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garik1379 [7]
2 years ago
5

Based on what you know about matter explain why the food coloring mixes at different rates

Chemistry
1 answer:
emmasim [6.3K]2 years ago
3 0

Answer:

Matter has different densities. Different food colorings have different densities. Those with a higher density will sink to the bottom and mix much easier than one with a lower density. The higher density also contains more particles making it easier to mix or give it a fuller color.

You might be interested in
Which gas law describes the relationship between volume and pressure at a constant temperature?
olasank [31]
Boyle law is a gas law stating the pressure and the volume of a gas have an inverse relationship when held at constant temperature.

5 0
3 years ago
If 5.1 l of antifreeze solution (specific gravity = 0.80) is added to 3.8 l of water to make a 8.9 l mixture, what is the specif
Maslowich
Density of the mixture = mass of the mixture / volume of the mixture

Mass of the mixture = mass of antifreeze solution + mass of water.

Mass of antifreeze solution = density of the antifreeze solution * volume

Mass of antifreeze solution = 0.8g/ml * 5.1 l * 1000 ml / l = 4,080 g

Mass of water = density of water * volume of water = 1.0 g/ml * 3.8 l * 1000 ml / l = 3,800 g

Mass of mixture = 4080 g + 3800 g= 7880 g

Volume of mixture = volume of antifreeze solution + volume of water

Volume of mixture = 5100 ml + 3800 ml = 8900 ml

Density of mixture = 7800 g / 8900 ml = 0.876 g/ml

Specific gravity of the mixture = density of the mixture / density of water = 0.876

Answer: 0.876
3 0
4 years ago
At 25 celsius, 1.00 ,ole of O2 was found to occupy a volume of 12.5 L at a pressure of 198 kPa. What value of the gas constant i
mr Goodwill [35]

Answer:

R=8.301\frac{L*kPa}{mol*K}

Explanation:

Hello,

In this case, assuming oxygen as an ideal gas, it is described by:

PV=nRT

Whereas the gas constant R is required, therefore, it is computed as shown below:

R=\frac{PV}{nT} =\frac{198kPa*12.5L}{1.00mol*(25+273.15)K} \\\\R=8.301\frac{L*kPa}{mol*K}

Best regards.

4 0
3 years ago
Can Some One Help pleasee
Dima020 [189]

Reactants are carbon dioxide and water, the products are glucose and oxygen. Don't know for sure what the yield is represented with, so, I can't help you with that.

7 0
3 years ago
A volume of 100 mL of 1.00 M HCl solution is titrated with 1.00 M NaOH solution. You added the following quantities of 1.00 M Na
s344n2d4d5 [400]

Answer:

a: before equivalence point

b: equivalence point

c: before equivalence point

d: after the eqivalence point

e: before equivalence point

f:  after the eqivalence point

Explanation:

Balanced equation of reaction:

NaOH +HCl =NaCl +H2O;

Volume of HCl is fixed and it 100ml and concentration is 1.0M

N1 and N2 normality of HCl and NaOH respectively;

V1 and V2 volume of HCl and NaOH respectively;

we have given molarity but we need normality;

Normality=molarity \times n-factor

<em>but in case of NaOH and HCl n-factor is 1 for each.</em>

hence

normality=molarity;

At equivalence point:  N_1V_1=N_2V_2

Before equivalence point : N_1V_1>N_2V_2

After the equivalence point: N_1V_1

N_1V_1=100\times1=100

case a:  5.00 mL of 1.00 M NaOH

N_2V_2=5\times1=5

N_1V_1>N_2V_2 hence it is before equivalence point

case b: 100mL of 1.00 M NaOH

N_2V_2=100\times1=100

N_1V_1=N_2V_2 hence it is equivalence point

case c:  10.0 mL of 1.00 M NaOH

N_2V_2=10\times1=10

N_1V_1>N_2V_2 hence it is before equivalence point

case d: 150 mL of 1.00 M NaOH

N_2V_2=150\times1=150

N_1V_1 hence it is after the eqivalence point

case e: 50.0 mL of 1.00 M NaOH

N_2V_2=50\times1=50

N_1V_1>N_2V_2 hence it is before equivalence point

case f: 200 mL of 1.00 M NaOH

N_2V_2=200\times1=200

N_1V_1 hence it is after the eqivalence point

7 0
3 years ago
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