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jonny [76]
2 years ago
11

Question 6

Chemistry
1 answer:
hodyreva [135]2 years ago
4 0

Answer:

0.00688 moles

Explanation:

The molarity ratio looks like this:

Molarity = moles / volume (L)

After you convert mL to L, you can plug the values into the equation and simplify to find moles.

27.5 mL / 1,000 = 0.0275 L

Molarity = moles / volume                         <----- Molarity ratio

0.250 M = moles / 0.0275 L                      <----- Insert values

0.00688 = moles                                       <----- Multiply both sides by 0.0275

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At room temperature, one of the molecules is a gas and one is a liquid. Identify which structure is a gas and which is a liquid.
scoundrel [369]
It starts at the bottom and then it goes up into the air as evaporation and that’s when the air gets cooler so gas is cooler and the liquid would be hotter
4 0
3 years ago
Mercury has a density of <br> 13.55 g/mL .What would be the volume of a 122 g sample?
Salsk061 [2.6K]

Answer:

V = 9.0037

Explanation:

V = M/D

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2 years ago
How many grams of P4O10 would be produced when 0.700 mole of phosphorus is burned?
AysviL [449]
We can solve the equation and show the solution below:

Oxygen atomic number is 16.
Phosphorus atomic number is 32.

We have the molecular weight:
Molecular weight = (31*4) + (16*10)
Molecular weight = 284 grams/mol

Solving for the grams:

0.4 mole (for P4) * (1 mol P4O10/1 mol P4) * (284 grams P4O10/1 mole P4O10)
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The answer is 113.6 grams.
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Porque a las vacunas los cientificos las llaman reactivos.​
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5 0
2 years ago
A 14.4-gg sample of granite initially at 86.0 ∘C∘C is immersed into 24.0 gg of water initially at 25.0 ∘C∘C. What is the final t
kari74 [83]

Answer:

The final temperature of both substances when they reach thermal equilibrium is 31.2 °C

Explanation:

Step 1: Data given

Mass of sample granite = 14.4 grams

Initial temperature = 86.0 °C

Mass of water = 24.0 grams

The initial temperature of water = 25.0 °C

The specific heat of water = 4.18 J/g°C

The specific heat of granite = 0.790 J/g°C

Step 2: Calculate the final temperature

Heat lost = heat gained

Qgranite = - Qwater

Q = m*c*ΔT

m(granite)*c(granite)*ΔT(granite) = -m(water)*c(water)*ΔT(water)

⇒with m(granite) = the mass of granite = 14.4 grams

⇒with c(granite) = The specific heat of granite = 0.790 J/g°C

⇒with ΔT⇒(granite) = the change of temperature of granite = T2 - T1 = T2 - 86.0 °C

⇒with m(water) = the mass of water = 24.0 grams

⇒with c(water) = The specific heat of water = 4.18 J/g°C

⇒with ΔT(water) = the change of temperature of granite = T2 - T1 = T2 -25.0°C

14.4 grams * 0.790 * (T2 - 86.0°C) = -24.0 *4.18 * (T2 - 25.0°C)

11.376T2 - 978.336 = -100.32T2 + 2508

111.696 T2 = 3486.336

T2 = 31.2 °C

The final temperature of both substances when they reach thermal equilibrium is 31.2 °C

4 0
3 years ago
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