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2 years ago

Fact check this please

Chemistry

2 answers:

kicyunya [14]2 years ago

8 0

Answer:

nhbknjkb hbffvjkernjvelfbbjvjc

Explanation:

olga_2 [115]2 years ago

4 0

.......the answer is B

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Which properties represent a metal ( select all that apply )

blagie [28]

Malleable, shiny and good conductors
A B E

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2 years ago

A substance has an atomic number of 80 and a mass number of 178. How many protons does this atom have? How many neutrons does it

Nitella [24]

Answer: idk

Explanation:

Idk

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2 years ago

How many pints of a 30% sugar solution must be added to a 5 pint of a 5% sugar solution to obtain a 20% sugar solution?

ser-zykov [4K]

You must add 7.5 pt of the 30 % sugar to the 5 % sugar to get a 20 % solution.

You can use a modified dilution formula to calculate the volume of 30 % sugar.

V_1×C_1 + V_2×C_2 = V_3×C_3

Let the volume of 30 % sugar = x pt. Then the volume of the final 20 % sugar = (5 + x ) pt

(x pt×30 % sugar) + (5 pt×5 % sugar) = (x + 5) pt × 20 % sugar

30x + 25 = 20x + 100

10x = 75

x = 75/10 = 7.5

5 0

3 years ago

What is the mass of 2.65 millimoles of sulfur?

Bas_tet [7]

Divide this problem in to two steps: 1. Convert millimoles to moles. 2. Convert moles to grams.

1.
2.65 mlmol x 1mol/1000mlmol=.00265mol

2.
.00265mol x 32.065g/1 mol=.0850g S

Let me know if you need more explanation on why I used the conversions that I did.

6 0

3 years ago

Urea (CH4N2O) is a common fertilizer that can be synthesized by the reaction of ammonia (NH3) with carbon dioxide as follows: 2N

Gnesinka [82]

Answer:

NH3 is the limiting reactant

The theoretical yield is 216.0 kg urea

The % for this reaction is 78.8 %

Explanation:

Step 1: Data given

Mass of ammonia = 122.5 kg

Mass of carbon dioxide = 211.4 kg

Mass of urea produced = 170.3 kg

Molar mass of ammnoia = 17.031 g/mol

Molar mass of carbon dioxide = 44.01 g/mol

Moalr mass of urea = 60.06 g/mol

Step 2: The balanced equation

2NH3(aq) + CO2(aq) --> CH4N2O(aq) + H2O(l)

Step 3: Calculate moles of NH3

Number of moles = mass / Molar mass

Moles NH3 = 122500 grams / 17.031 g/mol

Moles NH3 = 7192.77 moles

Step 4: Calculate moles of CO2

Moles CO2 = 211400 / 44.01 g/mol

Moles CO2 = 4803.45 moles

Step 5: Calculate limiting reactant

For 2 moles NH3 consumed, we need 1 moles of CO2 to produce 1 mole urea and 1 mole H2O

NH3 is the limiting reactant. It will completely be consumed (7192.77 moles).

CO2 is in excess. There will be consumed 7192.77/2 = 3596.4 moles

There will remain 4803.45 - 3596.4 = 1207.05 moles of CO2

Step 6: Calculate moles of urea produced:

For 2 moles NH3 consumed, we need 1 moles of CO2 to produce 1 mole urea and 1 mole H2O

For 7192.77 moles of NH3, we have 3596.4 moles of urea produced

Step 7: Calculate mass of urea

Mass urea = moles urea * molar mass urea

Mass urea = 3596.4 moles * 60.06 g/mol

Mass urea = 216000 grams = 216 kg = theoretical yield

Step 8: Calculate % yield

% yield = (actual yield / theoretical yield)*100%

% yield = (170.3 / 216) *100% = 78.8%

The % for this reaction is 78.8 %

3 0

3 years ago