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3 years ago

14

Did u see my bag ??? which song’s lyric is thiss?

2 answers:

Sedbober [7]3 years ago

3 0

Sorry I didn't see it my guy

julsineya [31]3 years ago

3 0

Answer:

its from Mic DRop

Explanation:

b*by watch ur mouth... i forgot what it says next lol

You might be interested in

A solution that contains less than the amount of solute that would be dissolved at equilibrium is considered to be

kirill115 [55]

Answer:

dilute solution

A solution containing less solute than the equilibrium amount is called a dilute solution. The solvent has a limited capacity to dissolve a solute.

Explanation:

5 0

2 years ago

A light strikes the boundary of a medium at a 45 degree angle. Instead of entering the second medium, the light reflects back in

Dahasolnce [82]

Answer:

D It must be equal to 1.00

Explanation:

The refraction in the second medium must be equal to 1. The refraction index is given as the ration of the angle of incidence to the angle of refraction. This is given as a fraction. In other words:

$\frac{sin\theta _{i} }{sin\theta _{r} } = n$

where Θ₁ and Θ₂ are angles of incidence an refraction, and n is the refractive index.

At a critical angle, the refraction is equal to the reflection inside the medium. This results in a phenomenon called total internal reflection where light is reflected internally in the medium.

6 0

3 years ago

How many grams of KOH do we need to make 250.0ml of a 0.75 M

ZanzabumX [31]

Answer:

10.52g KOH

Explanation:

250.0 ml X 1L/1000ml X 0.75 mol KOH/1L X 56.105gKOH/1 mol KOH =10.52g KOH

5 0

3 years ago

A certain alcohol contains only three elements, carbon, hydrogen, and oxygen. Combustion of a 60.00 gram sample of the alcohol p

marin [14]

Answer:

$C_2H_6O$

Explanation:

The first step is the <u>calculation of the moles</u> of $H_2O$ and $CO_2$ , so:

$114.6~g~CO_2\frac{1~mol~CO_2}{44~g~CO_2}=2.6~mol~of~CO_2$

$70.44~g~H_2O\frac{1~mol~H_2O}{18~g~H_2O}=~3.9~mol~H_2O$

Now, in 1 mol of CO2 we have 1 mol of C and in 1 mol of $H_2O$ we have 1 mol of H. Additionally, if we want to calculate the moles of oxygen we need to <u>calculate the grams of C and O</u> and then do the <u>substraction</u> form the initial amount, so:

$2.6~mol~CO_2\frac{1~mol~C}{1~mol~CO_2}\frac{12~g~C}{1~mol~C}=31.25~g~of~C$

$3.9~mol~H_2O\frac{2~mol~H}{1~mol~H_2O}\frac{1~g~H}{1~mol~H}=7.82~g~of~H$

$Total~grams=~31.25~+~7.82=39.08~g$

$grams~of~O=60.00~g-~39.08~g=20.92~g~of~O$

Now we can <u>convert the grams</u> of O to moles, so:

$20.92~g~of~O\frac{1~mol~O}{16~g~O}=1.30~mol~O$

The next step is to divide all the mol values by the <u>smallest one</u>:

$O=\frac{1.30~mol~O}{1.30~mol~O}=~1$

$C=\frac{2.6~mol~C}{1.30~mol~O}=~2$

$H=\frac{7.82~mol~H}{1.30~mol~O}=6$

Therefore the formula is $C_2H_6O$

6 0

3 years ago

If a temperature increase from 21.0 ∘c to 35.0 ∘c triples the rate constant for a reaction, what is the value of the activation

Airida [17]

Answer:

59.077 kJ/mol.

Explanation:

From Arrhenius law: <em>K = Ae(-Ea/RT)</em>

where, K is the rate constant of the reaction.

A is the Arrhenius factor.

Ea is the activation energy.

R is the general gas constant.

T is the temperature.

At different temperatures:

<em>ln(k₂/k₁) = Ea/R [(T₂-T₁)/(T₁T₂)]</em>

k₂ = 3k₁ , Ea = ??? J/mol, R = 8.314 J/mol.K, T₁ = 294.0 K, T₂ = 308.0 K.

ln(3k₁/k₁) = (Ea / 8.314 J/mol.K) [(308.0 K - 294.0 K) / (294.0 K x 308.0 K)]

∴ ln(3) = 1.859 x 10⁻⁵ Ea

∴ Ea = ln(3) / (1.859 x 10⁻⁵) = 59.077 kJ/mol.

4 0

3 years ago