<h3>
Answer:</h3>
2.41 g NaCl
<h3>
Explanation:</h3>
We are given the;
Molarity of NaCl = 0.150 M
Volume of the solution = 275.0 mL
We are required to calculate the mass of NaCl solute;
<h3>Step 1: Number of moles of NaCl solute </h3>
Moles are calculated by;
Moles = Molarity × Volume
Therefore;
Number of moles of NaCl = 0.150 M × 0.275 L
= 0.04125 moles
<h3>Step 2: Mass of NaCl solute </h3>
Mass of the compound is given by multiplying the number of moles by the molar mass of the compound.
Molar mass of NaCl = 58.44 g/mol
Therefore;
Mass of NaCl = 0.04125 moles × 58.44 g/mol
= 2.41 g
Thus, the mass of NaCl solute is 2.41 g
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
Answer:
2.1056L or 2105.6mL
Explanation:
We'll begin by calculating the number of mole in 10g of Na2CO3. This can be obtained as follow:
Molar mass of Na2CO3 = (23x2) + 12 + (16x3) = 106g/mol
Mass of Na2CO3 = 10g
Mole of Na2CO3 =.?
Mole = mass /molar mass
Mole of Na2CO3 = 10/106
Mole of Na2CO3 = 0.094 mole
Next, we shall determine the number of mole CO2 produced by the reaction of 0.094 mole of Na2CO3. This is illustrated below:
Na2CO3 + 2HCl —> 2NaCl + H2O + CO2
From the balanced equation above,
1 mole of Na2CO3 reacted to produce 1 mole of CO2.
Therefore, 0.094 mole of Na2CO3 will also react to 0.094 mole of CO2.
Next, we shall determine the volume occupied by 0.094 mole of CO2 at STP. This is illustrated below:
1 mole of a gas occupy 22.4L at STP. This implies that 1 mole CO2 occupies 22.4L at STP.
Now, if 1 mole of CO2 occupy 22.4L at STP, then, 0.094 mole of CO2 will occupy = 0.094 x 22.4 = 2.1056L
Therefore, the volume of CO2 produced is 2.1056L or 2105.6mL