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bagirrra123 [75]
3 years ago
5

What is a formula for sodium nitrate

Chemistry
1 answer:
Inessa [10]3 years ago
8 0

Answer:

NaNO3

Explanation:

sodium= Na

Nitrogen= N

and there's 3 oxygens = O3

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PLEASE HELP!! TIMED QUESTION.
BigorU [14]

Answer:

p=m/v

m=mass:402g

v=volume:385ml

p=density:?

divide 402 by 385 to get 1.04 the density is 1.04ml

Density (p)= mass (g)/volume (ml)

5 0
3 years ago
Compound Y has a distribution coefficient of 4.0 when extracted from water with chloroform, with Y being more soluble in chlorof
Aleks04 [339]

Answer:

There are needed 3 extractions to extract at least 95%

Explanation:

The distribution coefficient of a compound is defined as the ratio in concentration of the compound in the organic solvent and the concentration in the aqueous solution:

K = Concentration organic solvent / Concentration in water

Assuming as initial amount of the organic solvent: 100% and X as the amount of Y that is extracted

<em>First extraction:</em>

4 = X / 10mL / (100-X) / 50mL

4 = 50X / 1000-X

4000 - 4X = 50X

4000 = 54X

X = 74.1%

In the first extraction, 74.1% of Y is extracted

And will remain: 100 - 74.1 = 25.9%

<em>Second extraction:</em>

4 = X / 10mL / (25.9-X) / 50mL

4 = 50X / 259-X

1036 - 4X = 50X

1036 = 54X

X = 19.2%

In the second extraction, 19.2% of Y is extracted

And will remain: 25.9 - 19.2 = 6.7%

<em>Third extraction</em>

4 = X / 10mL / (6.7-X) / 50mL

4 = 50X / 67-X

268 - 4X = 50X

268 = 54X

X = 5.0%

In the first extraction, 5.0% of Y is extracted

And are extracted:

74.1% + 19.2% + 5.0% = 98.3%

That means there are needed 3 extractions to extract at least 95%

4 0
3 years ago
Is decay pool or flux
mash [69]

Answer:

A single carbon pool can often have several fluxes both adding and removing carbon simultaneously. For example, the atmosphere has inflows from decomposition (CO2 released by the breakdown of organic matter), forest fires and fossil fuel combustion and outflows from plant growth and uptake by the oceans.

Explanation:

7 0
2 years ago
How many grams of NH3 can be produced from 2.30 mol of N2 and excess H2.
MrRa [10]
<h3>Answer:</h3>

78.34 g

<h3>Explanation:</h3>

From the question we are given;

Moles of Nitrogen gas as 2.3 moles

we are required to calculate the mass of NH₃ that may be reproduced.

<h3>Step 1: Writing the balanced equation for the reaction </h3>

The Balanced equation for the reaction is;

   N₂(g) + 3H₂(g) → 2NH₃(g)

<h3>Step 2: Calculating the number of moles of NH₃</h3>

From the equation 1 mole of nitrogen gas reacts to produce 2 moles of NH₃

Therefore, the mole ratio of N₂ to NH₃ is 1 : 2

Thus, Moles of NH₃ = Moles of N₂ × 2

                                  = 2.3 moles × 2

                                  = 4.6 moles

<h3>Step 3: Calculating the mass of ammonia produced </h3>

Mass = Moles × molar mass

Molar mass of ammonia gas = 17.031 g/mol

Therefore;

Mass = 4.6 moles × 17.031 g/mol

         = 78.3426 g

         = 78.34 g

Thus, the mass of NH₃ produced is 78.34 g

3 0
3 years ago
Convert the composition of the following alloy from atom percent to weight percent (a) 44.9 at% of silver, (b) 46.3 at% of gold,
vesna_86 [32]

Answer:

Mass percentage of gold : 33.35%

Mass percentage of silver: 62.80%

Mass percentage of copper : 3.85%

Explanation:

N=n\times N_A

Where : N = Number of atoms

n = moles

N_A=6.022\times 10^{23} = Avogadro number

Let the total atoms present in the alloy be 100.

Atom percentage of silver = 44.9%

Atoms of of silver = 44.9% of 100 atoms = 44.9

Atomic weight of silver = 107.87 g/mol

Mass of 44.9 silver atoms = 107.87 g/mol ×  44.9 = 4,843.363 g/mol

Atom percentage of gold= 46.3%

Atoms of of gold = 46.3% of 100 atoms = 46.3

Atomic weight of gold = 196.97 g/mol

Mass of 44.9 gold atoms = 196.97 g/mol × 46.3 = 9,119.711 g/mol

Atom percentage of copper = 8.8 %

Atoms of of copper = 8.8% of 100 atoms = 8.8

Atomic weight of copper = 63.55 g/mol

Mass of 44.9 copper atoms = 63.55 g/mol ×  8.8 = 559.24 g/mol

Total mass of an alloy :4,843.363 g/mol + 9,119.711 g/mol +  559.24 g/mol

Mass percentage of gold :

\frac{4,843.363 g/mol}{4,843.363 g/mol + 9,119.711 g/mol +  559.24 g/mol}\times 100=33.35 \%

Mass percentage of silver:

\frac{9,119.711 g/mol}{4,843.363 g/mol + 9,119.711 g/mol +  559.24 g/mol}\times 100=62.80\%

Mass percentage of copper :

\frac{559.24 g/mol}{4,843.363 g/mol + 9,119.711 g/mol +  559.24 g/mol}\times 100=3.85\%

7 0
3 years ago
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