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tankabanditka [31]
3 years ago
7

I need help plzzzz science homework

Chemistry
1 answer:
fiasKO [112]3 years ago
5 0
Wait, do we do both of the places or only the types of elements because i don't understand...
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Given that the boiling points of ethanol and diethyl ether are 78.1°C and 34.6°C, respectively, which should have the higher vap
anyanavicka [17]
Diethyl ether, since 20 is closer to 34.6 it should start boiling faster than the other which is no where near it
7 0
3 years ago
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When something heats up, new energy is created, and when something cools down, energy is destroyed. agree disagree
olga nikolaevna [1]
Disagree because in the law of conservation of mass, it states energy cannot be created or destroyed
4 0
3 years ago
PLEASE HELP!!
lana66690 [7]

Answer:

c

Explanation:

3 0
3 years ago
Read 2 more answers
What is the molar concentration of a 2 liter solution containing 200 grams of glucose?
NARA [144]

The molar concentration is 1.11M.

<h3>What is molar concentration?</h3>

The phrase "molar concentration" (also known as "molarity," "amount concentration," or "substance concentration") refers to the amount of a substance per unit volume of solution and is used to describe the concentration of a chemical species, specifically a solute, in a solution. The most frequent measure of molarity in chemistry is the number of moles per liter, denoted by the unit symbol mol/L or mol/dm3 in SI units. A solution with a concentration of 1 mol/L is referred to as 1 molar, or 1 M.

<h3>Given : </h3>

Volume of the solution = 2L

Mass of glucose given = 200g

Concentration of glucose= ?

<h3>Formula use: </h3>

Molarity = no. of moles of solute / volume of the solution (L)

Moles of solute = given mass of solute / molar mass of the solute

<h3>Solution: </h3>

No. of moles of solute( glucose ) = 200 / 180 = 1.11 moles'

Molarity = 1.11 / 2 = 0.5555 mol L ^(-1)

Therefore, the molar concentration of glucose in the solution = 0.555 mol L ^(-1)

To learn more about molar concentration :

brainly.com/question/15532279

#SPJ4

8 0
2 years ago
For the equilibrium
Mamont248 [21]

Answer:

Equilibrium concentrations of the gases are

H_2S=0.596M

H_2=0.004 M

S_2=0.002 M

Explanation:

We are given that  for the equilibrium

2H_2S\rightleftharpoons 2H_2(g)+S_2(g)

k_c=9.0\times 10^{-8}

Temperature, T=700^{\circ}C

Initial concentration of

H_2S=0.30M

H_2=0.30 M

S_2=0.150 M

We have to find the equilibrium concentration of gases.

After certain time

2x number of moles  of reactant reduced and form product

Concentration of

H_2S=0.30+2x

H_2=0.30-2x

S_2=0.150-x

At equilibrium

Equilibrium constant

K_c=\frac{product}{Reactant}=\frac{[H_2]^2[S_2]}{[H_2S]^2}

Substitute the values

9\times 10^{-8}=\frac{(0.30-2x)^2(0.150-x)}{(0.30+2x)^2}

9\times 10^{-8}=\frac{(0.30-2x)^2(0.150-x)}{(0.30+2x)^2}

9\times 10^{-8}=\frac{(0.30-2x)^2(0.150-x)}{(0.30+2x)^2}

By solving we get

x\approx 0.148

Now, equilibrium concentration  of gases

H_2S=0.30+2(0.148)=0.596M

H_2=0.30-2(0.148)=0.004 M

S_2=0.150-0.148=0.002 M

3 0
3 years ago
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