All categories

3 years ago

The formation of a gas or a precipitate may be evidence for a ___________.

Chemistry

1 answer:

ololo11 [35]3 years ago

5 0

Reaction [taking place]

Hope that helps

You might be interested in

How to classify hemicetal and acetal

kumpel [21]

You might have meant hemiacetal, not hemicetal.
Acetals contain two –OR groups, one –R group and a –H atom. In hemiacetals, one of the –OR groups in acetals is replaced by a –OH group<span>.
</span>

3 0

3 years ago

When a phasor-domain circuit has dependent sources, you should not use a sequence of source transforms to find the thévenin and

Ierofanga [76]

Yes but the answer is A

4 0

2 years ago

Which bond is formed when a fatty acid is attached to a glycerol<br> molecule?<br> 2 of 8 QU

Mekhanik [1.2K]

Answer: Heyo Kenji Here! Here's your answer- In a fat molecule, the fatty acids are attached to each of the three carbons of the glycerol molecule with an ester bond through the oxygen atom. During the ester bond formation, three molecules are released. Since fats consist of three fatty acids and a glycerol, they are also called triacylglycerols or triglycerides.

Explanation: Hope this helps!

Have a nice day!

- Kenji ^^

4 0

2 years ago

Helppppppppppppppppp

vovangra [49]

Do not trust people that share links these people hack and put viruses. to your phone/computer’s

DO NOT TRUST THESE LINKS

7 0

2 years ago

Measurements show that unknown compound has the following composition: element mass 38.7 % calcium, 19.9 % phosphorus and 41.2 %

Mademuasel [1]

Answer:

D) empirical formula is: C₃P₂O₈

Explanation:

Given:

Mass % Calcium (Ca) = 38.7%

Mass % Phosphorus (P) = 19.9%

Mass % oxygen (O) = 41.2 %

This implies that for a 100 g sample of the unknown compound:

Mass Ca = 38.7 g

Mass P = 19.9 g

Mass O = 41.2 g

Step 1: Calculate the moles of Ca, P, O

Atomic mass Ca = 40.08 g/mol

Atomic mass P = 30.97 g/mol

Atomic mass O = 16.00 g/mol

$Moles\ Ca = \frac{38.7g}{40.08g/mol} =0.966\ mol\\\\Moles\ P = \frac{19.9g}{30.97g/mol} =0.643\ mol\\\\Moles\ O = \frac{41.2g}{16.00g/mol} =2.58\ mol$

Step 2: Calculate the molar ratio

$C = \frac{0.966}{0.643} =1.50\\\\P = \frac{0.643}{0.643} = 1.00\\\\O = \frac{2.58}{0.643} =4.00$

Step 3: Calculate the closest whole number ratio

C: P: O = 1.50 : 1.00 : 4.00

C : P : O = 3:2:8

Therefore, the empirical formula is: C₃P₂O₈

7 0

3 years ago