Consider the following comparison.

What do you think will happen to the Kool-Aid drink after it is boiled? CER(Claim,CITE evidence and reasoning) response THANK U

mafiozo [28]

Answer:

It is just sugar decomposing to a dark tar looking goo.

Explanation:

Cite: What happens after you boil kool-aid

4 0

3 years ago

Will GIVE BRAINLIEST --A student makes a standard solution of potassium hydroxide by adding 14.555 g to 500.0 mL of water. Answe

leva [86]

Answer:

0.5188 M or 0.5188 mol/L

Explanation:

Concentration is calculated as molarity, which is the number of moles per litre.

***Molarity is represented by either "M" or "c" depending on your teacher. I will use "c".

The formula for molarity is:

$c = \frac{n}{V}$

n = moles (unit mol)

V = volume (unit L)

Find the molar mass (M) of potassium hydroxide.

$M_{KOH} = \frac{39.098 g}{mol}+\frac{16.000 g}{mol}+\frac{1.008 g}{mol}$

$M_{KOH} = 56.106 \frac{g}{mol}$

Calculate the moles of potassium hydroxide.

$n_{KOH} = \frac{14.555 g}{1}*\frac{1mol}{56.106g}$

$n_{KOH} = 0.25941(9)mol$

Carry one insignificant figure (shown in brackets).

Convert the volume of water to litres.

$V = \frac{500.0mL}{1}*\frac{1L}{1000mL}$

$V = 0.5000L$

Here, carrying an insignificant figure doesn't change the value.

Calculate the concentration.

$c = \frac{n}{V}$

$c = \frac{0.25941(9)mol}{0.5000 L}$

$c = 0.5188(3) \frac{mol}{L}$ <= Keep an insignificant figure for rounding

$c = 0.5188 \frac{mol}{L}$ <= Rounded up

$c = 0.5188M$ <= You use the unit "M" instead of "mol/L"

The concentration of this standard solution is 0.5188 M.

7 0

3 years ago

PLEASE HELP!!! CHEMISTRY

Lubov Fominskaja [6]

Hydrogen bonds are typically stronger than Van der Waals forces bc they are based on permanent dipoles, that form when hydrogen comes in vicinity of a highly electronegative atom (like F, N, or O). These bonds are long-lasting and pretty strong.

3 0

3 years ago

The temperature of a sample of water changes from 10°C to 20°C when the water absorbs 100 calories of heat. What is the mass of

Vlad1618 [11]

Answer:

10 g

Explanation:

Right from the start, just by inspecting the values given, you can say that the answer will be

10 g

.

Now, here's what that is the case.

As you know, a substance's specific heat tells you how much heat is needed to increase the temperature of

1 g

of that substance by

1

∘

C

.

Water has a specific heat of approximately

4.18

J

g

∘

C

. This tells you that in order to increase the temperature of

1 g

of water by

1

∘

C

, you need to provide

4.18 J

of heat.

Now, how much heat would be required to increase the temperature of

1 g

of water by

10

∘

C

?

Well, you'd need

4.18 J

to increase it by

1

∘

C

, another

4.18 J

to increase it by another

1

∘

C

, and so on. This means that you'd need

4.18 J

×

10

=

41.8 J

to increase the temperature of

1 g

of water by

10

∘

C

.

Now look at the value given to you. If you need

41.8 J

to increase the temperature of

1 g

of water by

10

∘

C

, what mass of water would require

10

times as much heat to increase its temperature by

10

∘

C

?

1 g

×

10

=

10 g

And that's your answer.

Mathematically, you can calculate this by using the equation

q

=

m

⋅

c

⋅

Δ

T

, where

q

- heat absorbed/lost

m

- the mass of the sample

c

- the specific heat of the substance

Δ

T

- the change in temperature, defined as final temperature minus initial temperature

Plug in your values to get

418

J

=

m

⋅

4.18

J

g

∘

C

⋅

(

20

−

10

)

∘

C

m

=

418

4.18

⋅

10

=

10 g

5 0

3 years ago

A rigid container is filled with chlorine gas. The gas has a pressure of 2.75 bar. The tank is then cooled down to -20.0oC at wh

Gnom [1K]

Answer:

Original temperature (T1) = - 37.16°C

Explanation:

Given:

Gas pressure (P1) = 2.75 bar

Temperature (T2) = - 20°C

Gas pressure (P2) = 1.48 bar

Find:

Original temperature (T1)

Computation:

Using Gay-Lussac's Law

⇒ P1 / T1 = P2 / T2

⇒ 2.75 / T1 = 1.48 / (-20)

⇒ T1 = (2.75)(-20) / 1.48

⇒ T1 = -55 / 1.48

⇒ T1 = - 37.16°C

Original temperature (T1) = - 37.16°C

3 0

3 years ago