All categories

3 years ago

What is the percentage composition when 10g of magnesium combines with 4g of nitrogen?

1 answer:

7 0

The % composition when 10g of magnesium combine with 4g of nitrogen is 71.43% magnesium and 28.57 % nitrogen

calculation

% composition = mass of an element / total mass x100
mass of magnesium = 10 g
mass of nitrogen = 4g

calculate the total mass used

= 10g of Magnesium + 4 g of nitrogen = 14 grams

% composition for magnesium is therefore = 10/14 x100 = 71.43 %

% composition for nitrogen is therefore = 4 /14 x100 = 28.57 %

You might be interested in

Be sure to answer all parts. Determine the electron-group arrangement, molecular shape, and ideal bond angle for the following m

choli [55]

Answer:

trigonal planar

Explanation:

The molecule SO3 is of the type AX3. The molecule is symmetrical and non polar.

There are three regions of electron density in the molecule. This corresponds to a trigonal planar geometry. This means that the three oxygen atoms are arranged at the corners of a triangle. The bond angle is 120 degrees.

7 0

3 years ago

The structural form of the element Ge closely resembles the structure of A. C (diamond) B. N (diatomic) C. As (tetrahedral) D. S

Ksivusya [100]

The structural form of the element Ge closely resembles the structure of C (diamond).

Diamond is composed of a lattice structure in which atoms of carbon are held together in a face centered cubic lattice.

Germanium, an element in the same group as carbon also forms a face centered cubic lattice that is very similar to that of diamond.

Hence, the structural form of the element Ge closely resembles the structure of C (diamond).

Learn more: brainly.com/question/14578576

6 0

3 years ago

How many moles of silver can be produced from silver nitrate from 1 mole of zinc?

jasenka [17]

Answer:

Answer: 6.5 moles of silver metal is formed in the given chemical reaction. The moles of excess reagent left are 0.55 moles.

Explanation:

To calculate the moles of silver formed and the moles of excess reagent left after the reaction, we need to balance the equation first and need to find the limiting and excess reagent.

The balanced chemical equation is:

Zn + 2AgNO3 ---> Zn (NO3)2 +2Ag

By Stoichiometry:

2 moles of Silver nitrate reacts with 1 mole of Zinc metal

So, 6.5 moles of silver nitrate will react with = 1/2 x 6.5 = 3.25 moles of zinc metal

The required amount of zinc metal is less than the given amount of zinc metal, hence, it is considered as an excess reagent.

Therefore, silver nitrate is the limiting reagent because it limits the formation of product.

By Stoichiometry of the reaction:

2 moles of silver nitrate produces 2 moles of silver metal

So, 6.5 moles of silver nitrate will produce = 2/2 x 6.5 = 6.5moles of silver metal.

Number of moles of excess reagent left after the completion of reaction = (3.8 - 3.25)moles = 0.55 moles

Hence, 6.5 moles of silver metal is formed in the given chemical reaction. The moles of excess reagent left are 0.55 moles.

4 0

2 years ago

4 On a bright, sunny day, which time of day can

Marina CMI [18]

It has to be J. 12:00am cause at night that’s when everything

5 0

3 years ago

A student has a 2.19 L bottle that contains a mixture of O 2 , N 2 , and CO 2 with a total pressure of 5.57 bar at 298 K . She k

Sergeeva-Olga [200]

<u>Answer:</u> The partial pressure of oxygen gas is 2.76 bar

<u>Explanation:</u>

To calculate the number of moles, we use the equation given by ideal gas which follows:

$PV=nRT$

where,

P = pressure of the gas = 5.57 bar

V = Volume of the gas = 2.19 L

T = Temperature of the gas = 298 K

R = Gas constant = $0.0831\text{ L bar }mol^{-1}K^{-1}$

n = Total number of moles = ?

Putting values in above equation, we get:

$5.57bar\times 2.19L=n\times 0.0831\text{ L. bar }mol^{-1}K^{-1}\times 298K\\\\n=\frac{5.57\times 2.19}{0.0831\times 298}=0.493mol$

To calculate the mole fraction of carbon dioxide, we use the equation given by Raoult's law, which is:

$p_{A}=p_T\times \chi_{A}$ ........(1)

where,

$p_A$ = partial pressure of carbon dioxide = 0.318 bar

$p_T$ = total pressure = 5.57 bar

$\chi_A$ = mole fraction of carbon dioxide = ?

Putting values in above equation, we get:

$0.318bar=5.57bar\times \chi_{CO_2}\\\\\chi_{CO_2}=\frac{0.381}{5.57}=0.0571$

Mole fraction of a substance is given by:

$\chi_A=\frac{n_A}{n_A+n_B}$

We are given:

Moles of nitrogen gas = 0.221 moles

Mole fraction of nitrogen gas, $\chi_{N_2}=\frac{0.221}{0.493}=0.448$

Calculating the partial pressure of oxygen gas by using equation 1, we get:

Mole fraction of oxygen gas = (1 - 0.0571 - 0.448) = 0.4949

Total pressure of the system = 5.57 bar

Putting values in equation 1, we get:

$p_{O_2}=5.57bar\times 0.4949\\\\p_{O_2}=2.76bar$

Hence, the partial pressure of oxygen gas is 2.76 bar

6 0

4 years ago