All categories

3 years ago

If 75.0 g of a liquid has a volume of 62.4 mL ,calculate the liquid’s density

1 answer:

4 0

<span>To calculate the density of a liquid, you have to first know that density is the amount of substance per unit of volume. In this specific question, density will be found with units of g/mL. Now, the density can be found by dividing the amount of liquid, 75.0g, by the volume, 62.4mL. Doing this we get: 75.0g/62.4mL= 1.2 g/mL as the density of the liquid.</span>

You might be interested in

Show that the Joule-Thompson Coefficient is zero for ideal gas.

melomori [17]

Answer:

Joule-Thomson coefficient for an ideal gas:

$\mu_{J.T} = 0$

Explanation:

Joule-Thomson coefficient can be defined as change of temperature with respect to pressure at constant enthalpy.

Thus,

$\mu_{J.T} = \left [\frac{\partial T}{\partial P} \right ]_H$

Also,

$H= H (T,P)$

$Differentiating\ it,$

$dH= \left [\frac{\partial H}{\partial T}\right ]_P dT + \left [\frac{\partial H}{\partial P}\right ]_T dT$

Also, $C_p$ is defined as:

$C_p = \left [\frac{\partial H}{\partial T}\right ]_P$

$So,$

$dH= C_p dT + \left [\frac{\partial H}{\partial P}\right ]_T dT$

Acoording to defination, the ethalpy is constant which means $dH = 0$

$So,$

$\left [\frac{\partial H}{\partial P}\right ]_T = -C_p\times \left [\frac{\partial T}{\partial P}\right ]_H$

$Also,$

$\mu_{J.T} = \left [\frac{\partial T}{\partial P}\right ]_H$

$So,$

$\left [\frac{\partial H}{\partial P}\right ]_T =-\mu_{J.T}\times C_p$

For an ideal gas,

$\left [\frac{\partial H}{\partial P}\right ]_T = 0$

So,

$0 =-\mu_{J.T}\times C_p$

Thus, $C_p$ ≠0. So,

$\mu_{J.T} = 0$

6 0

3 years ago

If your client receives a letter from SilverScript requesting proof of Creditable Coverage, and the client has already submitted

KatRina [158]

If your client receives a letter from SilverScript requesting proof of Creditable Coverage, they can ignore the letter is a false statement

<h3>What is Creditable coverage?</h3>

Creditable coverage is known to be a form of an health insurance, that consist of prescription drug and also other kinds of health benefit plan that is known to meets the minimum amount of qualifications.

Note that the types of creditable coverage plans are group and individual health plans, student health plans, and others.

It is not right to ignore any letter sent to you concerning the case above as it may be vital and as such, the statement of If your client receives a letter from SilverScript requesting proof of Creditable Coverage, and the client has already submitted proof of creditable coverage with the enrollment application, they can ignore the letter is a wrong/false statement.

Learn more about Creditable Coverage from

brainly.com/question/14547867

#SPJ1

8 0

2 years ago

To identify a diatomic gas (X2), a researcher carried out the following experiment: She weighed an empty 3.6-L bulb, then filled

aleksandrvk [35]

Answer:

The gas was N₂

Explanation:

V = 3.6L

P = 2.0 atm

T = 24.0°C = 297K

R = 0.0821 L.atm/K.mol

m = 8.3g

M = molar mass = ?

Using ideal gas equation;

PV = nRT

n = no. Of moles = mass / molar mass

n = m/M

PV = m/M * RT

M = mRT / PV

M = (8.3*0.0821*297) / (2.0*3.6)

M = 28.10

Since X is a diatomic molecule

M = 28.10 / 2 = 14.05 g/mol

M = Nitrogen

X = N₂

5 0

2 years ago

I WILL GIVE 50 POINTS AND BRAINLIEST TO WHO EVER ANSWERS THIS RIGHT AND FAST!!! PLS!!!!

poizon [28]

The answer is D

I took the test and I got this answer, it was correct :)

I don’t care if I get brainliest I just want to help

4 0

3 years ago

Read 2 more answers

What is a gift offered to a god by killing or burning something? Plz answer I’ll give brainlist and 50 points for free

umka2103 [35]

Answer:

it is a sacrafice to god so the spirtit can go free

Explanation:

4 0

3 years ago

Read 2 more answers