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aleksklad [387]
2 years ago
12

Which describes any substance that shatters or breaks easily? malleable brittle volatile ductile

Chemistry
2 answers:
dimulka [17.4K]2 years ago
8 0

Answer:

B) brittle

Explanation: if you want to thank me just use dis                                               ♡  ∩_∩

(„• ֊ •„)♡

┏━∪∪━━━━┓

♡ Thank you 。 ♡

┗━━━━━━━┛

swat322 years ago
5 0

Answer:

b

Explanation:

ididthetest

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(1 pt) You have 250.0 mL of 0.136 M HCl. Using a volumetric pipet, you take 25.00 mL of that solution, place it in a 100.00 ml v
alexdok [17]

Answer:

0.034M HCl is the concentration of the diluted solution

Explanation:

You take, initially, 25.00mL of the 0.136M HCl. Then, you dilute the solution to 100.00mL. The solution is diluted:

100.00mL / 25.00mL = 4. The solution was diluted 4 times.

That means the concentration of the diluted solution is:

0.136M / 4 =

<h3>0.034M HCl is the concentration of the diluted solution</h3>
7 0
3 years ago
The blank solution used to calibrate the spectrophotometer is 10.0 mL of 0.2 M Fe(NO3)3 diluted to 25.0 mL with 0.1 M HNO3. Why
Sliva [168]
<span>FeNCS+ product...............thats how you do it i believe </span>
4 0
3 years ago
To what temperature should soup that contains cooked beef be reheated for hot-holding?
nirvana33 [79]

The temperature , <u>165 degree F</u> should soup that contains cooked beef be reheated for hot-holding.

Food should reach a temperature serving 135°F before it can be opened in the restaurant and warmed for hot holding. Within two hours, quickly reheat the food. If food is being served right away, it can be served anywhere at temperature as long that it as it was cooked and chilled appropriately.

TCS food must be totally chilled to 41°F or even below in 6 hours after being thoroughly cooled between 135°F to 70°F in 2 hours. Within four hours, TCS food made with ingredients at room temperature needs to be chilled to 41°F or lower.

Therefore, the temperature , <u>165 degree F</u> should soup that contains cooked beef be reheated for hot-holding.

To know more about temperature

brainly.com/question/24264586

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8 0
2 years ago
A nucleotide is composed of a(n) _____. phosphate group, a nitrogen-containing base, and a five-carbon sugar amino group, a nitr
lapo4ka [179]

A nucleotide is composed of a phosphate group, a nitrogen-containing base, and a five-carbon sugar amino group. A nucleotide is composed of a phosphate group, a nitrogen-containing base, and a five-carbon sugar amino group. A nucleotide is the building block or structural component of DNA and RNA. It consists of a base , that is one from adenine, thymine, guanine, and cytosine. and a molecule of sugar and one of phosphoric acid.


5 0
3 years ago
Which of the following has the strongest buffering capacity? A. H2O B. 0.1 M HCl C. 0.1 M carbonic/bicarbonate (H2CO3/HCO3-) at
enyata [817]

Explanation:

(A)   As we know that carbonic acid (H_{2}CO_{3}) and Sodium bicarbonate (NaHCO_{3}) forms an acidic buffer.

Therefore, pH of an acidic buffer is given by Hendeerson-Hasselbalch equation as follows.

               pH = pK_{a} + log(\frac{[Salt]}{[Acid]}) ........... (1)

So mathematically,  if [Salt] = [Acid]  then \frac{[Salt]}{[Acid]} = 1 .

And,  log (\frac{[Salt]}{[Acid]}) = 0

Therefore, equation (1) gives us the following.

         pH = pK_{a} (when acid and salt are equal in concentration)

Hence, pK_{a} of H_{2}CO_{3} (carbonic acid) is 6.35.

And, with this we have following results.

In (A) and (D) we have the case \frac{[NaHCO_{3}]}{[H_{2}CO_{3}]}[/tex] i.e. [Salt] = [Acid].

Hence, for the cases pH = pK_{a} = 6.35.

(B)    [NaHCO_{3}] = 0.045 M and,  [H_{2}CO_{3}] = 0.45 M

Hence,   pH = 6.35 + log([NaHCO_{3}][[H_{2}CO_{3}])

                     = 6.35 + log(\frac{0.045}{0.45})

                     = 6.35 + (-1)

                     = 5.35

Therefore, it means that this buffer will be most suitable buffer as it has pH on acidic side and addition of slight excess base will not affect much of its pH value.

(C)    [NaHCO_{3}] = 0.45 M [H_{2}CO_{3}]

                          = 0.045 M

So,       pH = 6.35 + log(\frac{[NaHCO_{3}]}{[H_{2}CO_{3}]})

                  = 6.35 + log(\frac{0.45}{0.045})

                  = 6.35 + (+1)

                 = 7.35

This means that pH on Basic side makes it no more acidic buffer.

5 0
3 years ago
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