Answer:
1×
Explanation:
To solve this, you will need to do multiple steps.
Firstly, you will have to convert to pOH from pH, which is done by simply subtracting 14-10 to get a pOH of 4.
Then, use the following formula to calculate the [OH-} concentration:
[OH-]= 
When converted to Scientific Notation<u>,</u> you get a resulting answer of
<u>1×</u>
<u>.</u>
Answer : The electrons in a bond between two iodine atoms are shared equally and non-polar.
Explanation :
Non-polar covalent bond : It is a type of bond that is present between the two similar and different atoms. This type of bond is formed when the electrons are shared equally between the atoms.
This type of bond is termed as non-polar because of the electronegativity difference between the atoms is negligible.
The given molecule
is formed by the equal sharing of electrons and the bond formed between the iodine-iodine is non-polar due to negligible electronegativity difference.
Therefore, the electrons in a bond between two iodine atoms are shared equally and non-polar.
TRUE OR FALSE
always find the important keyword in this case IDENTICAL which means: <span>similar in every detail; exactly alike.
FALSE because scientists all have different ways of thinking and they might observe things from a different perspective.</span><span />
Answer:
0.33moles
Explanation:
RAM of;
Na = 23, H=1, C=12, O=16
molar mass of NaHCO= 23+1+12+16= 52g/mol
mole= mass / molar mass
mole = 17.31/52
mole= 0.33moles
Answer: The number of moles of excess reagent remain unreacted will be, 6.004 moles.
Explanation : Given,
Moles of
= 1.332 mol
Mass of
= 6.504 mol
First we have to calculate the limiting and excess reagent.
The balanced chemical equation is:

From the balanced reaction we conclude that
As, 13 mole of
react with 1 mole of 
So, 6.504 moles of
react with
moles of 
From this we conclude that,
is an excess reagent because the given moles are greater than the required moles and
is a limiting reagent and it limits the formation of product.
Number of moles remain unreacted = 6.504 mol - 0.5003 mol = 6.004 mol
Therefore, the number of moles of excess reagent remain unreacted will be, 6.004 moles.