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nirvana33 [79]
3 years ago
11

Which of the following is NOT an example of a chemical reaction

Chemistry
1 answer:
Charra [1.4K]3 years ago
8 0
D water evaporating it is a physical change
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When sodium chloride is dissolved in water, the resulting solution is classified as a?
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A homogeneous mixture is formed when it is dissolved in water
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Rank each of the following gases in order of increasing urms assuming equivalent amounts and all gases are at the same temperatu
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68.3 grams of sodium hydroxide reacts with 78.3 grams of magnesium nitrate. ____ grams of magnesium hydroxide will form from thi
Vera_Pavlovna [14]

Answer:

30.8 grams of magnesium hydroxide will form from this reaction, and magnesium nitrate is the limiting reagent.

Explanation:

The reaction that takes place is:

  • 2NaOH + Mg(NO₃)₂ → 2NaNO₃ + Mg(OH)₂

Now we <u>convert the given masses of reactants to moles</u>, using their respective <em>molar masses</em>:

  • 68.3 g NaOH ÷ 40 g/mol = 1.71 mol NaOH
  • 78.3 g Mg(NO₃)₂ ÷ 148.3 g/mol = 0.528 mol Mg(NO₃)₂

0.528 moles of Mg(NO₃)₂ would react completely with (0.528 * 2) 1.056 moles of NaOH. There are more than enough NaOH moles, so NaOH is the reagent in excess and <em>Mg(NO₃)₂ is the limiting reagent.</em>

Now we <u>calculate how many Mg(OH)₂ are produced</u>, using the <em>moles of the limiting reagent</em>:

  • 0.528 mol Mg(NO₃)₂ * \frac{1molMg(OH)_2}{1molMg(NO_3)_2} = 0.528 mol Mg(OH)₂

Finally we convert Mg(OH)₂ moles to grams:

  • 0.528 mol Mg(OH)₂ * 58.32 g/mol = 30.8 g
7 0
3 years ago
What is the wavelength of the matter wave associated with an electron (me= 9.1 x 10-31 kg) moving with a speed of 2.5 x 107 m/s?
Novay_Z [31]
Lambda = h\ Mv
lambda = 6.624 x 10^-34 / 9.1 x 10^-31 x 2.5 x 10^7
lambda = 2.9 x 10^-11 is your wavelength
3 0
3 years ago
A total of 20.0 mL of sodium hydroxide (NaOH) was neutralized by 30.0 mL of 0.250 M hydrogen bromide (HBr). What was the concent
Karolina [17]

Answer:

0.375 M

Explanation:

NaOH(aq) + HBr(aq) ------------> NaBr(aq) + H2O(l)

Concetration of acid CA= 0.250M

Concentration of base CB= ????

Volume of acid VA= 30.0mL

Volume of base VB= 20.0mL

Number of moles of acid nA= 1

Number of moles of base nB= 1

CA VA/CB VB= nA/nB

CB= CAVAnB/VB nA

= 0.25× 30×1/20×1= 0.375 M

8 0
3 years ago
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