Density = 1.01 g/cm^3 or 1.01 kg/dm^3 or 1010 kg/m^3
Density = mass/volume = 1010 g/1000 cm^3 = 1.01 g/cm^3 = 1.01 kg/dm^3
= 1010 kg/m^3
Answer:
1.375%
Explanation:
Percent Error = measured value - accepted value/ accepted value x 100
Measured Value: 15.78
Accepted Value: 16.00
Work:
* 100
* 100 = -1.375
You cannot have a negative percentage. Therefore the answer is 1.375%
Light is helpful because it allows us to see things around us.
I hope this helps :)
The idea behind balancing chemical equations is that the number of atoms an element has on the reactants' side must be equal to the number of atoms it has on the products' side.
These atoms will become a part of different compounds once the reaction is completed, but they must always be in equal numbers on both sides.
So, look at iron first. One atom reacts, but two are produced - notice the 2 subscript iron has in Fe2O3. This means you must double the number of atoms on the reactants' side to reach an equality.
2Fe(s)+O2(g)→Fe2O3(s)
Now look at oxygen. Two atoms react, but three are produced. The trick here is to find a common multiple that will make the number of atoms equal on both sides.
The easiest way to do this is to multiply the atoms that react by 3, which will give you 6 oxygen atoms that react, and the atoms that are produced by 2 - this will get you 6 oxygen atoms produced.
2Fe(s)+3O2(g)→2Fe2O3(s)
However, notice that the iron atoms are unbalanced again. You have 2 that react, but 4 that are produced → multiply the atoms that react by 2 again, which will give you
4Fe(s)+3O2(g)→2Fe2O3(s)
Answer:
A) 0.1225 M
B) 100.4 g/mol
Explanation:
Step 1: Write the generic neutralization reaction
HA(aq) + NaOH(aq) ⇒ NaA(aq) + H₂O(l)
Step 2: Calculate the reacting moles of NaOH
17.73 mL of 0.1036 M NaOH react. The reacting moles are:
0.01773 L × 0.1036 mol/L = 1.837 × 10⁻³ mol
Step 3: Calculate the reacting moles of HA
The molar ratio of HA to NaOH is 1:1. The reacting moles of HA are 1/1 × 1.837 × 10⁻³ mol = 1.837 × 10⁻³ mol.
Step 4: Calculate the molar concentration of HA
1.837 × 10⁻³ moles of HA are in a 15.00 mL volume. The molar concentration is:
M = 1.837 × 10⁻³ mol / 0.01500 L = 0.1225 M
Step 5: Calculate the molar mass of HA
1.837 × 10⁻³ moles of HA weigh 0.1845 g. The molar mass of HA is:
0.1845 g / 1.837 × 10⁻³ mol = 100.4 g/mol
