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3 years ago

5

1 pc

1 answer:

Alex3 years ago

5 0

Answer:0.085

Explanation:

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Describe how you could use two LB/agar plates, some E. coli, and some ampicillin to determine how E. coli cells are affected by

KiRa [710]

Answer:

For this experiment we are going to take plate 1 as the control plate, so, in it there will be just E. coli in LB/agar; in plate 2, we are going to put E. coli in LB/agar and some ampicillin. Then, we have to wait for the E. coli colonies to form. After a while, the E. coli growth can be compared on both plates and determine if ampicillin affects or not the E. coli colonies.

Explanation:

If the ampicillin affects negatively E. coli colonies, we are going to observe that in plate 1 (control plate) there are E. coli colonies growing, but in plate 2, there is no E. coli colonies or, at least, there is a fewer number of colonies on it. If ampicillin doesn't affect E.coli, plate 1 (control) and plate 2 (ampicillin experiment) are going to be similar in number of colonies.

8 0

3 years ago

What is the unit of Avogadro's number?

kompoz [17]

Answer:

it may be electron, atom or ions depend on the nature of the substance and the character of reaction

like :- particles per mole, coulombs per electron

8 0

3 years ago

Silver chloride is virtually insoluble in water so that the reaction appears to go to completion. How many grams of solid NaCl m

masya89 [10]

Answer:

0.535 g

Explanation:

The reaction that takes place is:

NaCl + AgNO₃ → AgCl + NaNO₃

First we <u>calculate how many AgNO₃ moles are there in 25.0 mL of a 0.366 M solution</u>, using the <em>definition of molarity</em>:

Molarity = moles / liters

moles = Molarity * liters

<em>Converting 25.0 mL to L </em>⇒ 25.0 / 1000 = 0.025 L

moles = 0.366 M * 0.025 L = 0.00915 mol AgNO₃

Then we <u>convert AgNO₃ moles into NaCl moles</u>:

0.00915 mol AgNO₃ * $\frac{1molNaCl}{1molAgNO_3}$ = 0.00915 mol NaCl

Finally we<u> convert NaCl moles into grams</u>, using its <em>molar mass</em>:

0.00915 mol NaCl * 58.44 g/mol = 0.535 g

4 0

3 years ago

Calculate the [ OH − ] and the pH of a solution with an [ H + ] = 0.090 M at 25 °C . [ OH − ] = M pH = Calculate the [ H + ] and

Valentin [98]

Answer: a) $[OH^-]=1.09\times 10^{-13}$ and pH = 1.04

b) $[H^+]=1.02\times 10^{-11}$ and $pH=10.99$

c) $[H^+]=7.08\times 10^{-11}$ and $[OH^-]=1.41\times 10^{-4}$

Explanation:

pH or pOH is the measure of acidity or alkalinity of a solution.

pH is calculated by taking negative logarithm of hydrogen ion concentration.

$pH=-\log [H^+]$

$pOH=-\log [OH^-]$

$pH+pOH=14$

a) $[H^+]=0.090M$

$pH=-\log [0.090]=1.04$

$pOH=14-1.04=12.96$

$12.96=-log[OH^-]$

$[OH^-]=1.09\times 10^{-13}$

b) $[OH^-]=0.00098M$

$pOH=-\log [0.00098]=3.01$

$pH=14-3.01=10.99$

$10.99=-log[H^+]$

$[H^+]=1.02\times 10^{-11}$

c) $pH=10.15$

$10.15=-\log [H^+]$

$[H^+]=7.08\times 10^{-11}$

$pOH=14-10.15=3.85$

$3.85=-log[OH^-]$

$[OH^-]=1.41\times 10^{-4}$

3 0

3 years ago

Please help ASAP! Will mark brainliest.

Maslowich

Answer:C. Compounds are the smallest unit of an element that occur on the periodic table!!

Explanation:

3 0

4 years ago

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