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PtichkaEL [24]
2 years ago
15

How would you write the conversion reaction from oil to biodiesel

Chemistry
1 answer:
Rashid [163]2 years ago
7 0

Answer:

Oil + alcohol -> glycerin

Explanation:

hope that this is what you meant

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You have to make 500 mL of a 1.50 M BaCl2. You have 2.0 M barium chloride solution available. Determine how to make the needed d
Inga [223]

Answer:

You need 375 mL of BaCl2 solution.

Explanation:

M1V1=M2V2

Dilution formula. Substitute known values and solve for V1.

M1 = 2.0 M

M2 = 1.50 M

V2 = 500 mL

(2.0 M)(V1) = (1.50 M)(500 mL)

V1 = (1.50 M)(500 mL) / (2.0 M)

V1 = 375 mL

7 0
3 years ago
How many objects in the solar system have been confirmed to currently support life?
yulyashka [42]
B. One
I believe that earth is the only planet that has life that we know of on it.
6 0
3 years ago
Read 2 more answers
Equation for the reaction btn nitric acid and metal X whose valence is 3​
viva [34]
(a) Reaction of nitric acid with non-metal:
C+4HNO
3
​
⟶CO
2
​
+2H
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​
O+4NO
2
​

S+6HNO
3
​
⟶H
2
​
SO
4
​
+2H
2
​
O+6NO
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​

(b) Nitric acid showing acidic character:
K
2
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O+2HNO
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⟶2KNO
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+H
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O
ZnO+2HNO
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⟶Zn(NO
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)
2
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+H
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O
(c) Nitric acid acting as oxidizing agent
P
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+20HNO
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⟶4H
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hope that helps you please mark brainliest
3 0
3 years ago
In the Bronsted-Lowry model of acids and bases, a(n) _____ is a hydrogen donor and a(n) _____ is a hydrogen acceptor.
ExtremeBDS [4]
<span>In the Bronsted-Lowry model of acids and bases, a(n) _acid____ is a hydrogen donor and a(n) _base____ is a hydrogen acceptor.</span>
5 0
3 years ago
How many milliliters of 0.260 m na2s are needed to react with 35.00 ml of 0.315 m agno3?
allochka39001 [22]

The complete balanced chemical reaction is:

2 AgNO3 + Na2S --> 2 NaNO3 + Ag2S

 

First let us calculate the number of moles of AgNO3.

moles AgNO3 = 0.315 M * 0.035 L

moles AgNO3 = 0.011025 mol

 

From the reaction, 1 mole of Na2S is needed for every 2 moles of AgNO3 hence:

moles Na2S required = 0.011025 mol AgNO3 * (1 mol Na2S / 2 mol AgNO3)

moles Na2S required = 5.5125 x 10^-3 mol

 

Therefore volume required is:

volume Na2S = 5.5125 x 10^-3 mol / 0.260 M

<span>volume Na2S = 0.0212 L = 21.2 mL</span>

6 0
3 years ago
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