Answer:
Percentage error = 1.88 %
Solution:
Data Given:
Mass of Sample = 20.46 g
Volume of Sample = 43.0 mL - 40.0 mL = 3.0 mL
Formula Used:
Density = Mass / Volume
Putting values,
Density = 20.46 g / 3.0 mL
Density = 6.82 g.mL⁻¹
Percentage Error:
Experimental Value = 6.82 g.mL⁻¹
Accepted Value = 6.95 g.mL⁻¹
= 6.82 g.mL⁻¹ / 6.95 g.mL⁻¹ × 100 = 98.12 %
Percentage Error = 100 % - 98.12 %
Percentage error = 1.88 %
Answer:
Explanation:
We have to start with the <u>reaction</u>:
We have the same amount of atoms on both sides, so, we can continue. The next step is to find the <u>number of moles</u> that we have in the 110.0 g of carbon dioxide, to this, we have to know the <u>atomic mass of each atom</u>:
C: 12 g/mol
O: 16 g/mol
Mg: 23.3 g/mol
If we take into account the number of atoms in the formula, we can calculate the <u>molar mass</u> of carbon dioxide:
In other words: 
. With this in mind, we can calculate the moles:
Now, the <u>molar ratio</u> between carbon dioxide and magnesium carbonate is 1:1, so:
With the molar mass of 
(
. With this in mind, we can calculate the <u>grams of magnesium carbonate</u>:
I hope it helps!
Answer:
V= 12mL
Explanation:
you had the right idea with your Significant figures however, when we divide we see that it requires 2 significant figures as our least amount. this is because when looking at our division, 62 has 2 sig. fig. while 5.35 has a total 3. when looking at your answer we see that you had a total of 3 sig. figures. so in actuakity you had to round up to 12 and not to the tenths because the decimal makes .6 count as your third sig fig.
Answer:
A) The catalyzed reaction passes through C.
Explanation:
