Anyone solve this for me

What is the volume of a sample of liquid mercury that has a mass of 76.2g, given that the density of mercury is 13.6g/mL?

nignag [31]

Hey there!

D = m / V

13.6 = 76.2 / V

V = 76.2 / 13.6

V = 5.602 mL

8 0

3 years ago

How much does a ram pickup cost asking for a friend

nika2105 [10]

Answer:alot lol

Explanation:

4 0

3 years ago

I'm stuck on this assignment, +20 PTS and BRAINLIEST for step by step:

dolphi86 [110]

Answer:

2. 2.74 L

3. 488 K

4. 7.47 L

5. 38.6 L

6. 2.85 mol

7. 319 K

8. 3.43 kPa

Explanation:

Generally speaking, you want to convert units to SI units, but in this case, we are working with ratios. This makes up for using the units that wouldn't appropriate elsewhere.

2. Use the equation P₁V₁ = P₂V₂. Solve for V₂.

(3.05 L)(870 kPa) = (969 kPa)(V₂)

V₂ = 2.74 L.

3. Use the equation V₁/T₁ = V₂/T₂. Solve for T₂.

(3.32 L)/(360 K) = (4.50 L)/(T₂)

T₂ = 488 K.

4. Do the same as above, but for V₂.

(5.10 L)/(-56°C) = V₂/(-82°C)

V₂ = 7.47 L

5. Use the equation V₁/n₁ = V₂/n₂. Solve for V₂.

(37.2 L)/(0.750 mol) = (V₂)/(0.778 mol)

V₂ = 38.6 L

6. Do the same as above, but for n₂.

(86.0 L)/(2.65 mol) = (92.5 L)(n₂)

n₂ = 2.85 mol

7. Use the equation P₁/T₁ = P₂/T₂. Solve for T₂.

(3.00 atm)/(390 K) = (2.45 atm)/(T₂)

T₂ = 319 K

8. Do the same as above, but for P₂.

In this specific case, however you will need to convert units. Since both temperatures don't have the same sign, the ratio won't come out right. Convert to Kelvin. Add 273.15 to the temperature in Celsius to convert to Kelvin -12.3°C = 260.85 K 25°C = 298.15 K.

(3.00 kPa)/(260.85 K) = P₂/(298.15 K)

P₂ = 3.43 kPa

There is a lot in here... If you are confused about something, let me know!

6 0

3 years ago

It was calculated that 4.3mL of 0.417 M HCl is required to titrate 11.9 mL of 0.151 M Mg(OH)2. Show evidence 2 HCl(aq) + Mg(OH)2

Lapatulllka [165]

Answer:

See explanation.

Explanation:

Hello,

In this case, for the described chemical reaction:

2 HCl(aq) + Mg(OH)2(aq) → MgCl2(aq) + 2 H2O(l)

We can notice there is a 2:1 molar ratio between the moles of hydrochloric acid and magnesium hydroxide, therefore, at the equivalence point:

$n_{HCl}=2*n_{Mg(OH)_2}$

And in terms of volumes and concentrations we verify:

$V_{HCl}M_{HCl}=2*V_{Mg(OH)_2}M_{Mg(OH)_2}$

So we use the given data to proof it:

$4.3mL*0.417M=2*11.9mL*0.151M\\1.793=3.594$

Therefore, we can conclude the data is wrong by means of the 2:1 mole ratio that for sure was not taken into account. This is also supported by the fact that normalities are actually the same, but the nomality of magnesium hydroxide is the half of the hydrochloric acid normality since the acid is monoprotic and the base has two hydroxyl ions.

Best regards.

4 0

3 years ago

What are the charged particles that form when atoms gain or lose electrons? *

Sunny_sXe [5.5K]

Answer:

ions

not sure but I think that's what it is

4 0

2 years ago

Anyone solve this for me​

Anyone solve this for me