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ExtremeBDS [4]
3 years ago
10

How many grams of P4O10 would be produced when 0.700 mole of phosphorus is burned?

Chemistry
1 answer:
AysviL [449]3 years ago
8 0
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)
Total grams = 113.6

The answer is 113.6 grams.
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Travka [436]
3/4 cause its addition
8 0
2 years ago
A metal sample has a mass of 64.2 g. What is the density of this metal?
brilliants [131]

density= mass/volume

we need the volume of the metal to find the density,  in which case the answer would be 64.2/volume=density

8 0
3 years ago
What are groups 1, 2, and 3 examples of on the periodic table? A. Nonmetals B. Metals C. Noble gases D. Metalloids
tankabanditka [31]

The correct answer is:

Metals

They are all alkali and transition metals

Explanation:

The periodic table includes elements clustered into groups with comparable properties. Alkali metals are reactive, soft metals with low densities. Transition metals are unreactive metals that have many have common uses. Halogens are reactive non-metals that form glowing vapors.



3 0
3 years ago
Read 2 more answers
A 2 L container contains 5 moles of helium gas at 0.5 atm. What is the<br> temperature of the gas?
lara [203]

Answer:

2.44 K IS THE TEMPERATURE OF THE GAS

Explanation:

PV = nRT

P = 0.5 atm

V = 2 L

n = 5 moles

R = 0.082 L atm mol^-1 K^-1

T = ?

Substituting for T in the equation, we obtain:

T = P V / nR

T = 0.5 * 2 / 5 * 0.082

T = 1 / 0.41

T = 2.44 K

The temperature of the gas is 2.44 K

7 0
3 years ago
Question
madam [21]

Answer:

The specific heat of the metal is 2.09899 J/g℃.

Explanation:

Given,

For Metal sample,

mass = 13 grams

T = 73°C

For Water sample,

mass = 60 grams

T = 22°C.

When the metal sample and water sample are mixed,

The addition of metal increases the temperature of the water, as the metal is at higher temperature, and the  addition of water decreases the temperature of metal. Therefore, heat lost by metal is equal to the heat gained by water.

Since, heat lost by metal is equal to the heat gained by water,

Qlost = Qgain

However,

Q = (mass) (ΔT) (Cp)

(mass) (ΔT) (Cp) = (mass) (ΔT) (Cp)

After mixing both samples, their temperature changes to 27°C.

It implies that , water sample temperature changed from  22°C to 27°C and metal sample temperature changed from 73°C to 27°C.

Since, Specific heat of water = 4.184 J/g°C

Let Cp be the specific heat of the metal.

Substituting values,

(13)(73°C - 27°C)(Cp) = (60)(27°C - 22℃)(4.184)

By solving, we get Cp =

Therefore, specific heat of the metal sample is 2.09899 J/g℃.

5 0
3 years ago
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