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Margaret [11]
3 years ago
9

I am a little confused

Chemistry
2 answers:
Sonbull [250]3 years ago
7 0
The answer is D. I did that and i got it right.
vekshin13 years ago
3 0
The answer is what the person said above. Genes
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Calculate the ΔG for the following system. Then state if the system is spontaneous or not spontaneous.
Orlov [11]

Answer:

B

Explanation:

5 0
3 years ago
Which of the following elements is the most metallic
Marysya12 [62]
We found cesium, strontium, aluminum, sulfur, chlorine, and fluorine on the periodic table. Cesium is the farthest left and the lowest, while fluorine is the farthest right and the highest, so we know they have the highest metallic character and the lowest metallic character, respectively.
7 0
3 years ago
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6.0 moles of Na and 4.0 moles of Cl2 are mixed, how manu moles of NaCl in moles cane be made from this mixture
Sladkaya [172]

Answer:

Moles of NaCl formed is 6.0 moles

Explanation:

We are given the equation;

2 Na(s) + Cl₂(g) → 2 NaCl(s)

  • Moles of Na is 6.0 moles
  • Moles of Cl₂ is 4.0 moles

From the reaction;

2 moles of sodium reacts with 1 mole of chlorine gas to form 2 moles of NaCl

In this case;

6 moles of Na would require 3 moles of Cl₂, this means that chlorine gas is in excess.

Thus, the rate limiting reagent is sodium.

But, 2 moles of sodium reacts to form 2 moles of NaCl

Therefore;

Moles of NaCl = Moles of Na

                      = 6.0 moles

Thus, moles of sodium chloride produced is 6.0 moles

3 0
3 years ago
How many moles are present in 107 g of sodium?
cricket20 [7]

Answer:

i think 4.65 moles......

7 0
3 years ago
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(20 points) i) An absorption intensity of 1.00 (in arbitrary units) is observed for the maximum peak of the 1:2:1 triplet of the
laiz [17]

Answer:

Concentration of ethanol required =  48.476 M

Explanation:

Given that:

the absorption intensity = 1.00

Molarity of ethanol = 1M

NMR instrument used = 160 MHz

Temperature used = 300 K

The required concentration of ethanol can be determined as follows:

=  ( 1 \ M \times \dfrac{160\ MHz }{450 \ MHz}) \times \dfrac{300 \ K}{2.2\ K}

=  ( 1 \ M \times 0.3555 ) \times136.36}

= 48.476 M

5 0
3 years ago
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