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denis23 [38]
3 years ago
12

Water has a density of 1.0 g/cm3. Glycerin, a liquid used in some soaps, has a density of 1.26 g/cm3. If an object sinks in wate

r but floats at a constant depth in glycerin, its density is _____.
A. between 1.0 g/cm3 and 1.26 cm3
B.exactly 1.26 g/cm3
C.greater than 1.26 cm3
D.less than 1.0 g/cm3
Chemistry
2 answers:
sdas [7]3 years ago
5 0
For an object to sink in something, it's density has to be higher than the object it is in, so if it sinks in water the number has be higher than 1

And to float, and objects density has to be lower than that of the substance it is put in, so it has to be lower than 1.26

So between 1 and 1.26
emmainna [20.7K]3 years ago
5 0

Hello! Since the object will sink in the water that has a density as 1.0, then it obviously cannot be D. If it floats at the Glycerin level at 1.26, then it cannot be C. An object or a substance can float on the exact level of density as water or liquids. The answer is B. Exactly 1.26 g/cm3. I hope I helped! Please mark me brainliest if you find suitable! Have a fantastic day! :)

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Yuri [45]

Answer:

B=5522.33kg/h

C=478.11kg/h

Explanation:

Hi! It's a mass balance. First we have to determine the inflow.

mass flow rate = 36000kg / 6h = 6000kg / h

We define the input variable

- input flow (A) = 6000kg / h

-XgA = percentage of fat in A = 0.04

We define output variables.

- skim milk (B)

-creme (C)

-XgB = fat percentage at B = 0.0045

-XgC = percentage of fat in C = 0.45

Then we can start with the balance.

As a general rule, the mass balance is:

Input = Output

Balance sheet

1) A = B + C

Fat balance

2) A * XgA = B * XgB + C * XgC

Now we can solve.

We replace and clear B in equation 2

6000kg / h * 0.04 = B * 0.0045 + C * 0.45

B = (240kg / h) /0.045-C*0.45/0.0045

3) B = 53333.33kg / h-C * 100

We replace equation 3 in 1 and clear C

A = B + C

6000kg / h = 53333.33kg / h-C * 100 + C

C=(6000kg/h-53333.33kg/h)/(-99)

C=478.11kg/h

We replace C in equation 3 and calculate B

B = 53333.33kg / h-478.11kg/h * 100

B=5522.33kg/h

Then we have the values ​​of the outflows.

C=478.11kg/h

B=5522.33kg/h

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3 years ago
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So the ideal gas law is pv=nrt
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(c) O2 gas is transferred from a 3 L vessel containing oxygen at 4 atm to an evacuated 20 L vessel at a constant temperature of
Fudgin [204]

Answer:

After the transfer the pressure inside the 20 L vessel is 0.6 atm.

Explanation:

Considering O2 as an ideal gas, it is at an initial state (1) with V1 = 3L and P1 = 4 atm. And a final state (2) with V2 = 20L. The temperature remain constant at all the process, thus here applies the Boyle-Mariotte law. This law establishes that at a constant temperature an ideal gas the relationship between pressure and volume remain constant at all time:

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