=54.64ml$

Now we have to calculate the molarity of solution.

$Molarity=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{volume of solution}}=\frac{98.0g\times 1000}{98.079g/mole\times 54.64ml}=18.29mole/L$

Now we have to calculate the molality of the solution.

$Molality=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Mass of solvent}}=\frac{98.0g\times 1000}{98.079g/mole\times 2g}=499.59mole/Kg$

Therefore, the molarity and molality of the solution is, 18.29 mole/L and 499.59 mole/Kg respectively.

7 0

3 years ago

An object was measured by a worker as 35.6 cm long, however, the manufacturer specifications list the length of the object at 35

Anon25 [30]

The answer for this question is 1.71%

7 0

3 years ago