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dezoksy [38]
3 years ago
13

Harim placed 5mL of ethanol into a container that weighs 1 gram using a dropper. He already knew the density of ethanol is 0.78

g/mL.
What is the mass of the ethanol, not including the container?
Chemistry
1 answer:
Aleks04 [339]3 years ago
8 0

Answer:

there it is fella tried on ma own consciousness

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Please help me!!!!! I REALLY NEED HELP!!! What is my carbon footprint and how can I reduce it?
vitfil [10]

Answer:

your carbon footprint is how often you use a car of vehical you can decreese it by riding a bike or walking.

Explanation:

7 0
3 years ago
Read 2 more answers
How many moles are in 442.8 g of strontium carbonate (SrCO3)?
GuDViN [60]

Answer:

3 moles

Explanation:

SrCO3

Mass = 442.8g

Molar mass = (87.6 + 12 * [16*3]) = 147.6g/mol

Number of moles = mass / molar mass

Number of moles = 442.8 / 147.6

Number of moles = 3

7 0
3 years ago
For the following reaction, 38.3 grams of sulfuric acid are allowed to react with 33.5 grams of calcium hydroxide sulfuric acid(
Likurg_2 [28]

Answer:

What is the maximum amount of calcium sulfate that can be formed? 53.1 grams CaSO4

What is the FORMULA for the limiting reagent? H2SO4

What amount of the excess reagent remains after the reaction is complete? 4.59 grams of Ca(OH)2

Explanation:

Step 1: Data given

Mass of sulfuric acid = 38.3 grams

Molar mass of H2SO4 = 98.08 g/mol

Mass of calcium hydroxide = 33.5 grams

Molar mass of Ca(OH)2 = 74.09 g/mol

Step 2: The balanced equation

H2SO4 + Ca(OH)2 → CaSO4 + 2H2O

Step 3: Calculate moles of H2SO4

moles H2SO4 = mass H2SO4 / molar mass H2SO4

moles H2SO4 = 38.3 grams / 98.08 g/mol

moles H2SO4 = 0.390 moles

Step 4: Calculate moles of Ca(OH)2

moles Ca(OH)2 = 33.5 grams / 74.09 g/mol

moles Ca(OH)2 =0.452 moles

Step 5: Calculate limiting reactant

For 1 mol H2SO4, we need 1 mol of Ca(OH)2 to produce, 1 mol of CaSO4 and 2 mol of H2O

H2SO4 is the limiting reactant. It will completely be consumed (0.390 moles).

Ca(OH)2 is in excess. There will be consumed 0.390 moles

There will remain 0.452 - 0.390 = 0.062 moles

This is 0.062 * 74.09 g/mol = 4.59 grams

Step 6: Calculate moles of calcium sulfate

For 1 mol H2SO4, we need 1 mol of Ca(OH)2 to produce, 1 mol of CaSO4 and 2 mol of H2O

For 0.390 moles of H2SO4, there will be produced 0.390 moles of CaSO4

Step 7: Calculate mass of CaSO4

Mass CaSO4 = moles CaSO4 * molar mass CaSO4

Mass CaSO4 = 0.390 moles * 136.14 g/mol

Mass of CaSO4 = 53.1 grams

7 0
3 years ago
If 23.6 g of hydrogen gas reacts with 28.3 g of nitrogen gas, what is the maximum amount of product that can be produced?
Aleonysh [2.5K]

Answer:

34.3 g NH3

Explanation:

M(H2) = 2*1 = 2 g/mol

M(N2) = 2*14 = 28 g/mol

M(NH3) = 14 + 3*1 = 17 g/mol

23.6 g H2* 1 mol/2 g = 11.8 mol H2

28.3 g N2 * 1 mol/28 g = 1.01 mol N2

                                 3H2 + N2 ------> 2NH3

from reaction         3 mol    1 mol

given                   11.8 mol    1.01 mol

We can see that H2 is given in excess, N2 is limiting reactant.

                                 3H2 + N2 ------> 2NH3

from reaction                     1 mol         2 mol

given                                 1.01 mol      x

x = 2*1.01/1= 2.02 mol NH3

2.02 mol * 17g/1 mol ≈ 34.3 g NH3

8 0
3 years ago
Can we be friends ????
Svet_ta [14]
Yes of course!!!! :)
5 0
3 years ago
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