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BaLLatris [955]
3 years ago
12

How many moles of calcium ion 40.0 g of calcium

Chemistry
1 answer:
VladimirAG [237]3 years ago
8 0
Look on this website! 
http://lawr.ucdavis.edu/classes/ssc100/CEC_Answers02.htm

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Question 3 of 11
daser333 [38]

Answer:

c

Explanation:

b and d are out, the variables are changed. a would be a repetition, not a replication. c uses the same method and variables with a different control group

6 0
3 years ago
An oxygen atom has a mass of 2.66 x 10^-23 g and a glass of water has a mass of 0.050kg. Use this information to answer the ques
fomenos
<h2>ANSWER OF EACH PART ARE GIVEN BELOW</h2>

Explanation:

A)

We know, each mole contains N_A= 6.023 \times 10^{23} atoms.

It is given that mass of one oxygen atom is m= 2.66\times 10^{-23}\ g.

Therefore, mass of one mole of oxygen, M=m\times N_A.

Putting value of n and N_A,

M=2.66\times 10^{-23}\times 6.023\times 10^{23} \ gm\\M=16.0\ gm

B)

Given,

Mass of water in glass=0.050 kg = 50 gm.

From above part mass of one mole of oxygen atoms = 16.0 gm.

Therefore, number of mole of oxygen equivalent to 50 gm oxygen=\dfrac{50}{16}=3.1 \ moles.

LEARN MORE :

Avogadro's number

brainly.com/question/12902286

3 0
3 years ago
4. Consider the following half-reactions: MnO4–(aq) + 8H+(aq) + 5e– → Mn+2(aq) + 4H2O(l) NO3–(aq) + 4H+(aq) + 3e– → NO(g) + 2H2O
jeyben [28]

Explanation:

The chemical reaction given in the question is as follows -

MnO₄⁻ (aq) + 8H⁺ (aq) + 5e⁻ → Mn²⁺ (aq) + 4H₂O (l)

NO₃⁻ (aq) + 4H⁺ (aq) + 3e⁻ → NO (g) + 2H₂O (l)

As we know , the value for reduction potential are -

Mn²⁺ = + 1.51  V

NO₃⁻  =  +0.96 V

From , the data given above , the value of the reduction potential of NO₃⁻ is less than the reduction potential of Mn²⁺ .

Hence ,

NO₃⁻  can not oxidize Mn²⁺ .

5 0
3 years ago
Lucia wants to develop a process for dyeing shirts so that the color will not fade when the shirt is washed. She proceeds with t
Firlakuza [10]
I cant because it not full
3 0
3 years ago
Read 2 more answers
I begin the reaction with 0.45 g of beryllium. If my actual yield of beryllium chloride (mm = 79.91 g/mol) was 3.5 grams, what w
trasher [3.6K]

The percentage of yield was 777.78%

<u>Explanation:</u>

We have the equation,

Be [s]  +  2 HCl [aq]  →  BeCl 2(aq]  + H 2(g]  ↑ Be (s]  + 2 HCl [aq]  →  BeCl 2(aq]  + H 2(g] ↑

To find the percent yield we have the formula

Percentage of Yield= what you actually get/ what you should theoretically get  x 100

                                   =3.5 g/0.45 g 100

                                    = 777.78 %

The percentage of yield was 777.78%

5 0
3 years ago
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