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mart [117]
3 years ago
9

Early life arose in an oxygen-free environment, but if any of these microbes had somehow come in contact with oxygen, the most l

ikely effect would have been ________.
Chemistry
1 answer:
Leya [2.2K]3 years ago
7 0

Answer:

the effect of oxygen on these types of microbes is it will kill them.

Explanation:

When oxygen present in the environment come in contact with anaerobe bacteria it kill them because oxygen in air act as excited oxygen singlet molecule which will react with the water present in the cell of bacteria and convert it into hydrogen peroxides and bacteria do not have any defense system from hydrogen peroxide and ultimately it kill the bacteria.

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An analytical chemist weighs out 0.188 g of an unknown triprotic acid into a 250 mL volumetric flask and dilutes to the mark wit
Anon25 [30]

<u>Answer:</u> The molar mass of unknown triprotic acid is 97.66 g/mol

<u>Explanation:</u>

To calculate the molarity of acid, we use the equation given by neutralization reaction:

n_1M_1V_1=n_2M_2V_2

where,

n_1,M_1\text{ and }V_1 are the n-factor, molarity and volume of triprotic acid

n_2,M_2\text{ and }V_2 are the n-factor, molarity and volume of base which is NaOH.

We are given:

n_1=3\\M_1=?M\\V_1=250mL\\n_2=1\\M_2=0.0600M\\V_2=95.9mL

Putting values in above equation, we get:

3\times M_1\times 250=1\times 0.0600\times 95.9\\\\M_1=0.0077M

To calculate the molecular mass of solute, we use the equation used to calculate the molarity of solution:

\text{Molarity of the solution}=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}

We are given:

Molarity of solution = 0.0077 M

Given mass of triprotic acid = 0.188 g

Volume of solution = 250 mL

Putting values in above equation, we get:

0.0077M=\frac{0.188\times 1000}{\text{Molar mass of triprotic acid}\times 250}\\\\\text{Molar mass of triprotic acid}=97.66g/mol

Hence, the molar mass of unknown triprotic acid is 97.66 g/mol

7 0
2 years ago
You are given a solution that is 518 mM lactose. You need to make up 4.5 L of 16.7 mM solution. What volume do you need to trans
OLga [1]

Answer:

The volume you need to transfer from the stock solution is 0.145 l

Explanation:

Since the number of moles of lactose in the volume of stock solution that you transfer will be the same as the number of moles of lactose in the final solution, you can use this expression:

number of moles in volume to transfer = number of moles in the final solution

Since number of moles = concentration * volume (if the concentration is expressed in molarity), then:

Ci * Vi = Cf * Vf

where:

Ci = concentration of the stock solution.

Vi = volume of the stock solution to be transferred.

Cf = concentration of the final solution

Vf = volume of the final solution

Then, replacing with the data:

518 mM * Vi = 16.7 mM * 4.5 l

Vi = 16.7 mM * 4.5 l / 518 mM

<u>Vi = 0.145 l or 145 ml</u>

Notice that any concentration unit can be used, as long as the units of the concentration of the stock and final solution are the same.

4 0
3 years ago
Identify the base in this acid-base reaction:
yarga [219]

Answer:

\boxed{\text{NaOH}}

Explanation:

\rm \underbrace{\hbox{NaOH }}_{\hbox{base}} + \underbrace{\hbox{HCl}}_{\hbox{acid}} \longrightarrow \, NaCl + H$_{2}$O

A hydroxide of a metal is a base.

\text{The base is }\boxed{\textbf{NaOH}}

B is wrong. HCl is an acid.

C is wrong. NaCl is a salt.

D is wrong. Water is neutral.

8 0
2 years ago
Read 2 more answers
What conversion factor would you use to convert correctly from the mass of a given
statuscvo [17]

Answer: 1 mole of the substance is equal to its molar mass or vice versa.

Explanation: Example: 18 g H2O x 1 mole H2O / 18g H2O = 1 mole H2O

7 0
2 years ago
How many moles of lithium hydroxide are required to react with 19.2 mol CO2, the average amount exhaled by a person each day?
sladkih [1.3K]

Answer:

38.4 mol LiOH

Explanation:

Step 1: Write the balanced neutralization equation

2 LiOH + CO₂ ⇒ Li₂CO₃ + H₂O

Step 2: Establish the appropriate molar ratio

According to the balanced equation, the molar ratio of LiOH to CO₂ is 2:1.

Step 3: Calculate how many moles of lithium hydroxide are required to react with 19.2 mol CO₂

19.2 mol CO₂ × 2 mol LiOH / 1 mol CO₂ = 38.4 mol LiOH

3 0
3 years ago
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