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lana [24]
3 years ago
5

Suppose a car sitting on a hydraulic lift exerts a downward force of 1,750 N on a piston with an area of 0.6m2. How much pressur

e does the car exert on the piston?
A. 1,050 Pa

B. 2,916.7 Pa

C. 3.43 Pa
Chemistry
2 answers:
soldier1979 [14.2K]3 years ago
6 0
<span>P = F/A = 1750/0.6 = 2,916.7 Pa</span>
34kurt3 years ago
3 0

Answer:

2,916.7 Pa

Explanation:

got it from the other person lol

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What is the mass of one gram of calcium?
Anvisha [2.4K]

pls follow me

Explanation:

The mass of one atom of calcium is the same as the molar mass of the calcium. Also, the atomic mass of calcium is 40.078 g/mol.

8 0
3 years ago
What would happen if you added more than 5ml of H2O2 to the 5ml of yeast
anzhelika [568]

Hydrogen peroxide breaks down into water and oxygen.


2H₂O₂ → 2H₂O + O₂


The yeast present contains an enzyme called catalase which catalyses the reaction.


More the amount of the catalyst added, faster will be the decomposition of the hydrogen peroxide.


Thus if we added more than 5 mL of yeast solution to the 2H₂O₂, the breakdown would occur faster. Thus the bubbles and the accompanying fizz would be much more.


Read more on Brainly.com - brainly.com/question/10670134#readmore

7 0
3 years ago
Read 2 more answers
A square chunk of plastic has a length of 5 cm, width of 5 cm and height of 5 cm. It has a mass of 200 g. What is it’s density
vichka [17]

Answer:

1.6g/mL

Explanation:

Density equation is D=m/v

Density = g/mL

m=mass of sample in grams

v = volume of sample in mL

The volume of a square can be calculated by V=l*w*h.

In this case it is 5cm*5cm*5cm = 125cm^3

Since we know that 1cm^3 ~ 1mL we can convert the volume to mL as so:

125cm^3 (1mL/(1cm^3)) = 125mL

Then simply plug into the density equation:

D=200g/125mL = 1.6g/mL

3 0
3 years ago
What is the molarity of a solution in which 0.732 moles of hcl are dissolved in 0.975 liters of solution?
posledela
Molarity is moles divided by liters so do .732 divided by .975 liters.
6 0
3 years ago
quizlet a basement is successfully sealed so that more radon may not enter the space, but 5.7×107 radon atoms are already trappe
12345 [234]

After 25 days, it remains radon 5.9x10^5 atoms.

Half-life is the time required for a quantity (in this example number of radioactive radon) to reduce to half its initial value.

N(Ra) = 5.7×10^7; initial number of radon atoms

t1/2(Ra) = 3.8 days; the half-life of the radon is 3.8 days

n = 25 days / 3.8 days

n = 6.58; number of half-lifes of radon

N1(Ra) = N(Ra) x (1/2)^n

N1(Ra) = 5.7×10^7 x (1/2)^6.58

N1(Ra) = 5.9x10^5; number of radon atoms after 25 days

The half-life is independent of initial concentration (size of the sample).

More about half-life: brainly.com/question/1160651

#SPJ4

6 0
2 years ago
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