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loris [4]
3 years ago
9

Where is most of the mass of an atom? •

Chemistry
1 answer:
yaroslaw [1]3 years ago
6 0
B. inside the nucleus
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Use a magnet to separate the iron from the sand.
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What is a potential risk of genetically modified food?
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It would be D since it could be unhealthy for you.

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Determine the limiting reactant (lr) and the mass (in g) of nitrogen that can be formed from 50.0 g n2o4 and 45.0 g n2h4. some p
Licemer1 [7]
                                                   N2O4(l) + 2 N2H4(l) → 3 N2(g) + 4 H2O(g)
1) to calculate the limiting reactant you need to pass grams to moles.
<span> moles is calculated by dividing mass by molar mass
</span>
mass of N2O4: 50.0 g 
molar mass of <span>N2O4 = 92.02 g/mol
</span><span>molar mass of N2H4 = 32.05 g/mol.
</span>mass of N2H4:45.0 g

moles N2O4=50.0/92.02 g/mol= 0,54 mol of N2O4
moles N2H4= 45/32.05 g/mol= 1,40 mol of <span><span>N2H4

</span> 2)</span>
By looking at the balanced equation, you can see that 1 mol of N2O4 needs 2 moles of N2H4 to fully react . So to react  0,54 moles of N2O4, you need 2x0,54 moles of <span>N2H4 moles
</span><span>N2H4 needed = 1,08 moles.
You have more that 1,08 moles </span><span>N2H4, so this means the limiting reagent is not N2H4, it's </span>N2O4. The molecule that has molecules that are left is never the limiting reactant.

3) 1 mol of N2O4 reacting, will produce 3 mol of N2 (look at the equation)
There are 0,54 mol of N2O4 available to react, so how many moles will produce of N2?
1 mol N2O4------------3 mol of N2
0,54 mol N2O4--------x
x=1,62 mol of N2

4) the only thing left to do is convert the moles obtained, to grams.
We use the same formula as before, moles equal to mass divided by molar mass.
moles= \frac{grams}{molar mass}             (molar mass of N2= 28)
1,62 mol of N2= mass/ 28
mass of N2= 45,36 grams

4 0
3 years ago
At an ACME plant in Scottsdale Arizonia methanol is produced commercially
patriot [66]

Answer:

thats nice to know i will send Wile E. Coyote there for his supplys

Explanation:

4 0
3 years ago
How many grams of iron(II) chloride are needed to produce 44.3 g iron(II) phosphate in the presence of excess sodium phosphate?
zalisa [80]

Answer:

47.2 g

Explanation:

Let's consider the following double displacement reaction.

3 FeCl₂ + 2 Na₃PO₄ → Fe₃(PO₄)₂ + 6 NaCl

The molar mass of Fe₃(PO₄)₂ is 357.48 g/mol. The moles corresponding to 44.3 g are:

44.3 g × (1 mol / 357.48 g) = 0.124 mol

The molar ratio of Fe₃(PO₄)₂ to FeCl₂ is 1:3. The moles of FeCl₂ are:

3 × 0.124 mol = 0.372 mol

The molar mass of FeCl₂ is 126.75 g/mol. The mass of FeCl₂ is:

0.372 mol × (126.75 g/mol) = 47.2 g

5 0
2 years ago
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