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suter [353]
4 years ago
9

I need help with this question please Chemistry

Chemistry
1 answer:
ivolga24 [154]4 years ago
5 0
I cannot see it can you make it bigger by any chance dm me on Instagram
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Match each diagram above with its correct description. Diagrams will be used once.
Zinaida [17]
16 is C
17 is E
18 is B
19 is A
20 is D

hope this helps :)
6 0
3 years ago
If a 0.2g of oil consumed 1ml of sodium thiosulphate, calculate its iodine value and classify the oil?
marta [7]

Iodine value is a measure of the degree of unsaturation in fats and oils. It is essentially the number of grams of iodine consumed by 100 g of fat. If the iodine number is in the range of 0-70 then it is a fat, any value above 70 is considered an oil.

Formula:

Iodine number = (ml of 0.1 N Thiosulphate blank- ml of 0.1N thiosulphate test) * 12.7 *100/1000* wt of sample

vol of thiosulphate required to titrate test sample (given oil) = 1 ml

wt of sample = 0.2 g

Information on the volume of thiosulphate required to titrate the blank solution is essential for calculation.

Iodine number = (X-1.0) * 12.7 * 100/1000* 0.2 = (X-1.0)*6.35

6 0
4 years ago
How do scientist categorize the electromagnetic spectrum?
sdas [7]

Answer:

Into seven sections based on frequency

Explanation:

There are 7 em spectrums from radio waves, to infrared waves, to gamma rays listest in order from lowest to highest frequency.

8 0
3 years ago
How many liters of NH 3 are needed to react completely with 30.0 L of NO (at STP)?
aleksklad [387]

Answer:

20.8 Lof NH₃ are needed for the reaction

Explanation:

This is the reaction:

4 NH₃  +  6 NO  →  5N₂  +   6H₂O

and the info. we need

Ammonia density = 0,00073 g/mL

Let's determine the moles of NO at STP by the Ideal Gases Law equation

P . V = n . R .T

1atm . 30L = n . 0.082 . 273K

(1atm . 30L) / (0.082 . 273K) = n → 1.34 moles of NO

Let's find out the amount of ammonia that should react

6 mol of NO react with 4 mol of ammonia

1.34 mol of NO will react with (1.34  .4)/6 = 0.893 moles of ammonia

Molar mass NH₃ = 17 g/m

0.893 mol . 17 g/m = 15.19 g of ammonia

Ammonia density = 0,00073 g/mL = NH₃ mass / NH₃ volume

0,00073 g/mL = 15.19 g / NH₃ volume

NH₃ volume = 15.19 g / 0,00073 g/mL → 20805.5 mL ⇒ 20.8 L

6 0
3 years ago
H¹⁺ & F¹⁻ ionic compounds
kenny6666 [7]

Answer:

h1.

Explanation:

nikaa ninja nikka cuh suwhoop

8 0
3 years ago
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