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Evgen [1.6K]
3 years ago
11

Please?????????? ????

Chemistry
1 answer:
natka813 [3]3 years ago
3 0
The causes of mass extinction
As there might be marks on fossils showing the suffering that happened to it!
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What must be true of the net charge for any sample of sodium chloride
Ede4ka [16]

Answer:

Zero (0)

Explanation:

Since Na^+ has a charge of +1 and Cl^- has a charge of —1 in NaCl

Then the net = +1 —1 = 0

4 0
3 years ago
Read 2 more answers
What is the mass of 8.0 x 10^26 UF6 molecules?
gulaghasi [49]
First, find the number of moles of UF6
Avagadro's number = 6.023 x 10^23

Number of moles = 8.0 x 10^26 / Avagadro's number = 8.0 x 10^26 / 6.023 x 10^23 = 1.328 x 10³ moles

Molecular weight of UF6 = Molecular weight of U (238.02891) + Molecular weight of F6 (6 x 18.9984032) = 238.02891 + 113.9904192 = 352.0193292 g/mol

Therefore mass of 8.0 x 10^26 UF6 molecules = 352.0193292 g/mol x 1.328 x 10³ moles = 467.481669 x 10³ grams




3 0
3 years ago
श्रीरयाफल्सन यसका लागि तियरल<br>प्रतिस्थापन स्तर को के हो​
jeka94
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3 0
3 years ago
If an atom of zinc has a mass of 64 it has how many neutrons
sveta [45]
Answer:

no of neutron = atomic mass - atomic number

Explanation:

here
atomic mass = 64
atomic number = 30

no of neutron = <span>64−30</span>

no of neutron = 34


3 0
3 years ago
If 28 ml of 5.8 m h2so4 was spilled, what is the minimum mass of nahco3 that must be added to the spill to neutralize the acid?
ipn [44]

First we have to refer to the reaction between the acid and the base: <span>

H2SO4 + 2 NaHCO3 ---> 2 H2O + 2 CO2 + Na2SO4 

From this balanced equation we can see that for every 1 mol of acid (H2SO4), we need 2 mol of base (NaHCO3) to neutralize it. Given 28 ml of 5.8 M acid, we need to find out how many mols of acid that is: 

<span>28mL * (1L/1000mL) * 5.8 mol/L =  0.1624 mol H2SO4</span></span>

<span>
Since we need 2 mol of base per mol of acid, we need:</span>

<span> 2*0.1624 mol = 0.3248 mol NaHCO3 </span><span>

MolarMass of NaHCO3 is 84.01 g/mol 

<span>0.3248 mol*(84.01g/mol) = 27.29 g NaHCO3</span></span>

3 0
3 years ago
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