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3 years ago

What is a control group?

Chemistry

2 answers:

alexira [117]3 years ago

8 0

Answer:

The group that tests the variable.

Explanation:

The group that tests the variable is a control group.

Juli2301 [7.4K]3 years ago

3 0

Answer:

C when testing the group put the varible in control

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What is the percent yield of the purification system

Jlenok [28]

Answer: 90.3

Explanation:

3 0

3 years ago

If 0.50 mol/L of butane is added to the original equilibrium mixture and the system shifts to a new equilibrium position, what i

Ostrovityanka [42]

Consider the isomerization of butane with equilibrium constant is 2.5 .The system is originally at equilibrium with :

[butane]=1.0 M , [isobutane]=2.5 M

If 0.50 mol/L of butane is added to the original equilibrium mixture and the system shifts to a new equilibrium position, what is the equilibrium concentration of each gas?

Answer:

The equilibrium concentration of each gas:

[Butane] = 1.14 M

[isobutane] = 2.86 M

Explanation:

Butane ⇄ Isobutane

At equilibrium

1.0 M 2.5 M

After addition of 0.50 M of butane:

(1.0 + 0.50) M -

After equilibrium reestablishes:

(1.50-x)M (2.5+x)

The equilibrium expression will wriiten as:

$K_c=\frac{[Isobutane]}{[Butane]}$

$2.5=\frac{2.5+x}{(1.50-x)}$

x = 0.36 M

The equilibrium concentration of each gas:

[Butane]= (1.50-x) = 1.50 M - 0.36M = 1.14 M

[isobutane]= (2.5+x) = 2.50 M + 0.36 M = 2.86 M

3 0

3 years ago

Scientists utilize models for a variety of different purposes, but each type of scientific model has limitations. What might be

Nezavi [6.7K]

They are based in current knowledge

6 0

4 years ago

Calculate the energy, in joules, required to ionize a hydrogen atom when its electron is initially in the n =2 energy level. The

qaws [65]

Answer:

$E_{ionization}=5.45\times 10^{-19}\ J$

Explanation:

$E_n=-2.18\times 10^{-18}\times \frac{1}{n^2}\ Joules$

For transitions:

$Energy\ Difference,\ \Delta E= E_f-E_i =-2.18\times 10^{-18}(\frac{1}{n_f^2}-\frac{1}{n_i^2})\ J=2.18\times 10^{-18}(\frac{1}{n_i^2} - \dfrac{1}{n_f^2})\ J$

$\Delta E=2.18\times 10^{-18}(\frac{1}{n_i^2} - \dfrac{1}{n_f^2})\ J$

So, $n_i=2$ and $n_f=\infty$ (As the hydrogen has to ionize)

Thus,

$\Delta E=2.18\times 10^{-18}(\frac{1}{2^2} - \dfrac{1}{{\infty}^2})\ J$

$\Delta E=2.18\times 10^{-18}(\frac{1}{2^2})\ J$

$E_{ionization}=5.45\times 10^{-19}\ J$

4 0

3 years ago

Answer this ASAP please!! Question: How many magnesium atoms does magnesium oxide need? (Science) (ONLY ANSWER IF YOU KNOW WHAT

aliina [53]

Answer:

40.3 for the unit that represents the minimum amount of magnesium oxide I beilieve.

Explanation:

Magnesium atoms are heavier than oxygen atoms, so we expect more than 50% of magnesium in the weight composition. Taking just one atom of Mg and one atom of O you will get a mass of 16.0 + 24.3 = 40.3.

Hope this helps! Please tell me if you have any questions :))

7 0

3 years ago