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Luba_88 [7]
3 years ago
12

Read the statement.

Chemistry
1 answer:
Rudiy273 years ago
6 0

Answer:

300 gr.

Explanation:

if 204gr/100gr=solubility, then according to the condition solubility=612/x.

It means, 204/100=612/x, where x=612*100/204=300 gr.

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Compute the sugar content in an 8 oz sample of a soft drink. If the sugar content as per label on the product =10g per 100ml.​
Nataly_w [17]

Answer:

m_{sugar}=23.7g\ sugar

Explanation:

Hello,

In this case, we can first compute the volume of the sample in mL from the ounces:

8oz*\frac{29.5735mL}{1oz} =236.6mL

Thus, with the volume of the sample, we can compute the amount of sugar given the 10 g of sugar per 100 mL of soft drink as shown below:

m_{sugar}=236.6mL*\frac{10g\ sugar}{100mL}\\ \\m_{sugar}=23.7g\ sugar

Best regards.

6 0
3 years ago
On the basis of the information below, the dissolution of FeF2(s) in acidic solution is A. thermodynamically favorable, because
antoniya [11.8K]

Since K3 is greater than 1, the reaction is thermodynamically favorable.

Data;

  • FeF2 is in an acidic solution
  • K2 > 1 is thermodynamically favorable
  • k3 > 1 is not thermodynamically favorable
  • K1<1 is not thermodynamically favorable

<h3>Hess Law</h3>

This states that the heat change in a given chemical reaction is the same irrespective of the number of process in which the reaction is effected.

Using Hess Law;

rxn_3 = rxn_1 * 2rxn_2\\k_3 = k_1 * k_2^2\\k_3 = 2*10^-^6 * (1*10^3)^2\\k_3 = 2

substituting this using Gibbs law,

\delta G^0_3 = -RT\ln K_3\\\delta G^0_3 < 0

Since K3 is greater than 1, the reaction is thermodynamically favorable.

Learn more on Hess law here;

brainly.com/question/12895682

8 0
2 years ago
An atom with 13 total electrons and 3 valence electrons has
zloy xaker [14]

Answer:

2

Explanation:

6 0
3 years ago
Nitrogen has two isotopes. One has an atomic mass of 14.003 amu and a relative abundance of 99.63% while the other isotope has a
Slav-nsk [51]

Answer:

Average atomic mass = 14.0067  amu.

Explanation:

Given data:

Abundance of 1st isotope  = 99.63%

Atomic mass of 1st isotope = 14.003 amu

Abundance of 2nd isotope  = 0.37%

Atomic mass of 2nd isotope = 15.000 amu

Solution:  

Average atomic mass = (abundance of 1st isotope × its atomic mass) +(abundance of 2nd isotope × its atomic mass)  / 100

Average atomic mass = (99.63×14.003)+(0.37×15.000) /100

Average atomic mass =  1395.119 + 5.55 / 100

Average atomic mass  = 1400.67 / 100

Average atomic mass = 14.0067  amu.

3 0
3 years ago
What element has an atomic number of 13
julia-pushkina [17]

Answer:

Al - Aluminum

Explanation:

3 0
3 years ago
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