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PIT_PIT [208]
3 years ago
5

What do methanol, ethanoic acid, and glucose all have in common?

Chemistry
1 answer:
Nady [450]3 years ago
3 0

Answer:

They are all organic compounds

Explanation:

Methanol, ethanoic acid and glucose are all organic compound that can be found in living organisms.

Glucose is a product of photosynthesis in plants. It is also produced from digestion of food in animals.

Ethanoic acid is a product of cellular respiration that occurs where there is insufficient oxygen.

Methanol is a waste often found in urine.

 These compounds can all be traced to living sources. They are called organic compounds.

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Consider the three statements below. Which numbered response contains all the statements that are true and no false statements?
vladimir1956 [14]

Answer:

a) I, II, and III

Explanation:

For the first statement;

Solvation, is the process of attraction and association of molecules of a solvent with molecules or ions of a solute. if the solvent is water, we call this process hydration.

This means the statement is TRUE.

For the second statement;

The negatively-charged side of the water molecules are attracted to  positively-charged ions. In the case of water, the oxygen end is the negatively charged side of water. This means the statement is TRUE.

For the third statement;

The positively-charged side of the water molecules are attracted to the negatively-charged chloride ions. In the case of water, the hydrogen end is the positively charged side of water. This means the statement is TRUE.

Going through the options, we can tell that the correct option is option A.

6 0
3 years ago
C. 49.8 grams of ki is dissolved in enough water to make 1.00 l of solution. what is the molarity?
taurus [48]

Hey There!:


Molar Mass KI => 166.003 g/mol


* number of moles:


n = mass of solute / molar mass


n = 49.8 / 166.003


n = 0.3 moles KI


Therefore:


M = n / V


M = 0.3 / 1.00


M = 0.3 mol/L


hope this helps!

7 0
3 years ago
Read 2 more answers
The compound Xe(CF3)2 decomposes in afirst-order reaction to elemental Xe with a half-life of 30. min.If you place 7.50 mg of Xe
Irina-Kira [14]

Answer:

t=147.24\ min

Explanation:

Given that:

Half life = 30 min

t_{1/2}=\frac{\ln2}{k}

Where, k is rate constant

So,  

k=\frac{\ln2}{t_{1/2}}

k=\frac{\ln2}{30}\ min^{-1}

The rate constant, k = 0.0231 min⁻¹

Using integrated rate law for first order kinetics as:

[A_t]=[A_0]e^{-kt}

Where,  

[A_t] is the concentration at time t

[A_0] is the initial concentration

Given that:

The rate constant, k = 0.0231 min⁻¹

Initial concentration [A_0] = 7.50 mg

Final concentration [A_t] = 0.25 mg

Time = ?

Applying in the above equation, we get that:-

0.25=7.50e^{-0.0231\times t}

750e^{-0.0231t}=25

750e^{-0.0231t}=25

x=\frac{\ln \left(30\right)}{0.0231}

t=147.24\ min

6 0
3 years ago
train travel from hong kong to beijing it traveled at an average speed of 160 km/m in the first 4 hours. after that it traveled
Nonamiya [84]

Answer:

7 hours

Explanation: step 1. 160x4=640

step 2. 1180-640=540

step 3. 540÷180=3

step 4. 3+4=7

5 0
3 years ago
Read 2 more answers
In the Polyacrylamide and Agarose Gels lab, when working with your protein or DNA sample you must wear:
blagie [28]
It’s a. All of these, you’re welcome
6 0
3 years ago
Read 2 more answers
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