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vredina [299]
3 years ago
8

Which statement about a chemical reaction is true

Chemistry
2 answers:
Zanzabum3 years ago
8 0
The answer to the question is c
liberstina [14]3 years ago
5 0
The answer to this question will be C
You might be interested in
Will a precipitate of magnesium fluoride form when 300. mL of 1.1 × 10 –3 M MgCl 2 are added to 500. mL of 1.2 × 10 –3 M NaF? [K
Tju [1.3M]

Answer:

No precipitate is formed.

Explanation:

Hello,

In this case, given the dissociation reaction of magnesium fluoride:

MgF_2(s)\rightleftharpoons Mg^{2+}+2F^-

And the undergoing chemical reaction:

MgCl_2+2NaF\rightarrow MgF_2+2NaCl

We need to compute the yielded moles of magnesium fluoride, but first we need to identify the limiting reactant for which we compute the available moles of magnesium chloride:

n_{MgCl_2}=0.3L*1.1x10^{-3}mol/L=3.3x10^{-4}molMgCl_2

Next, the moles of magnesium chloride consumed by the sodium fluoride:

n_{MgCl_2}^{consumed}=0.5L*1.2x10^{-3}molNaF/L*\frac{1molCaCl_2}{2molNaF} =3x10^{-4}molMgCl_2

Thus, less moles are consumed by the NaF, for which the moles of formed magnesium fluoride are:

n_{MgF_2}=3x10^{-4}molMgCl_2*\frac{1molMgF_2}{1molMgCl_2}=3x10^{-4}molMgF_2

Next, since the magnesium fluoride to magnesium and fluoride ions is in a 1:1 and 1:2 molar ratio, the concentrations of such ions are:

[Mg^{2+}]=\frac{3x10^{-4}molMg^{+2}}{(0.3+0.5)L} =3.75x10^{-4}M

[F^-]=\frac{2*3x10^{-4}molMg^{+2}}{(0.3+0.5)L} =7.5x10^{-4}M

Thereby, the reaction quotient is:

Q=(3.75x10^{-4})(7.5x10^{-4})^2=2.11x10^{-10}

In such a way, since Q<Ksp we say that the ions tend to be formed, so no precipitate is formed.

Regards.

6 0
3 years ago
Alkynes have what in<br> common with alkenes ?
jonny [76]

Answer:

Alkanes, alkenes and alkynes are simple hydrocarbon chains with no functional groups. Alkanes are identified because the carbon chain has only single bonds. Alkenes have at least one double bond and alkynes have at least one triple bond. The most common alkyne is ethyne, better known as acetylene.

8 0
3 years ago
Ammonium nitrate dissociates in water according to the following equation:
Viktor [21]

Answer:

Explanation:

NH₄NO₃ = NH₄⁺ +NO₃⁻

heat released  by water = msΔ T

m is mass , s is specific heat and ΔT is fall in temperature

= 50  x 4.18 x ( 22 - 16.5 )  ( mass of 50 mL is 50 g )

= 1149.5 J .

This heat will be absorbed by the reaction above .

q for the reaction = + 1149.5 J

2 )

molecular weight of NH₄NO₃ = 80

No of moles reacted = 5/80 = 1 / 16 moles.

3 )  

5 g absorbs 1149.5 J

80 g absorbs 1149.5 x 16 J

= 18392 J

= 18.392 kJ.

= + 18.392 kJ

ΔH =  18.392 kJ / mol

6 0
3 years ago
The average score for games played in the NFL is 21.1 and the standard deviation is 8.9 points. 46 games are randomly selected.
kaheart [24]

Answer:

A) P(21.4317 < ¯x < 22.7561) = 0.2975

B) Q1 for the ¯x distribution = 21.9844

Explanation:

The Central Limit theorem allows us to say that

Sample mean = Population mean = 21.1 points

Mean of sampling distribution = σₓ = (σ/√n)

σ = population standard deviation = 8.9 points

n = sample size = 46

σₓ = (8.9/√46) = 1.3122334098 = 1.3122

A) P(21.4317 < ¯x < 22.7561) =

This is a normal distribution problem

To find this probability, we will use the normal probability tables

We first normalize/standardize 21.4317 and 22.7561.

The standardized score of any value is that value minus the mean divided by the standard deviation.

For 21.4317

z = (x - μ)/σ = (21.4317 - 21.1)/1.3122 = 0.25

For 22.7561

z = (x - μ)/σ = (22.7561 - 21.1)/1.3122 = 1.26

The required probability

P(21.4317 < ¯x < 22.7561) = P(0.25 < z < 1.26)

Checking the tables

P(21.4317 < ¯x < 22.7561) = P(0.25 < z < 1.26)

= P(z < 1.26) - P(z < 0.25)

= 0.89617 - 0.59871

= 0.29746 = 0.2975 to 4 d.p.

B) Q1 for the distribution is the first quartile. The first quartile is greater than 25% of the distribution.

P(x > Q1) = 0.25

Let the z-score that corresponds to Q1 be z'

P(x > Q1) = P(z > z') = 0.25

But P(z > z') = 1 - P(z ≤ z') = 0.25

P(z ≤ z') = 1 - 0.25 = 0.75

From the normal distribution tables,

z' = 0.674

z' = (Q1 - μ)/σ

0.674 = (Q1 - 21.1)/1.3122

Q1 = 0.674×1.3122 + 21.1 = 21.9844228 = 21.9844 to 4 d.p.

Hope this Helps!!!

7 0
3 years ago
How many electrons can be described by the following quantum numbers: n=2, 1=-1, m-1.
SIZIF [17.4K]

Answer:

6 electrons

Explanation:

The principal quantum number n; describes the energy level and here, it is 2.

The azimuthal or secondary quantum number l is given as 1, we are describing the p-orbital

The magnetic quantum number is -1, and we have 3 sub-orbitals here:

   It is a 2p-suborbital

So; in each sub orbital we have 2 electrons

The number of electrons that can be described here  =  2 x 3 = 6 electrons.

8 0
2 years ago
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