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Serggg [28]
4 years ago
12

F 15 g of C₂H₆ reacts with 60.0 g of O₂, how many moles of water (H₂O) will be produced? (Hint: use what you know about limiting

reactant) *
10 points
Chemistry
1 answer:
Mars2501 [29]4 years ago
4 0

Answer:

1.5 mol H2O

Explanation:

2C2H6 + 7O2 -> 4CO2 + 6H2O

The limiting reactant is C2H6

30 g C2H6 ->1 mol C2H6

15 g C2H6   -> x                      x= 0.5 mol C2H6

2 mol C2H6 -> 6 mol H2O

0.5 mol c2H6 -> x                    x = 1.5 mol H2O

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What is similar about the outer shell of electrons in the alkali metals family
Nadusha1986 [10]

Answer:

1 electron

Explanation:

These metals have a single electron in the outer shell

4 0
3 years ago
Assume that Na+ is being transported across a membrane via facilitated diffusion. Which of the following conditions would allow
lesya [120]

Answer:

<h2>4. Na+ diffusing toward the side of the membrane with Cl− and 50% less Na+.</h2>

Explanation:

Facilitated diffusion is a type of transport mechanism in which the special proteins are involved and play an important role in the transport of the atoms, ions or molecules. This mechanism is based on the electrochemical gradient differences. When this difference increase, then the transport of the sodium takes place because sodium ions are chemically attracted by chloride ions. In a facilitated diffusion process, no energy requirement takes place. This process occurs along the concentration gradient.

3 0
4 years ago
What mass of chromium would be produced from the reaction of 57.0 g of potassium with 199 g of chromium(II) bromide according to
fredd [130]

Answer:

Mass of Chromium produced = 37.91 grams

Explanation:

2K + CrBr₂  →  2KBr + Cr

2mole     1 mole                1 mole

mass of Potassium = 57.0 grams

molar mass of Potassium = 39.1 g/mol

no of moles of Potassium = 57.0 / 39.1 = 1.458 moles

mass of CrBr₂= 199 grams

molar mass of CrBr₂ = 211.8 gram/mole

no of moles of CrBr₂ = 199 / 211.8 = 0.939 mole

From chemical equation

1 mole of CrBr₂ = 2 moles of K

∴ 0.939 moles of CrBr₂ = ?

   ⇒ 0.939 x 2/1 = 1.878 moles of K

1.878 moles of K is needed, but there is 1.458 moles of K. So, Potassium is completed first during the reaction . Hence, Potassium is limiting reagent. and CrBr₂ is excess reagent .

From chemical equation

2 moles of K = 1 mole of Cr

∴ 1.458 moles of K = ?

   ⇒ 1.458 x 1/ 2 = 0.729 moles of Cr

no of moles of Cr formed = 0.729 moles

molar mass of Cr = 52.0 g/mol

mass of one mole of Cr = 52.0 grams

mass of 0.729 moles of Cr = 52.0 x 0.729 = 37.908 grams

mass of Chromium produced = 37.91 grams

6 0
3 years ago
Which scientists developed a model of the atom that looked like a nucleus surrounded by electrons
Kobotan [32]
Nicki Minaj did no cap
4 0
3 years ago
Under certain conditions the rate of this reaction is zero order in ammonia with a rate constant of 0.0089 M.s^-1:
masya89 [10]

Answer:

Explanation:

The rate of reaction will not depend upon concentration of reactant . It will be always constant and equal to .0089M s⁻¹.

Initial moles of reactant = 400 x 10⁻³ mole in 5 L

molarity = 400 x 10⁻³ /5 M

= 80 x 10⁻³ M .

= .08M

no of moles reacted in 2 s  = .0089 x 2

= .0178 M

concentration left = .08 - .0178 M

= .0622 M .

No of moles left in 5 L

= 5 x .0622 = .31 moles .

5 0
4 years ago
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