Answer:
endo takes energy in and exo releases it out
Explanation:
Answer:
The answer is: <u>Al2O3</u>
Explanation:
The data they give us is:
To find the empirical formula without knowing the grams of the compound, we find it per mole:
- 0.545 g Al * 1 mol Al / 27 g Al = 0.02 mol Al
- 0.485 g O * 1 mol O / 16 g O = 0.03 mol O
Then we must divide the results obtained by the lowest result, which in this case is 0.02:
- 0.02 mol Al / 0.02 = 1 Al
- 0.03 mol O / 0.02 = 1.5 O
Since both numbers have to give an integer, multiply by 2 until both remain integers:
Now the answer is given correctly:
These problems are a bit interesting. :)
First let's write the molecular formula for ammonium carbonate.
NH4CO3 (Note! The 4 and 3 are subscripts, and not coefficients)
17.6 gNH4CO3
Now to convert to mol of one of our substances we take the percent composition of that particular part of the molecule and multiply it by our starting mass. This is what it looks like using dimensional analyse.
17.6 gNH4CO3 * (Molar Mass of NH4 / Molar Mass of NH4CO3)
Grab a periodic table (or look one up) and find the molar masses for these molecules! Well. In this case I'll do it for you. (Note: I round the molar masses off to two decimal places)
NH4 = 14.01 + 4*1.01 = 18.05 g/mol
NH4CO3 = 14.01 + 4*1.01 + 12.01 + 3*16.00 = 78.06 g/mol
17.6 gNH4CO3 * (18.05 molNH4 / 78.06 molNH4CO3)
= 4.07 gNH4
Now just take the molar mass we found to convert that amount into moles!
4.07 gNH4 * (1 molNH4 / 18.05 gNH4) = 0.225 molNH4
The [OH⁻] of the solution is 5.37×10⁵ M.
<h3 /><h3>What is pOH?</h3>
This is the negative logarithm to base 10 of hydroxy ion [OH⁻] concentration.
To calculate the hydroxy ion [OH⁻] concentration we use the formula below.
Note:
- pOH = 14-pH
- pOH = 14-9.77
- pOH = 4.27
Formula:
- [OH⁻] = 1/
................. Equation 1
Given:
Substitute the value into equation 1
- [OH⁻] = 1/
- [OH⁻] = 5.37×10⁵
Hence, The [OH⁻] of the solution is 5.37×10⁵ M.
Learn more about hydroxy ion concentration here: brainly.com/question/17090407
