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Natalija [7]
4 years ago
12

A hydrogen fuel cell is more efficient than a combustion reaction because very little of the released energy is released as ___.

When used at home, a fuel cell eliminates the need for___
1) A. Heat
B. Electrical energy
C. Chemical energy

2) A. Oxygen
B. Hydrogen
C. An electrical grid
Chemistry
2 answers:
Studentka2010 [4]4 years ago
8 0
It would be A. heat and C. an electric grid.

Hope This Helps You!
Good Luck Studying :)
Klio2033 [76]4 years ago
7 0
A hydrogen fuel cell is more efficient than a combustion reaction because very little of the released energy is released as "Heat". <span>When used at home, a fuel cell eliminates the need for "An Electrical grid"

Hope this helps!</span>
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Sixty-three grams of copper reacted with 32 grams of sulfur. this chemical reaction produced the compound copper sulfide. what w
Nuetrik [128]

The mass of, Cu_2S, compound formed is 77.9g

62 grams of copper reacted with 32grams of sulfur to form copper sulfide.

Cu  +   S_2 \rightarrow  CuS_2

stoichiometry of Cu to S is 2:1

We need to find the limiting reactant

Molar mass of copper = 63.5 g/mol

Molar mass of Sulfur = 32 g/mol

Number of moles of Copper =  \dfrac{mass} {molar mass } = 0.99mole

Number of moles of Sulfur = \dfrac{mass} {molar mass} = \dfrac{32g} {32g/mol} = 1 mole

Since copper have lesser number of moles, therefore the limiting reagent is copper so the amount of product formed depends on amount of Cu present

stoichiometry of Copper to Cu_2S is 2:1

0.99 mol of Copper forms = \dfrac{0.99} {2}  = 0.49 mol of Cu_2S

Mass of Cu_2S produced = Number of moles  \times Molar mass                

Mass of Cu_2S produced = 0.49 mol \times 159 g/mol = 77.9g

Learn more about limiting reagent here-

brainly.com/question/9913056

#SPJ4

6 0
2 years ago
Calculate the mass of chloroform that contains 1.00X10^12 molecules of chloroform
Brums [2.3K]

<span>Molar mass of chloroform (CHCl3)
C = 1 * 12 = 12 a.m.u
H = 1 * 1 = 1 a.m.u
Cl = 3 * 35.5 = 106.5 a.m.u
 Molar Mass (CHCl3) = 12 + 1 + 106.5 = 119.5 g / mol
 
Knowing that: The value of Avogadro's constant corresponds to approximately </span>6,02*10^{23} molecules, if chloroform contains 1,00 * 10^{12} molecules. <span> 
We have:
 
</span>1,00 * 10^{12} * 6,02 * 10^{23} = 6,02 * 10 ^ {35} molecules<span>
 
Now, how many grams are there in the chloroform molecule?

grams -------- molecules</span><span>
119,5 </span>→ 6,02 * 10^{23}
x → 6,02 * 10^{35}

6,02 * 10 ^ {23} * x = 119,5 * 6,02 * 10 ^{35}
6,02 * 10 ^ {23}x = 719,39 * 10^{35}
x =  \frac{719,39}{6,02}*10^{35-23}
\boxed{x = 119,5*10^{12}grams}
7 0
3 years ago
Give an example of organic and inorganic compound
Olenka [21]
Organic: sugar
inorganic: salt
5 0
4 years ago
Calculate the number of grams in 4.56 x 1026 atoms of sodium phosphate. Be sure to balance the charges of sodium phosphate. Help
zalisa [80]

Answer:

124225.91 g of Na₃PO₄

Explanation:

From the question given above, the following data were obtained:

Number of atoms of Na₃PO₄ = 4.56×10²⁶ atoms

Mass of Na₃PO₄ =?

From Avogadro's hypothesis,

6.02×10²³ atoms = 1 mole of Na₃PO₄

Next, we shall determine the mass of 1 mole of Na₃PO₄. This can be obtained as follow:

1 mole of Na₃PO₄ = (23×3) + 31 + (16×4)

= 69 + 31 + 64

= 164 g

Thus,

6.02×10²³ atoms = 164 g of Na₃PO₄

Finally, we shall determine the mass of Na₃PO₄ that contains 4.56×10²⁶ atoms. This can be obtained as follow:

6.02×10²³ atoms = 164 g of Na₃PO₄

Therefore,

4.56×10²⁶ atoms = (4.56×10²⁶ × 164)/6.02×10²³

4.56×10²⁶ atoms = 124225.91 g of Na₃PO₄

Therefore, 124225.91 g of Na₃PO₄ contains 4.56×10²⁶ atoms

4 0
3 years ago
Which has a greater amount of particles, 1.00 mole of hydrogen (H) or 1.00 moles of oxygen(0)?
Bas_tet [7]

Answer:

Both have the same amount of particles.

Explanation:

From Avogadro's hypothesis, we understood that 1 mole of any substance contains 6.02×10²³ particles.

This implies that 1 mole of Hydrogen contains 6.02×10²³ particles. Also, 1 mole of oxygen contains 6.02×10²³ particles.

Thus, 1 mole of Hydrogen and 1 mole of oxygen contains the same number of particles.

7 0
3 years ago
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