Percent error is the difference between the measured and known value, divided by the known value, multiplied by 100%.
So first, we take our measured value, .299 cm, minus our known value, .225 cm.
.299 cm - .225 cm=.004 cm
Next, we divide that by our known value

Finally, multiply your answer by 100
.0177777778 x 100= 1.77777778 %
Round to three significant figures, and you're done.
=1.78 % error
Answer:
it forms :
1. Gold ( Au )
2. Zinc nitrate ( Zn(NO3)2 )
Explanation:
When a chunk of zinc is added to a solution of gold (III) nitrate to extract the gold. The reaction forms Gold and Zinc nitrate .
it's a single displacement reaction,
here's the balanced equation for above reaction :
3 Zn + 2 Au(NO3)3 =》3 Zn(NO3)2 + 2 Au
Answer:

Explanation:
Hello,
In this case, the undergoing chemical reaction is:

Next, we identify the limiting reactant by computing the available moles of ethane and the moles of ethane consumed by 60.0 grams of oxygen:

Thus, we notice there are less available moles, for that reason, the ethane is the limiting reactant. Finally, we can compute the produced moles of water by:

Best regards.
Answer: The ionic formula of compound is
and the name is cobalt phosphide.
Explanation:
For formation of a neutral ionic compound, the charges on cation and anion must be balanced. The cation is formed by loss of electrons by metals and anions are formed by gain of electrons by non metals.
Here element Co is having an oxidation state of +3 called as
cation and phosphprous forms
anion with oxidation state of -3. Thus they combine and their oxidation states are exchanged and written in simplest whole number ratios to give neutral
The nomenclature of ionic compounds is given by:
1. Positive is written first followed by the oxidation state of metal in roman numerals in square brackets.
2. The negative ion is written next and a suffix is added at the end of the negative ion. The suffix written is '-ide'.
Thus the name of
is cobalt phosphide.
Answer:
in the periodic table we can see that the ions of Cl is greater than the ions in Na