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Harlamova29_29 [7]
3 years ago
12

What type of mixture is this salad dressing

Chemistry
2 answers:
lakkis [162]3 years ago
6 0
As there are more than two substance in this thing, (salad dressing), it would be heterogeneous.

In short, Your Answer would be "Heterogeneous"

Hope this helps!
melisa1 [442]3 years ago
5 0
I believe that would be oil and vinegar.

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what is the mass of a gold bar with the following dimensions" 30 cm X 10 cm X 10 cm if the density of gold is 19.3g/cm^3?
Lapatulllka [165]

Answer:

Volume of the bar=10 X 4 X 5=200 cm^3

Mass=200 X 19.3=3,860 grams

Mass=3.86 Kilograms

Explanation:

8 0
3 years ago
If two gases are present in a container, the total pressure in the container is equal to
qwelly [4]

Answer: sum of the pressure of the two gases present.

Explanation:

According to Dalton's law, the total pressure of a mixture of gases is the sum of individual pressures exerted by the constituent gases.

p_{total}=p_A+p_B

For example if there are there are two gases hydrogen and oxygen with individual pressure of 30 and 20 atm each. Then the total pressure in the container will be:

p_{total}=p_{H_2}+p_{O_2}

Thus p_{total|=30atm+20atm

p_{total}=50atm

Thus if two gases are present in a container, the total pressure in the container is equal to sum of the pressure of the two gases present.

5 0
3 years ago
1-butanol yields 1-bromobutane in the presence of concentrated sulfuric acid and an excess of sodium bromide. CH3CH2CH2CH2OH (l)
Zanzabum

Answer:

The answer to your question is: yield = 56.27%

Explanation:

Data

             CH3CH2CH2CH2OH (l) → CH3 CH2CH2CH2Br

                18.54 ml 1-butanol            15.65 g of 1-bromobutane

% yield = ?

density = 0.81 g/ml

MM = 74 g  1- butanol

MM = 137 g 1-bromobutane

Process

Calculate mass of 1- butanol

              density = mass/volume

              mass = density x volume

              mass = 0.81 x 18.54

              mass = 15.02 g of 1-butanol

Theoretical yield

              74 g of 1- butanol -----------------  137 g of 1-bromobutane

              15.02 g of 1- butanol -------------   x

              x = (15.02 x 137) / 74

              x = 27.81 g of 1-bromobutane

% yield = experimental yield / theoretical yield x 100

% yield = 15.65 / 27.81 x 100

% yield = 56.28

6 0
4 years ago
The pressure of a 250 mL sample of gas is 105 kPa. What would be the pressure if the volume were increased to 375 ml?
Arte-miy333 [17]

Answer:

<h2>The answer is 70 kPa</h2>

Explanation:

In order to find the pressure if the volume were increased to 375 ml we use Boyle's law

That's

P_1V_1 = P_2V_2

where

P1 is the initial pressure

P2 is the final pressure

V1 is the initial volume

V2 is the final volume

Since we are finding the final pressure

P_2 =  \frac{P_1V_1}{V_2}  \\

From the question

P1 = 105 kPa = 105 , 000 Pa

V1 = 250 mL

V2 = 375 mL

So we have

P_2 =  \frac{105000 \times 250}{375}  =  \frac{26250000}{375}  \\  = 70000

We have the final answer as

<h3>70 kPa</h3>

Hope this helps you

8 0
4 years ago
___________ carry oxygen throughout the human body.
SOVA2 [1]
Red blood cells carry oxygen throughout the human body
3 0
4 years ago
Read 2 more answers
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