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Mkey [24]
3 years ago
9

really need this answer

Chemistry
1 answer:
scoundrel [369]3 years ago
3 0

Answer:

B 100,000%

Explanation:

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Convert 12.64 g of NaOH to moles
Luden [163]
.316 moles just divide by the molar mass of naoh which is 39.997. So 12.64 divived by 39.997 is .3160237. that is 3.16x10-1
8 0
3 years ago
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What law states that the volume of a gas is proportional to the miles of the gas when pressure and temperature are kept constant
Rufina [12.5K]

ideal gas law. but you are talking about moles of gas not miles

5 0
3 years ago
H2(g) + Co(g) _CH3OH<br> balance the equation <br>​
Tom [10]

Answer:

CO + 2H2 = CH3OH

Explanation:

1. Label Each Compound With a Variable

  aCO + bH2 = cCH3OH

2. Create a System of Equations, One Per Element

  C: 1a + 0b = 1c

  O: 1a + 0b = 1c

  H: 0a + 2b = 4c

3. Solve For All Variables (using substitution, gauss elimination, or a calculator)

  a = 1

  b = 2

  c = 1

4. Substitute Coefficients and Verify Result

  CO + 2H2 = CH3OH

      L R

  C: 1 1 ✔️

  O: 1 1 ✔️

  H: 4 4 ✔️

8 0
2 years ago
I didn’t study for this lol if your answer correct I’ll mark ur answer brainliest
kotykmax [81]

Answer:

3

Explanation:

3 is the answer is did this a little while ago

4 0
3 years ago
If 842 grams of sodium hydroxide reacts with 750.0 grams of aluminum, how many grams of aluminum hydroxide should theoretically
Phantasy [73]

548.55 grams of aluminum hydroxide should theoretically form.

Explanation:

Balanced equation for the reaction:

3 NaOH + Al ⇒ Al(OH)3 +3 Na

DATA GIVEN:

mass of NaOH = 842 grams, atomic mass =39.9 grams/mole

mass of Al = 750 grams, atomic mass = 26.9 grams/mole

aluminum hydroxide theoretical yield = ?

Moles of NaOH reacted

number of moles = \frac{mass}{atomic mass of 1 mole}

putting the values in the equation

NaOH = \frac{842}{39.9}

           = 21.1 MOLES OF NaOH

Al = \frac{750}{26.9}

   = 27.8 moles

from the equation

 from 3 moles of NaOH 1 mole of Al(OH)3 is produced

21.1 moles of NaOH will react to give x moles of Al(OH)3

\frac{1}{3} = \frac{x}{21.1}

7.03 moles of Al(OH)3 is formed.

and

1 mole of Al(OH)3 is formed from 1 mole of Al in the reaction

so, 27.8 Moles will react to give give 27.8 moles of Al(OH)3 limiting reagent of the given reaction is NaOH

mass of Al(OH)3 =7.03 x 78 (atomic mass of Al(OH)3)

          = 548.55 grams

theoretical  yield from the given data is 548.55 grams

3 0
3 years ago
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