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marishachu [46]
3 years ago
7

Fill in the blanks. SPELLING COUNTS!!

Chemistry
1 answer:
Alex787 [66]3 years ago
8 0

Answer:

cocooooooooooooooooooooooooooo

Explanation:

3n\hj 3ihhjcvyh 6]0u70-uh]h ]]huc7]]uh uhu-hu -u h-u7-u37-]u3- -7u7uh5h57h0u70y c0yuuh-u-]u]3u3 cuyuycuy0nuy0yc6udy8056uy08

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How many moles of copper atoms are in 150g of copper metal
djyliett [7]
The answer to this question would be: 2.36 mol

To answer this question, you need to know the molecular weight of copper. Molecular weight determines how much the weight of 1 mol of a molecule has. Copper molecular weight about 63.5g/mol. Then, the amount of mol in 150g copper should be: 150g / (63.5g/mol)= 2.36 mol
3 0
3 years ago
What is the stoichiometric ratio between BaCl2 and NaCl
bixtya [17]
<span>BaCl2+Na2SO4---->BaSO4+2NaCl There is 1.0g of BaCl2 and 1.0g of Na2SO4, which is the limiting reagent? "First convert grams into moles" 1.0g BaCl2 * (1 mol BaCl2 / 208.2g BaCl2) = 4.8 x 10^-3 mol BaCl2 1.0g Na2SO4 * (1 mol Na2SO4 / 142.04g Na2SO4) = 7.0 x 10^-3 mol Na2SO4 (7.0 x 10^-3 mol Na2SO4 / 4.8 x 10^-3 mol BaCl2 ) = 1.5 mol Na2SO4 / mol BaCl2 "From this ratio compare it to the equation, BaCl2+Na2SO4---->BaSO4+2NaCl" The equation shows that for every mol of BaCl2 requires 1 mol of Na2SO4. But we found that there is 1.5 mol of Na2SO4 per mol of BaCl2. Therefore, BaCl2 is the limiting reagent.</span>
7 0
3 years ago
A mixture of 10.0 g of Ne and 10.0 g Ar have a total pressure of 1.6 atm. What is the partial pressure of Ne?
Natasha2012 [34]
If u add u will get ur answer
7 0
3 years ago
Read 2 more answers
2H2 + O2 + 2H20
kondaur [170]

Answer:

Hello

The answer is 25gr H2O

Explanation:

50grH2×1 molH2/2 gr H2×2mol H2O/2molH2=25gr H2O

8 0
3 years ago
A cylindrical tube 13.5 cm high and 4.5 cm in diameter is used to collect blood samples. How many cubic decimeters (dm3) of bloo
Aleks [24]

Answer:

The cylindrical tube can hold 0.215 cubic decimeters of blood.

Explanation:

Diameter of the cylindrical tube = d = 4.5 cm

r = d/2 ,r = 2.25 cm

Height of the cylindrical tube = h = 13.5 cm

Volume of the cylinder = V = \pi r^2h

V=3.14\times (2.25 cm)^2\times 13.5 cm

V=214.6 cm^3=0.2146 dm^3\approx 0.215 dm^3

1 cm^3=0.001 dm^3

The cylindrical tube can hold 0.215 cubic decimeters of blood.

3 0
3 years ago
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