Answer:
The answer is "W-4 form".
Explanation:
The W-4 form is part of the Internal Revenue Service tax form, which is used to submitted by the U.S. government form the customer to information and his or her tax situation. This form informs the customer of the right amount of tax to deduct from his or her salary.
That's why she filled out a W-4 form and know-how often tax he will subtract in the payroll depending on provisions dealing, dependents, potential tax deductions, and credits, etc.
True
Enzymes help the body in digestion and other bodily functions which involve chemicals.
When carbon reacts with oxygen it forms CO2. This can depicted by the below equation.
C + O2→ CO2
It has been mentioned that when 14.4 g of C reacts with 53.9 g of O2, then 15.5 g of O2 remains unreacted. <u>This indicates that Carbon is the limiting reagent and hence the amount of CO2 produced is based on the amount of Carbon burnt.</u>
C + O2→ CO2
In the above equation , 1 mole of carbon reacts with 1 mole of O2 to produce 1 mole of CO2.
In this case 14.4 g of Carbon reacts with 53.9 of O2 to produce "x"g of CO2.
<u>No of moles = mass of the substance÷molar mass of the substance</u>
No of moles of carbon = 14.4 /12= 1.2 moles
No of moles of O2 = Mass of reacted O2/Molar mass of O2.
No of moles of O2 = (Total mass of O2 burned - Mass of unreacted O2)/32
No of moles of O2 = (53.9-15.5) ÷ 32 = 1.2 moles.
Hence as already discussed 1 mole of Carbon reacts with 1 mole of O2 to produce 1 mole of CO2. In this case 1.2 moles of carbon reacts with 1.2 moles of O2 to produce 1.2 moles of CO2.
Moles of carbon dioxide = Mass of CO2 produced /Molar mass of CO2
Mass of CO2 produced(x) = Moles of CO2 ×Molar mass of CO2
Mass of CO2 produced(x) = 1.2 x 44 = 52.8 g
<u>Thus 52.8 g of CO2 is produced.</u>
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
7.32 moles of Chromium is present in 4.41 × 10²⁴ atoms.
<h3>How to find the number of moles ?</h3>
Number of moles = 
<h3>What is Avogadro's Number ?</h3>
Avogadro's number is the number of particles in one mole of substance. 6.022 × 10²³ is known as Avogadro's Constant / Avogadro's Number.
Avogadro's Number = 6.022 × 10²³
Now put the values in above formula we get
Number of moles =
= 
= 7.32 moles
Thus from the above conclusion we can say that 7.32 moles of Chromium is present in 4.41 × 10²⁴ atoms.
Learn more about the Avogadro's Number here: brainly.com/question/1581342
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