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Reil [10]
2 years ago
13

Silver (ag) has a molar mass of 107.8682 g, nitrogen has a molar mass of 14.0067 g, and oxygen has a molar mass of 15.9994 g. wh

at is the molar mass of silver nitrate, agno3?
Chemistry
2 answers:
alina1380 [7]2 years ago
4 0

Answer : The molar mass of silver nitrate (AgNO_3) is, 169.87 g

Explanation :

Molar mass : It is defined as mass of 1 mole of a substance.

Given:

Molar mass of Sliver (Ag) = 107.8682 g

Molar mass of nitrogen (N) = 14.0067 g

Molar mass of oxygen (O) = 15.9994 g

Now we have to calculate the molar mass of silver nitrate (AgNO_3)

Molar mass of silver nitrate (AgNO_3) = Molar mass of Sliver (Ag) + Molar mass of nitrogen (N) + (3 × Molar mass of oxygen (O))

Molar mass of silver nitrate (AgNO_3) = 107.8682 g + 14.0067 g + 3(15.9994 g)

Molar mass of silver nitrate (AgNO_3) = 169.87 g

Therefore, the molar mass of silver nitrate (AgNO_3) is, 169.87 g

ASHA 777 [7]2 years ago
3 0
AgNO3=
(ag + n + o3) =  \\  (107.8682) + (14.0067) + (15.9994 \times 3) \\  = 169.8731g \: per \: mole
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Review the list of common titration errors. Determine whether each error would cause the calculation for moles of analyte to be
kakasveta [241]

Answer and Explanation:

<em>A funnel is in the top of the buret and a beaker is positioned underneath the buret:</em> This is correct and is necessary to fill the buret, but the funnel and the beaker has to be removed before the titration starts. The calculation for moles of analyte does not affect.

<em>A solution is being poured from a bottle into the buret via the funnel:</em> Using a funnel helps to fill the burette but it must be removed to filling the buret at 0.0 mL. In this case, the calculation for moles of analyte do not affect.

<em>Adding titrant past the color change of the analyte solution</em>: In this case, an excess of titrant is added, thus the calculation for moles of anality will be higher than it should be.

<em>Recording the molarity of titrant as 0.1 M rather than its actual value of 0.01 M</em>: In this case, the titrant is considered more concentrated than it is hence, the calculation for moles of anality will be higher than it should be.

<em>Spilling some analyte out of the flask during the titration</em>: The excess of titrant spilled out of the flask higher up the volume of titrant measured. Therefore, the calculation for moles of anality will be higher than it should be.

<em>Starting the titration with air bubbles in the buret</em>: The air inside the burette occupies measured volume, thus the volume of titrant measured will be higher than the real volume spilled in the flask. Hence the calculation for moles of anality will be higher than it should be.

<em>Filling the buret above the 0.0 mL volume mark</em>: Some volume of titrant will be spilled inside the flask but will no be measured since the buret measures the titrant below the 0.0mL mark, thus the calculation for moles of anality will be lower than it should be.

3 0
2 years ago
A chemist adds 180.0 mL of a 1.42M sodium carbonate (Na CO,) solution to a reaction flask. Calculate the millimoles of sodium ca
svet-max [94.6K]

Answer: The millimoles of sodium carbonate the chemist has added to the flask are 256

Explanation:

Molarity is defined as the number of moles dissolved per liter of the solution.

To calculate the number of moles for given molarity, we use the equation:

\text{Molarity of the solution}=\frac{\text{milli moles of solute}}{\text{Volume of solution in ml}}     .....(1)

Molarity of BaCl_2 solution = 1.42 M

Volume of solution = 180.0 mL

Putting values in equation 1, we get:

1.42M=\frac{\text{milli moles of }BaCl_2}{180.0ml}\\\\\text{milli moles of }BaCl_2}={1.42M\times 180.0ml}=256milli mol

Thus the millimoles of sodium carbonate the chemist has added to the flask are 256.

6 0
2 years ago
James was given a sample of sodium bicarbonate NaHCO3 in a weighing dish to analyze. His teacher asked him to find the percent o
gtnhenbr [62]
This question could be answered easily if the results of the abundance of the other elements are given. You will just have to subtract the sum of all their abundances to 100. Since it's not given, the solution would just be:

Na = 23 g/mol* 1 = 23 g
H = 1 g/mol * 1 = 1 g
C = 12 g/mol * 1 = 12 g
O = 16 g/mol * 3 = 48 g
Total = 84 g

% O = 48/84 * 100 = <em>57.14%</em>

5 0
2 years ago
Read 2 more answers
mastering chem If the experiment below is run for 60 s, 0.16 mol A remain. Which of the following statements is or are true? At
Dvinal [7]

Answer: All of the statements are true.

Explanation:

(a) Considering the system mentioned in the equation:-

The sum of total moles in the flask will always be equal to 1 which leads to confirmation of this statement as for 60 secs= 0.16 mol A and 0.84 mol B

(b) 0<t< 20s,  mole A got reduced from 1 mole to 0.54 moles while at 40s to 60s A got decreased from 0.30 moles to 0.16 moles.

0 to 20s is 0.46 (1 - 0.54 = 0.46)mol whereas,

40 to 60s is 0.14 (0.30-.16 = 0.14) mol

(0.46 > 0.14) mol leading this statement to be true as well.

(c) Average rate from t1 = 40 to t2 = 60 s is given by:

\delta moles/\delta time  = 0.30-0.16/60-40 = 0.007 Mol/s which is true as well

7 0
2 years ago
If the concentration of a saturated solution at 0∘C is 12.5 gCuSO4/100 g soln, what mass of CuSO4⋅5H2O would be obtained? [Hint:
Andrew [12]

Answer:

Mass of CuSO4.H2O obtained: m_{total}=19.52 g

Explanation:

The molecular weight of the salt is: M=159.5 g/mol

<u>In the solution</u>: 12.5 g of CuSO4

In moles: n=\frac{12.5 g}{159.5 g/mol}=0.078 mol

<u>Mass of wate</u>r:

5 moles of water per mol of salt: m=\frac{5mol}{1mol}*0.078mol*\frac{18g}{mol}=7.02 g

Mass of CuSO4.H2O obtained: m_{total}=12.5 g + 7.02 g=19.52 g

7 0
2 years ago
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