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muminat
3 years ago
7

The element in the top left corner of the periodic table is ______________.answer:

Chemistry
1 answer:
rodikova [14]3 years ago
5 0
The answer is hydrogen
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What evidence of a chemical reaction might you see in the following cases?
Ad libitum [116K]
For bleaching a stain, there will be an identifiable odor and it will easily absorb colors and/or stains. Burning a match also has odor involved, but the smoke that is released is from a hot fire that is chemically produced by the formula to make the tip of the match.
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3 years ago
Differentiate between sandy soil, clayey soil and loamy soil on the basis of the growth of plants????
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The mixture of rock particle sand humus is called the soil.
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8 0
3 years ago
The number of electrons in n=1 and n=2 shells of aluminum are
harina [27]

Answer:

n=1 holds two electrons and n=2 holds eight electrons.

Explanation:

Hello

In this case, since the atomic number of aluminum is 13, its electron configuration is:

Al^{13}: 1s^2,2s^2,2p^6,3s^2,3p^1

In such a way, we can see that the level n=1 is filled with two electrons since the subshell s is able to hold two electrons and the level n=2 is also filled but with eight electrons as s holds two whereas p holds 6. Moreover, n=3 is holding three electrons.

Best regards.

5 0
3 years ago
Determine the approximate amount of potassium hydrogen phthalate, KHP, that you will need to neutralize 6.00 ml of 0.100 M NaOH.
Sveta_85 [38]

Answer:

potassium hydrogen phthalate KHP MOLAR MASS = 204.233 glmol

to get 1000 ml

Molar concentration = Mass concentration/Molar Mass

mass concentration = molar concentration x molar mass

mass concentration=0.1 M,

molar mass= 204.233 g/mol

so to get 1L

mass conc = 204.233 x 0.1

= 20.4233g  for 1L or 1000 ml

to get 6.00 ml

if 20.4233g is for 1000ml

then to 6.00 ml

= 20.4233 x 6 / 1000

= 0.123g for 6.00 ml

according to the equation below

NaOH(aq) + KHC8H4O4(aq) --> KNaC8H4O4(aq) + H2O(l)

number of moles of NaOH is equal to that of KHP

so the same amount will be needed too, which is

= 0.123g

6 0
3 years ago
A vial containing radioactive selenium-75 has an activity of 3.0 mCi/mL. If 2.6 mCi are required for a leukemia test, how many m
oksian1 [2.3K]

Answer : The 866.66\mu L must be administered.

Solution :

As we are given that a vial containing radioactive selenium-75 has an activity of 3.0mCi/mL.

As, 3.0 mCi radioactive selenium-75 present in 1 ml

So, 2.6 mCi radioactive selenium-75 present in \frac{2.6mCi}{3.0mCi}\times 1ml=0.86666ml\times 1000=866.66\mu L

Conversion :

(1ml=1000\mu L)

Therefore, the 866.66\mu L must be administered.

4 0
3 years ago
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