All categories

3 years ago

Crossover formula of potassium oxide.

Chemistry

1 answer:

Pie3 years ago

8 0

Answer:

K2O

Explanation:

It carries the chemical formula K2O. Potassium cannot be found free because it is too reactive. It has valency +1 and combines readily with oxygen atoms forming K2O.

hope it helps

You might be interested in

Which is a component of John Dalton’s atomic theory?

s344n2d4d5 [400]

I believe its a...................

7 0

3 years ago

Read 2 more answers

The reaction P4 +502P4010 + 712 kcal is

igor_vitrenko [27]

Answer:

The reaction is exothermic (option 4)

Explanation:

P4 + 5O2 → P4O10 + 712 kcal

In chemical reactions heat can be absorbed or released:

⇒in the first case, when heat is absorbed, this is called an endothermic reaction. The products have more energy than the reactants. The reaction requires or absorbs energy from it's surroundings. That means that in this reaction energy , in the form of heat, will be absorbed by the reactants.

⇒ when heat is released, this is called an exothermic reaction. The reactants have more energy than the products. The reaction gives or releases energy to it's surroundings. That means that in this reaction energy , in the form of heat, will be released by the reactants.

in the case of P4 + 5O2 → P4O10 + 712 kcal

We notice that on the right side, which is the product side, there is a positive amount of energy. This means that the energy is released by the the reactants, in this reaction. The reaction is exothermic.

4 0

3 years ago

Read 2 more answers

For the reaction shown, calculate how many grams of oxygen form when each quantity of reactant completely reacts.

Nikolay [14]

Explanation:

Molar mass of $KClO_{3}$ = 39.1 + 35.5 + 3(16.0) = 122.6 g

Molar mass of KCl = 39.1 + 35.5 = 74.6 g

Molar mass of $O_{2}$ = 32.0 g

According to the equation, 2 moles of $KClO_{3}$ reacts to give 3 moles of oxygen.

Therefore, 2 (122.6) = 245.2 g of $KClO_{3}$ will give 3 (32.0) = 96.0 g of oxygen. Thus, 245.2 g of $KClO_{3}$ gives 96.0 g of oxygen.

(a) Calculate the amount of oxygen given by 2.72 g of $KClO_{3}$ as follows.

$\frac {96g}{245.2g} \times 2.72 g = 1.06 g$ of $O_{2}$

(b) Calculate the amount of oxygen given by 0.361 g of $KClO_{3}$ as follows.

$\frac {96g}{245.2g} \times 0.361 g = 0.141 g$ of $O_{2}$

c) Calculate the amount of oxygen given by 83.6 kg $KClO_{3}$ as follows.

$\frac {96g}{245.2g} \times 83.6 g = 32.731 kg$ of $O_{2}$

Convert kg into grams as follows.

$\frac{32.731 kg}{1 kg} \times 1000 g$ = 32731 g of $O_{2}$

(d) Calculate the amount of oxygen given by 22.5 mg of $KClO_{3}$ as follows.

$\frac {96g}{245.2g} \times 22.5 mg = 8.79 mg$

Convert mg into grams as follows.

$\frac{8.79 mg}{1 mg}\times 10^{-3} g = 8.79 \times 10^{-3} g$ of $O_{2}$

8 0

3 years ago

Put these numbers from greatest to least precision 0.10 g/mL 0.0447g/mL 0.0700g/mL

horrorfan [7]

I am going to go with,

0.10 g/mL
0.0700 g/mL
0.0447 g/mL

I don't know if this is the correct answer, but I am 80% sure that it may be.
:) :)

6 0

3 years ago

It takes 495.0 kJ of energy to remove 1 mole of electron from an atom on the surface of sodium metal. How much energy does it ta

Zigmanuir [339]

Answer:

$\lambda=241.9\ nm$

Explanation:

The work function of the sodium= 495.0 kJ/mol

It means that

1 mole of electrons can be removed by applying of 495.0 kJ of energy.

Also,

1 mole = $6.023\times 10^{23}\ electrons$

So,

$6.023\times 10^{23}$ electrons can be removed by applying of 495.0 kJ of energy.

1 electron can be removed by applying of $\frac {495.0}{6.023\times 10^{23}}\ kJ$ of energy.

Energy required = $82.18\times 10^{-23}\ kJ$

Also,

1 kJ = 1000 J

So,

Energy required = $82.18\times 10^{-20}\ J$

Also, $E=\frac {h\times c}{\lambda}$

Where,

h is Plank's constant having value $6.626\times 10^{-34}\ Js$

c is the speed of light having value $3\times 10^8\ m/s$

So,

$79.78\times 10^{-20}=\frac {6.626\times 10^{-34}\times 3\times 10^8}{\lambda}$

$\lambda=\frac{6.626\times 10^{-34}\times 3\times 10^8}{82.18\times 10^{-20}}$

$\lambda=\frac{10^{-26}\times \:19.878}{10^{-20}\times \:82.18}$

$\lambda=\frac{19.878}{10^6\times \:82.18}$

$\lambda=2.4188\times 10^{-7}\ m$

Also,

1 m = 10⁻⁹ nm

So,

$\lambda=241.9\ nm$

6 0

3 years ago