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Komok [63]
3 years ago
12

What is the mass of 9.67x10^23 molecules of water?

Chemistry
1 answer:
vazorg [7]3 years ago
4 0
100000000000000 I think):
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What is the molar mass of nickel (ll) chloride(NiCl2)
goldenfox [79]

Answer:

Explanation:

yt

6 0
3 years ago
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Each incomplete reaction is a neutralization reaction. Match each equation with the salt formed.
ddd [48]
Based on the incomplete neutralization reaction above, the corresponding salts that each reaction would form are as follows:
1. H2SO4 + 2NH4OH----(NH4)2SO4 with water as a byproduct
2. 2NaOH + H2CO3 ----Na2CO3 with water as a byproduct
3. HNO3 + KOH ----KNO3 with water as a byproduct
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2 years ago
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Which of the following is the correct formula for sodium oxide? NaO2 Na­O Na2O3 Na3O4 Na2O
olga nikolaevna [1]
Since Na has a 1+ charge and O has a -2 charge, by reversing the charges and placing them as subscripts for the other atoms the formula is Na2O1 or simply Na2O.
7 0
3 years ago
How many protons, neutrons, and electrons could an isotope of this atom contain?
Dmitry_Shevchenko [17]

Answer:

The atomic number on the Periodic Table identifies the number of protons in any atom of that element. Copper, atomic number 29, has 29 protons. Finding the atomic number of an element reveals the number of protons.

To find the number of neutrons in the atom, subtract the atomic number from the atomic mass.

8 0
3 years ago
Balanced nuclear equation forces the alpha decay Po-210
Serjik [45]

Answer:

_{84}^{210}\text{Po} \longrightarrow \, _{82}^{206}\text{Pb} \, + \, _{2}^{4}\text{He}

Explanation:

Your unbalanced nuclear equation is:

_{84}^{210}\text{Po} \longrightarrow \, _{x}^{y}\text{Z} + \, _{2}^{4}\text{He}

The main point to remember in balancing nuclear equations is that the sums of the superscripts and the subscripts must be the same on each side of the equation.  

Then

 84 = x + 2, so x =  84 - 2 =   82

210 = y + 4, so y = 210 - 4 = 206

Element 82 is lead, so the nuclear equation becomes

_{84}^{210}\text{Po} \longrightarrow \, _{82}^{206}\text{Pb} \, + \, _{2}^{4}\text{He}

8 0
3 years ago
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