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sammy [17]
2 years ago
11

Is MgCl₂ an empirical or a molecular formula for magnesium chloride? Explain.

Chemistry
1 answer:
Masteriza [31]2 years ago
4 0

MgCl₂ is an empirical formula for magnesium chloride.

<h3>What is empirical formula?</h3>

The empirical formula of a chemical compound in chemistry is the simplest whole number ratio of atoms in a compound. Two simple instances of this concept are the empirical formulas of sulfur monoxide (SO) and disulfur dioxide (S2O2). As a result, the empirical formulas for the sulfur and oxygen compounds sulfur monoxide and disulfur dioxide are the same. However, they do not share the same molecular formulas, which indicate the number of atoms in each molecule of a chemical compound.

An empirical formula does not mention the arrangement or number of atoms. It is typical for many macromolecules, including ionic compounds like calcium chloride and silicon dioxide (SiO2) (CaCl2).

To learn more about empirical formula visit:

brainly.com/question/14044066

#SPJ4

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What is the energy required to remove the electron from a hydrogen atom in the n = 11 state?
zaharov [31]

Based on the data given, the energy required to remove an electron from a hydrogen atom in the n = 11 state is -0.112 eV

<h3>What is ionization energy?</h3>

Ionization energy is the energy requiredto remove an electron from a neutral atom or a cation in its gaseous state.

To calculate the energy required to remove the electron from a hydrogen atom in the n = 11 state, the formula below is used:

  • E_n = \frac{E_0 × Z^{2}}{n^{2}}

where

E_0  \: is  \: ground  \: state  \: energy  \: of \:  hydrogen \:  atom \:  = -13.6 eV \\

  • Z = 1
  • n = 11

substituting the values:

E_n = \frac{ - 13.6 × 1^{2}}{11^{2}} \:  =  - 0.112 \: eV

Therefore, the energy required to remove an electron from a hydrogen atom in the n = 11 state is -0.112 eV

Learn more about ionization energy at: brainly.com/question/1445179

5 0
2 years ago
A chemist is making 200 l of a solution that is 62% acid. he is mixing an 80% acid solution with a 30% acid solution. how much o
Savatey [412]
Answer: 72L of 30% and 128L of 80%

You can determine the weight of the acid by multiplying the concentration with the volume. Let say v1 is the volume of 30% solution needed and v2 is the volume of 80% solution.
The weight of acid from the used solution should be equal to the product. You can get this equation
final solution= solution1 + solution2
200l * 62%= v1 * 30% + v2*80%
124L= 0.3v1 + 0.8v2
124L- 0.3v1= 0.8v2
v2=155L- 0.375v1

The total volume of both should be 200l. If you use the previous equation, you can calculate:
v1+v2=200L
v1+ (155L- 0.375v1)= 200L
0.625v1= 200L - 155L
v1= 45/ 0.625= 72L


v1+v2=200L
v2= 200L- 72L= 128L
5 0
3 years ago
How much heat is absorbed when 90.5 g of ice is heated from -11.0 °C to 145.0 °C?
Nadusha1986 [10]

Answer:

Q(total) = 283Kj

Explanation:

5 Heat Transitions …

Specific Heats => c(s) = 0.50cal/g∙⁰C,  c(l) = 1.0 cal/g∙⁰C, c(g) = 0.48 cal/g∙⁰C

Phase Transition Constants => ΔHᵪ = Heat of Fusion = 80 cal/g; ΔHᵥ = Heat of Vaporization = 540cal/g

Note => Phase change regions => no temp. change occurs when 2 phases are in contact (melting and evaporation). Only when single phase substance exists (s, l or g) does temperature change occur. See heating curve for water diagram. The increasing slopes are temperature change regions and heat flow is given by Q =mcΔT. The horizontal slopes are phase changes ( melting & evaporation) and heat flow for each of those regions is given by Q = m·ΔH. Each transition energy is calculated individually (see below) and added to obtain the total heat flow needed.

Q = mcΔT for temperature change regions of the heating curve (single phase only)

Q = m∙ΔH for phase transition regions of the heating curve (2 phases in contact)

Solid (ice) => Melting Pt  => Q(s) = mcΔT = (90.5g)(0.50cal/g∙⁰C)(11⁰C) = 478 cal

Melting (s/l) => Liquid (water) =>   Q(s/l) = m∙ΔHᵪ = (90.5g)(80cal/g) = 7240 cal

Liquid (water) => Boiling Pt => Q(l) = mcΔT = (90.5g)(1.0cal/g∙⁰C)(100⁰C) = 9050 cal

Boiling (l/g) => Gas (steam) => Q(l/g) = m∙ΔHᵥ = (90.5g)(540cal/g) = 48,870 cal

Gas (steam) => Steam @ 145⁰C => Q(g = mcΔT = (90.5g)(0.48cal/g∙⁰C)(45⁰C) = 2036 cal

Total Heat Transfer (Qᵤ) = Q(s) + Q(s/l) + Q(l) + Q(l/g) + Q(g)  

                                 = 478cal +7240cal + 9050 cal + 48,870cal + 2036cal

                                 = 67,674 cal x 4.184 j/cal = 283,148 joules = 283 Kj

4 0
4 years ago
The total mass of the atmosphere is about 5.00 x 1018 kg. How many moles each of air, O2, and CO2 are present in the atmosphere?
n200080 [17]

<u>Answer:</u> The moles of oxygen and carbon dioxide in air is 3.63\times 10^{19}mol and 7.18\times 10^{16}mol respectively

<u>Explanation:</u>

To calculate the number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}

Given mass of atmosphere = 5.00\times 10^{18}kg=5.00\times 10^{21}g

Average molar mass of atmosphere = 28.96 g/mol

Putting values in above equation, we get:

\text{Moles of atmosphere}=\frac{5.00\times 10^{21}g}{28.96g/mol}=1.73\times 10^{20}mol

We know that:

Percent of oxygen in air = 21 %

Percent of carbon dioxide in air = 0.0415 %

Moles of oxygen in air = \frac{21}{100}\times 1.73\times 10^{20}=3.63\times 10^{19}mol

Moles of carbon dioxide in air = \frac{0.0415}{100}\times 1.73\times 10^{20}=7.18\times 10^{16}mol

Hence, the moles of oxygen and carbon dioxide in air is 3.63\times 10^{19}mol and 7.18\times 10^{16}mol respectively

6 0
3 years ago
In a chromatography experiment, chlorophyll pigments are separated using paper. What is the stationary phase in this experiment?
DaniilM [7]
The answer in this case would be Paper. 
3 0
3 years ago
Read 2 more answers
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