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3 years ago

Table salt, NaCl

Chemistry

2 answers:

elena-14-01-66 [18.8K]3 years ago

5 0

NaCl = Sodium(Na) Chlorine(Cl)

Answer = Chlorine,Sodium (Alphabetically)

Explanation:

Sodium chloride, also identified as salt, is an ionic compound with the chemical formula NaCl, drawing a 1:1 ratio of sodium and chloride ions. Sodium chloride also has numerous different uses for seasoning and storing foods. Hospitals use an intravenous sodium chloride solution. Sodium Chloride Injection 0.9% can also be practiced as a dry irrigation solution.

Elis [28]3 years ago

3 0

NaCl = Sodium(Na) Chlorine(Cl)

Answer = Chlorine,Sodium (Alphabetically)

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The blood cells that directly attack germs are: *

irga5000 [103]

Answer:

lymphocytes

Explanation:

it consists of the b and the t cells

4 0

3 years ago

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Ammonia and oxygen react to form nitrogen monoxide and water. Construct your own balanced equation to determine the amount of NO

stiv31 [10]

Answer:

NO would form 65.7 g.

H₂O would form 59.13 g.

Explanation:

Given data:

Moles of NH₃ = 2.19

Moles of O₂ = 4.93

Mass of NO produced = ?

Mass of produced H₂O = ?

Solution:

First of all we will write the balance chemical equation,

4NH₃ + 5O₂ → 4NO + 6H₂O

Now we will compare the moles of NO and H₂O with ammonia from balanced chemical equation:

NH₃ : NO NH₃ : H₂O

4 : 4 4 : 6

2.19 : 2.19 2.19 : 6/4 × 2.19 = 3.285 mol

Now we will compare the moles of NO and H₂O with oxygen from balanced chemical equation:

O₂ : NO O₂ : H₂O

5 : 4 5 : 6

4.93 : 4/5×4.93 = 3.944 mol 4.93 : 6/5 × 4.93 = 5.916 mol

we can see that moles of water and nitrogen monoxide produced from the ammonia are less, so ammonia will be limiting reactant and will limit the product yield.

Mass of water = number of moles × molar mass

Mass of water = 3.285 mol × 18 g/mol

Mass of water = 59.13 g

Mass of nitrogen monoxide = number of moles × molar mass

Mass of nitrogen monoxide = 2.19 mol × 30 g/mol

Mass of nitrogen monoxide = 65.7 g

4 0

3 years ago

A compound consisting of B, N, and H undergoes elemental analysis. The % composition by mass is found to be 40.28% B, 52.20% N,

swat32

Empirical formula is the simplest ratio of components making up a compound.

The percentage composition of each element has been given

therefore the mass present of each element in 100 g of compound is

B N H

mass 40.28 g 52.20 g 7.53 g

number of moles

40.28 g / 11 g/mol 52.20 g / 14 g/mol 7.53 g / 1 g/mol

= 3.662 mol = 3.729 mol = 7.53 mol

divide the number of moles by the least number of moles, that is 3.662

3.662 / 3.662 3.729 / 3.662 7.53 / 3.662

= 1.000 = 1.018 = 2.056

the ratio of the elements after rounding off to the nearest whole number is

B : N : H = 1 : 1 : 2

therefore empirical formula for the compound is B₁N₁H₂

that can be written as BNH₂

3 0

3 years ago

Lead forms two compounds with oxygen. One contains 2.98g of lead and 0.461g of oxygen. The other contains 9.89g of lead and 0.76

aliina [53]

The given question is incomplete, the complete question is:

Lead forms two compounds with oxygen. One contains 2.98g of lead and 0.461g of oxygen. The other contains 9.89g of lead and 0.763g of oxygen. For a given mass of oxygen, what is the lowest whole-number mass ratio of lead in the two compounds that combines with a given mass of oxygen?

Answer:

The lowest whole-number mass ratio in the two compounds is 1:2.

Explanation:

There is a need to find the mole ratio between lead and oxygen atoms in order to find the whole-number mass ratio of lead in the two compounds. In the first compound, the given mass of lead is 2.98 grams, the molar mass of lead is 207.2 gram per mole.

The no. of moles can be determined by using the formula,

moles = mass/molecular mass

moles = 2.98 g/207.2 g/mol

= 0.0144 moles

The mass of oxygen in the compound I is 0.461 grams, the molecular mass of oxygen is 16 gram per mol.

moles = 0.461 g /16 g/mol

= 0.0288 moles

The ratio between the lead and oxygen in the compound I is 0.0144/0.0288 = 1:2

On the other hand, in the compound II, the mass of lead given is 9.89 grams, therefore, the moles of lead in compound II is,

moles = 9.89 g / 207.2 g/mol

= 0.0477 moles

The mass of oxygen given in compound II is 0.763 grams, the moles of oxygen present in the compound II is,

moles = 0.763 g / 16 g

= 0.0477 moles

The ratio between the lead and oxygen in the compound II is, 0.0477 moles lead /0.0477 moles oxygen = 1:1

Hence, of the two compounds, the lowest ratio is found in the compound I, that is, 1:2.

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3 years ago

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pav-90 [236]

<span>Thin, flattened, and typically curved bones, such as the ribs or sternum, form falt bones.

The two main functions of flat bones are protection or provide a large surface for musular attachment.

These are the flat bones of the human body: in the skull, parietal, frontal, occipial, lacrimal, nasal and vomer; scapula; sternum; rib and hip.

</span>

4 0

3 years ago

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