3 KOH + H3PO3 = K3PO4 + 3 H2O
moles KOH = 26.5 g/56.1 g/mol=0.472
moles K3PO4 = 0.472/3 =0.157
Answer: B
Explanation:
To find the amount of heat required, you would use q=mcΔT.
q=284.2 J
*Please ignore the capital A in the equation. I can't find a way to type in the degree sign into the equation without the A appearing.
First, we have to get how many grams of C & H & O in the compound:
- the mass of C on CO2 = mass of CO2*molar mass of C /molar mass of CO2
= 0.5213 * 12 / 44 = 0.142 g
- the mass of H atom on H2O = mass of H2O*molar mass of H / molar mass of H2O
=0.2835 * 2 / 18 = 0.0315 g
- the mass of O = the total mass - the mass of C atom - the mass of H atom
= 0.3 - 0.142 - 0.0315 = 0.1265 g
Convert the mass to mole by divided by molar mass
C(0.142/12) H(0.0315/2) O(0.1265/16)
C(0.0118) H(0.01575) O(0.0079) by dividing by the smallest value 0.0079
C1.504 H3.99 O1 by rounding to the nearst fraction
C3/2 H4/1 )1/1 multiply by 2
∴ the emprical formula C3H8O2
Charcoal with a carbon-14 activity of 0.60 compared to new wood has less than 5,730 years.
<h3>What is a radioactive isotope?</h3>
A radioactive isotope is an element in nature that emit radioactivity in a given period of time (e.g., the half-life for C14 is equal to 5,730 years).
Radioactive dating is a technique to measure the age of an element by measuring its radioactive activity.
In conclusion, charcoal with a carbon-14 activity of 0.60 compared to new wood has less than 5,730 yr.
Learn more about radioactive dating here:
brainly.com/question/8831242
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Answer: So if you had 570 cm of ribbon, then 570%2F8.5=67.05 which means that about 67 students can do the experiment (round down to the nearest whole number).
Explanation: If you had 8.5 cm of ribbon, then only 8.5%2F8.5=1 student can do the experiment. If you had 17 cm of ribbon, then 17%2F8.5=2 students can do the experiment.
