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12345 [234]
3 years ago
14

How many grams are in a mole of fluorine

Chemistry
1 answer:
andre [41]3 years ago
7 0

Question: How many grams are in a mole of fluorine?

Answer: 18.9984032 grams

question answered by

(jacemorris04)


You might be interested in
Metric dimensional analysis worksheet problem: 645 kg to g
Andrews [41]

Answer:

645000 grams my dude

3 0
3 years ago
When carbon is burned in air, it reacts with oxygen to form carbon dioxide. When 14.4 g of carbon were burned in the presence of
Yanka [14]

Answer:

Mass of carbon dioxide produced = 52.8 g

Explanation:

Given data:

Mass of carbon react = 14.4 g

Mass of oxygen = 56.5 g

Mass of oxygen left = 18.1 g

Mass of carbon dioxide produced = ?

Solution:

C + O₂     →      CO₂

Number of moles of C:

Number of moles = mass/ molar mass

Number of moles = 14.4 g/ 12 g/mol

Number of moles = 1.2 mol

18.1 g of oxygen left it means carbon is limiting reactant.

Now we will compare the moles of C with CO₂.

                       C             :         CO₂

                        1             :          1

                      1.2           :          1.2

Mass of CO₂:

Mass = number of moles × molar mass

Mass = 1.2  mol × 44 g/mol

Mass = 52.8 g

8 0
3 years ago
A 1.2 kg block of iron at 32 ∘C is rapidly heated by a torch such that 12 kJ is transferred to it. What temperature would the bl
lbvjy [14]

Answer:

For iron

Final temperature = 54,22°C

For copper

Final Temperature = 63.67 °C

Explanation

Hello,

You are using a torch to warm up a block of iron that has an initial temperature of 32°C.

The first you have to know is that the "heat capacity" could simply define as the heat required to go from an initial temperature to a final temperature.

So you need to use the heat capacity equation as follow in the paper.

The equation has to have all terms in the same units, so:

q = 12000 J

s = 0.450 J / g °C

m = 1200 g

Ti = 32 °C

Download odt
3 0
3 years ago
Ammonia NH3 may react with oxygen to form nitrogen gas and water.4NH3 (aq) + 3O2 (g) \rightarrow 2 N2 (g) + 6H2O (l)If 2.15g of
bagirrra123 [75]

Answer:

NH3 is the limiting reactant

The % yield is 36.1 %

Explanation:

<u>Step 1: </u>Data given

Mass of NH3 = 2.15 grams

Mass of O2 = 3.23 grams

Molar mass of NH3 = 17.03 g/mol

Molar mass of O2 = 32 g/mol

volume of N2 produced = 0.550 L

Temperature = 295 K

Pressure = 1.00 atm

<u>Step 2:</u> The balanced equation:

4NH3 (aq) + 3O2 (g) → 2 N2 (g) + 6H2O (l)

<u>Step 3:</u> Calculate moles of NH3

Moles NH3 = Mass NH3 / Molar Mass NH3

Moles NH3 = 2.15 grams / 17.03 g/mol

Moles NH3 = 0.126 moles

<u>Step 4:</u> Calculate moles of O2

Moles O2 = 3.23 grams / 32 g/mol

Moles O2 = 0.101 moles

<u>Step 5: </u>Calculate the limiting reactant

For 4 moles NH3 consumed, we need 3 moles of O2 to produce, 2 moles of N2 and 6 moles of H2O

NH3 is the limiting reactant. It will completely be consumed ( 0.126 moles).

O2 is in excess, there will be 3/4 * 0.126 = 0.0945 moles consumed

There will remain 0.101 - 0.945 = 0.0065 moles of O2

<u>Step 6:</u> Calculate moles of N2

For 4 moles NH3 consumed, we need 3 moles of O2 to produce, 2 moles of N2 and 6 moles of H2O

For 4 moles NH3 , we'll have 2 moles of N2 produced

For 0.126 moles NH3 consumed, we'll have 0.063 moles of N2 produced.

<u>Step 7</u>: Calculate volume of N2 produced

p*V = n*R*T

⇒ with p = the pressure of the gas = 1.00 atm

⇒ with V = the volume = TO BE DETERMINED

⇒ with n = the number of moles N2 = 0.063 moles

⇒ with R = the gasconstant = 0.08206 L*atm/K*mol

⇒ with T = the temperature = 295

V = (nRT)/p

V = (0.063*0.08206*295)/1

V = 1.525 L = theoretical yield

<u>Step 8:</u> Calculate the % yield

% yield = actual yield / theoretical yield

% yield = (0.550 L / 1.525 L)*100%

% yield = 36.1 %

4 0
3 years ago
19.
solmaris [256]

Answer:

В.  No, because the mass of the reactants is less than the mass of the products.

Explanation:

Chemical equation:

NaBr + Cl₂      →      2NaCl + Br₂

The given equation is not balanced because number of moles of sodium and bromine atoms are less on reactant side while more on the product side.

There are one mole of sodium and one mole of bromine atom on left side of equation while on right side there are 2 moles of bromine and 2 moles of sodium atom are present. The number of moles of chlorine atoms are balanced.

Balanced chemical equation:

2NaBr + Cl₂      →      2NaCl + Br₂

Now equation is balanced. Number of moles of sodium , chlorine and bromine atoms are equal on both side.

5 0
3 years ago
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