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TEA [102]
2 years ago
8

What is the solution and solvent in Kool-Aid?​

Chemistry
2 answers:
alexandr402 [8]2 years ago
5 0

In this solution the solvent is water and the solutes are sugar, artificial flavor and artificial color. Another interesting property of solutions is that different concentrations of solute can be made. As all of you are aware, you can make very sweet Kool Aid and less sweet Kool Aid.

Andrej [43]2 years ago
4 0

Answer:

<em>In this solution the solvent is water and the solutes are sugar, artificial flavor and artificial color. Another interesting property of solutions is that different concentrations of solute can be made. As all of you are aware, you can make very sweet Kool Aid and less sweet Kool Aid.</em>

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Enter your answer in the provided box.
Lina20 [59]

Answer:

5.06atm

Explanation:

Using the combined gas law equation;

P1V1/T1 = P2V2/T2

Where;

P1 = initial pressure (atm)

P2 = final pressure (atm)

V1 = initial volume (Litres)

V2 = final volume (Litres)

T1 = initial temperature (K)

T2 = final temperature (K)

According to the information provided in this question;

P1 = 1.34 atm

P2 = ?

V1 = 5.48 L

V2 = 1.32 L

T1 = 61 °C = 61 + 273 = 334K

T2 = 31 °C = 31 + 273 = 304K

Using P1V1/T1 = P2V2/T2

1.34 × 5.48/334 = P2 × 1.32/304

7.34/334 = 1.32P2/304

Cross multiply

334 × 1.32P2 = 304 × 7.34

440.88P2 = 2231.36

P2 = 2231.36/440.88

P2 = 5.06

The final pressure is 5.06atm

6 0
3 years ago
Which element(s) is being reduced in the redox reaction below? 6 nh4clo4 (s) + 10 al(s) 5 al2o3 (g) + 6 hcl (g) + 3 n2 (g) + 9 h
vitfil [10]
The answer is Cl because its oxidation number goes from a +7 down to a -1.
4 0
3 years ago
How many moles of silane gas (SiH4) are present in 8.68 mL<br> measured at 18 0C and 1.50 atm?
makvit [3.9K]

Answer:

5.45*10⁻⁴ moles of silane gas (SiH₄) are present in 8.68 mL  measured at 18°C and 1.50 atm.

Explanation:

An ideal gas is a theoretical gas that is considered to be composed of point particles that move randomly and do not interact with each other. Gases in general are ideal when they are at high temperatures and low pressures.

An ideal gas is characterized by three state variables: absolute pressure (P), volume (V), and absolute temperature (T). The relationship between them constitutes the ideal gas law, an equation that relates the three variables if the amount of substance, number of moles n, remains constant and where R is the molar constant of the gases:

P * V = n * R * T

In this case:

  • P= 1.5 atm
  • V= 8.68 mL= 0.00868 L (being 1000 mL= 1 L)
  • n= ?
  • R= 0.082 \frac{atm*L}{mol*K}
  • T= 18 C= 291 K (being 0 C= 273 K)

Replacing:

1.5 atm* 0.00868 L= n* 0.082\frac{atm*L}{mol*K} *291 K

Solving:

n=\frac{1.5 atm*0.00868 L}{0.082 \frac{atm*L}{mol*K}*291 K}

n= 5.45*10⁻⁴ moles

<u><em>5.45*10⁻⁴ moles of silane gas (SiH₄) are present in 8.68 mL  measured at 18°C and 1.50 atm.</em></u>

8 0
2 years ago
Help plz I ONLY GOT 2 mins left
Hoochie [10]
I need more information... what was the experiment??
5 0
3 years ago
Read 2 more answers
The density of air under ordinary conditions at 25°C is 1.19 g/L. How many kilograms of air are in a room that measures 10.0 ft
Lunna [17]

Answer: The mass of air present in the room is 37.068 kg

Explanation :  Given,

Length of the room = 10.0 ft

Breadth of the room = 11.0 ft

Height of the room = 10.0 ft

To calculate the volume of the room by using the formula of volume of cuboid, we use the equation:

V=lbh

where,

V = volume of the room

l = length of the room

b = breadth of the room

h = height of of the room

Putting values in above equation, we get:

V=10.0ft\times 11.0ft\times 10.0ft=1100ft^3=31148.53L

Conversion used : 1ft^3=28.3168L

Now we have to calculate the mass of air in the room.

Density=\frac{Mass}{Volume}

1.19g/L=\frac{Mass}{31148.53L}

Mass=37066.7507g=37.068kg

Conversion used : (1 kg = 1000 g)

Therefore, the mass of air present in the room is 37.068 kg

3 0
3 years ago
Read 2 more answers
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