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NemiM [27]
2 years ago
13

If your client receives a letter from SilverScript requesting proof of Creditable Coverage, and the client has already submitted

proof of creditable coverage with the enrollment application, they can ignore the letter.
Chemistry
1 answer:
KatRina [158]2 years ago
8 0

If your client receives a letter from SilverScript requesting proof of Creditable Coverage, they can ignore the letter is a false statement

<h3>What is Creditable coverage?</h3>

Creditable coverage is known to be a form of an health insurance, that consist of prescription drug and also other kinds of health benefit plan that is known to meets  the minimum amount of qualifications.

Note that  the types of creditable coverage plans are group and individual health plans, student health plans,  and others.

It is not right to ignore any letter sent to you concerning the case above as it may be vital and as such, the statement of If your client receives a letter from SilverScript requesting proof of Creditable Coverage, and the client has already submitted proof of creditable coverage with the enrollment application, they can ignore the letter is a wrong/false statement.

Learn more about Creditable Coverage from

brainly.com/question/14547867

#SPJ1

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Answer: c

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What happens to the matter in the body of an animal after it dies
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Three mixtures were prepared from three very narrow molar mass distribution polystyrene samples with molar masses of 10,000, 30,
8_murik_8 [283]

Answer:

(a). 46,666.7 g/mol; 78,571.4 g/mol

(b). 86950g/mol; 46,666.7 g/mol.

(c). 86950g/mol; 43,333.33 g/mol

Explanation:

So, we are given the molar masses for the three samples as: 10,000, 30,000 and 100,000 g mol−1.

Thus, the equal number of molecule in each sample = ( 10,000 + 30,000 + 100,000 ) / 3 = 46,666.7 g/mol.

The average molar mass = [ ( 10,000)^2 + (30,000)^2 + 100,000)^2] ÷ 10,000 + 30,000 + 100,000 = 78,571. 4 g/mol.

(b). The equal masses of each sample = 3/[ ( 1/ 10,000) + (1/30,000 ) + (1/100,000) ] = 20930.23 g/mol.

Average molar mass = ( 10,000 + 30,000 + 100,000 ) / 3 = 46,666.7 g/mol.

(c). Equal masses of the two samples = (0.145 × 10,000) + (0.855 × 100,000)/ 0.145 + 0.855 = 86950g/mol.

The weight average molar mass = 1.7 + 10,000 + 100,000/ 1.7 + 1 = 43,333.33 g/mol.

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Suppose 200.0 mL of a 2.50 M solution of sodium hydroxide is combined with 400.0 mL of a 1.50 M solution of iron(III) nitrate. W
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Explanation:

The detailed solution is shown in the image attached. First we must work out the balanced reaction equation because accurate solution of the problem must be based on the stoichiometry of the reaction. From the given concentration and volume of reactants, we calculate the amount of substance reacted hence identify the limiting reactant. Lastly we use simple proportion to obtain the theoretical yield of the precipitate. This is now used to calculate the actual yield as shown in the solution attached.

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