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Olegator [25]
2 years ago
11

How many protons, neutrons, and electrons are in an atom of beryllium with a mass number of 9?

Chemistry
2 answers:
antiseptic1488 [7]2 years ago
8 0

Answer:

D) 4 protons, 4 electrons, 5 neutrons

ki77a [65]2 years ago
7 0

Answer:

c / a

correct me if im wrong

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Which of the following you do NOT need to pay attention while using a centrifuge?
Phoenix [80]

Answer:

The radius of the centrifuge.

Explanation:

Hello,

Since the radius of the centrifuge is just a design parameter, it wouldn't be a cause of failure because it is used to know how many tubes could be fitted in into the centrifuge. On the other hand, keeping you attention away from other factors could turn into a failure as long as the sample could be poured down or just turn out inadequate for the expected results.

Best regards.

8 0
3 years ago
Write the balanced equation for the reaction of acetic acid (CH3COOH) and sodium bicarbonate (NaHCO3). What type of reaction(s)
stealth61 [152]

Answer:CH3COOH + NaHCO3 > H2O + CO2(g) + CH3COONa

Explanation:acid and base neutralize creating water and CO2 gas along with a salt

5 0
3 years ago
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mel-nik [20]

Answer:

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7 0
3 years ago
What is the situation when I might travel high velocity but low acceleration
Anna007 [38]
Cruising at 35,000 feet in an airliner, straight toward the east, 
at 500 miles per hour.  

Since your speed is not changing and you're moving in a straight line,
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6 0
3 years ago
The radioisotope phosphorus-32 is used in tracers for measuring phosphorus uptake by plants. The half-life of phosphorus-32 is 1
Harman [31]

Answer:

54 days

Explanation:

We have to use the formula;

0.693/t1/2 =2.303/t log Ao/A

Where;

t1/2= half-life of phosphorus-32= 14.3 days

t= time taken for the activity to fall to 7.34% of its original value

Ao=initial activity of phosphorus-32

A= activity of phosphorus-32 after a time t

Note that;

A=0.0734Ao (the activity of the sample decreased to 7.34% of the activity of the original sample)

Substituting values;

0.693/14.3 = 2.303/t log Ao/0.0734Ao

0.693/14.3 = 2.303/t log 1/0.0734

0.693/14.3 = 2.6/t

0.048=2.6/t

t= 2.6/0.048

t= 54 days

3 0
3 years ago
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