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Natalka [10]
2 years ago
15

For social science.

Chemistry
2 answers:
sergejj [24]2 years ago
7 0
I would say A because all the other answers seen wrong
Pie2 years ago
6 0

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HELP IT’S ABOUT TO BE DUE!
Vinil7 [7]
The answer is b
Explanation:
7 0
2 years ago
Arunner sprints 550 feet in 100 seconds. If the runner can hold this pace for an hour
Alex
<h3>Answer:</h3>

Distance traveled is 9.055 km

<h3>Explanation:</h3>

Speed is given by dividing distance by time.

The distance given is 550 ft and the time is 100 seconds.

Therefore; Speed = 550 ft ÷ 100 sec

                              = 5.5 ft/sec

We can calculate the distance the runner covered in one and half hour.

Time = 1.5 hours

But 1 hour = 3600 sec

1.5 hours = (1.5 × 3600)

               = 5400 sec

Distance is given by multiplying speed and time

Thus; Distance = 5.5 ft/sec × 5400 sec

                         = 29700 ft

But; 1 m = 3.28 ft

1000 m = 1 km

Therefore; 29700 ft will be equivalent to

 = 29700 ft/3.28 ft

= 9054.878 M

but 1000 m = 1 km

Thus, distance = 9054.878 ÷ 1000 m

                        = 9.055 km

6 0
2 years ago
How many atoms of hydrogen are present in one molecule of the following compounds: (NH),SO, NH, NH CONH,​
iren [92.7K]

Answer:

Explanation:

1)6

2)0

3)2

4)2

r the no. of atoms present in the molecules

3 0
2 years ago
Can anyone plz provide me 30 element with the list of ::
iragen [17]

Electron Configuration table:-

\boxed {\begin{array}{|c|c|c|c|cccc|}\cline {1-8}\sf Element & \sf Symbol & \sf Valency & \sf{Atomic\:Number} & & \sf{Electron} & \sf Configuration & \\ \cline {5-8} &&&&\bf K&\bf L &\bf M&\bf N \\ \cline {1-8}\rm Hydrogen &\sf H & 1 & 1 & 1 &&& \\ \cline {1-8}\rm Helium &\sf He & 0 & 2 & 2 &&& \\ \cline {1-8}\rm Lithium &\sf Li & 1& 3 &2 &1 && \\ \cline {1-8}\rm Beryllium &\sf Be & 2 & 4 & 2 & 2 && \\ \cline {1-8}\rm Boron &\sf B & 3 & 5 & 2 & 3 && \\ \cline {1-8}\rm Carbon & \sf C & 4 & 6 & 2 & 4 && \\ \cline {1-8}\rm Nitrogen &\sf N & 3 & 7 & 2 & 5 && \\ \cline {1-8}\rm Oxygen &\sf O & 2 & 8 & 2 & 6 && \\ \cline {1-8}\rm Fluorine &\sf F & 1 & 9 & 2 & 7 && \\ \cline {1-8}\rm Neon & \sf Ne & 0 & 10 & 2 & 8 && \\ \cline {1-8}\rm Sodium &\sf Na & 1 & 11 & 2 & 8 & 1 &\\ \cline {1-8}\rm Magnesium &\sf Mg & 2 & 12 & 2 & 8 & 2 & \\ \cline {1-8}\rm Aluminium &\sf Al & 3 & 13 & 2 & 8 & 3 & \\ \cline{1-8}\rm Silicon &\sf Si & 4 & 14 & 2 & 8 & 4 & \\ \cline {1-8}\rm Phosphorus &\sf P & 3 & 15 & 2 & 8 & 7 & \\ \cline {1-8}\rm Sulphur &\sf S & 2 & 16 & 2 & 8 & 8 & \\ \cline{1-8}\rm Chlorine &\sf Cl & 1 & 17 & 2 & 8 & 8 &1 \\ \cline {1-8}\rm Argon &\sf Ar & 0 & 18 & 2 & 8 & 8 & 2 \\ \cline {1-8}\rm Potassium &\sf K & 1 & 19 & 2 & 8 & 8 &3 \\ \cline {1-8}\rm Calcium &\sf Ca & 2 & 20 & 2 & 8 & 8 & 4 \\ \cline{1-8} \end {array}}

3 0
3 years ago
Calculate the ph of a solution that is 0.26 m in hf and 0.12 m in naf.
DochEvi [55]
When a mixture consists of a weak acid, such as HF, and its corresponding salt, NaF, this is an acid buffer. So, we can use the equation as written below:

pH = pKa + log(salt/acid)

For HF, the pKa = 3.17

Thus,
pH = 3.17 + log(0.12/0.26)
<em>pH = 2.83</em>
5 0
2 years ago
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