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il63 [147K]
3 years ago
8

What is the formula for rubidium nitride?

Chemistry
2 answers:
denis-greek [22]3 years ago
8 0
The formula for rubidium nitride is RbNO3.
irakobra [83]3 years ago
8 0
*rubidium nitrade*

Rb<span>NO3</span>. Rubidium nitrade is not produced from the earth. It is man-made.It is an alkali metal nitrate. It dissolves in water, kind-of like salt and sugar do.

Hope it helps!

I cant wait to be in Chemistry to learn all of this stuff! It sounds so lit! :)
You might be interested in
How many molecules are in 1kg of water
Mila [183]

Answer:

334.2× 10²³ molecules

Explanation:

Given data:

Mass of water = 1 Kg ( 1000 g )

Number of molecules = ?

Solution:

Number of moles of water:

Number of moles = mass/ molar mass

Number of moles = 1000 g/ 18 g/mol

Number of moles = 55.5 mol

1 mole contain 6.022× 10²³ molecules

55.5 mol×6.022× 10²³ molecules

334.2× 10²³ molecules

8 0
3 years ago
A compound is 42.9% C, 2.4% H, 16.7% N, and 38.1% O, by mass. Addition of 6.45 g of this compound to 50.0 mL benzene, lowers the
Romashka [77]

This is an incomplete question, here is a complete question.

A compound is 42.9% C, 2.4% H, 16.7% N and 38.1% O by mass. Addition of 6.45 g of this compound to 50.0 mL benzene, C₆H₆ (d= 0.879 g/mL; Kf= 5.12 degrees Celsius/m), lowers the freezing point from 5.53 to 1.37 degrees Celsius. What is the molecular formula of this compound?

Answer : The molecular of the compound is, C_6H_4N_2O_4

Explanation :

First we have to calculate the mass of benzene.

\text{Mass of benzene}=\text{Density of benzene}\times \text{Volume of benzene}

\text{Mass of benzene}=0.879g/mL\times 50.0mL=43.95g

Now we have to calculate the molar mass of unknown compound.

Given:

Mass of unknown compound (solute) = 6.45 g

Mass of benzene (solvent) = 43.95 g  = 0.04395 kg

Formula used :  

\Delta T_f=K_f\times m\\\\\Delta T_f=K_f\times\frac{\text{Mass of unknown compound}}{\text{Molar mass of unknown compound}\times \text{Mass of benzene in Kg}}

where,

\Delta T_f = change in freezing point  = 5.53-1.37=4.16^oC

\Delta T_s = freezing point of solution

\Delta T^o = freezing point of benzene

Molal-freezing-point-depression constant (K_f) for benzene = 5.12^oC/m

m = molality

Now put all the given values in this formula, we get

4.16^oC=(5.12^oC/m)\times \frac{6.45g}{\text{Molar mass of unknown compound}\times 0.04395kg}

\text{Molar mass of unknown compound}=180.6g/mol

If percentage are given then we are taking total mass is 100 grams.

So, the mass of each element is equal to the percentage given.

Mass of C = 42.9 g

Mass of H = 2.4 g

Mass of N = 16.7 g

Mass of O = 38.1 g

Molar mass of C = 12 g/mole

Molar mass of H = 1 g/mole

Molar mass of N = 14 g/mole

Molar mass of O = 16 g/mole

Step 1 : convert given masses into moles.

Moles of C = \frac{\text{ given mass of C}}{\text{ molar mass of C}}= \frac{42.9g}{12g/mole}=3.575moles

Moles of H = \frac{\text{ given mass of H}}{\text{ molar mass of H}}= \frac{2.4g}{1g/mole}=2.4moles

Moles of N = \frac{\text{ given mass of N}}{\text{ molar mass of N}}= \frac{16.7g}{14g/mole}=1.193moles

Moles of O = \frac{\text{ given mass of O}}{\text{ molar mass of O}}= \frac{38.1g}{16g/mole}=2.381moles

Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.

For C = \frac{3.575}{1.193}=2.99\approx 3

For H = \frac{2.4}{1.193}=2.01\approx 2

For N = \frac{1.193}{1.193}=1

For O = \frac{2.381}{1.193}=1.99\approx 2

The ratio of C : H : N : O = 3 : 2 : 1 : 2

The mole ratio of the element is represented by subscripts in empirical formula.

The Empirical formula = C_3H_2N_1O_2

The empirical formula weight = 3(12) + 2(1) + 1(14) + 2(16) = 84 gram/eq

Now we have to calculate the molecular formula of the compound.

Formula used :

n=\frac{\text{Molecular formula}}{\text{Empirical formula weight}}

n=\frac{180.6}{84}=2

Molecular formula = (C_3H_2N_1O_2)_n=(C_3H_2N_1O_2)_2=C_6H_4N_2O_4

Therefore, the molecular of the compound is, C_6H_4N_2O_4

3 0
3 years ago
The thermochemical equation for the reaction of carbon dioxide with sulfur dioxide is shown below. How can the reaction be descr
prisoha [69]

Answer: d. has high activation energy

Activation energy is the energy that a system requires to start a certain process. Also, it <u>is the minimum energy necessary for a given chemical reaction to occur</u>. For a reaction to occur between two molecules, they must collide in the correct orientation and have a minimum amount of energy equal to the activation energy.

As the molecules approach, their electron clouds repel, so energy is required for the collision to occur and therefore the reaction. The activation energy comes from the heat of the system, that is, the translational, vibrational energy, etc. of each molecule. However, if this energy is not enough, the reaction will not be spontaneous.

<u>A reaction between two molecules can be favored by supplying energy to the system.</u> In the case raised in the question, <u>energy equal to 1104 kJ is provided to the system to favor the next reaction </u>

CO2 (g) + 2SO2 (g) → CS2 (g) + 3O2 (g)

<u>Since the energy equal to 1104 kJ is included in the reactants, it can be deduced that it is the energy that is provided to the system for the reaction to occur. </u>However, from the value of this energy it can not be said whether the system is endothermic or exothermic since it is a kinetic variable and the variables of this type do not allow predicting the thermodynamic behavior of a system.

Furthermore, it can be seen that the value of this energy is considerably high, therefore the reaction described has a high activation energy.

3 0
2 years ago
Select all the signs of a CHEMICAL change from the list below. (You may need to select more than one for full points)
wlad13 [49]

The answers are B, C, and D.

4 0
3 years ago
Read 2 more answers
I am having trouble predicting this equation. "Coke will reduce hot arsenious oxide to pure arsenic (As4). With the 4 as the sub
Rudik [331]
Well, Coke is just carbon in chemical reaction, so I think you get: 3 C + 2 As2O3--> 4 As + 3 CO2. Hope this helps! :)
7 0
3 years ago
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