Answer:
The answer to your question is: 234.7 cans
Explanation:
data
caffeine concentration = 3.55 mg/oz
10.0 g of caffeine is lethal
there are 12 oz of caffeine in a can
Then
3.55 mg ----------------- 1 oz
x mg -----------------12 oz (in a can)
x = 42.6 mg of caffeine in a can
Convert it to grams 42,6 mg = 0.0426 g of caffeine in a can
Finally
0.0426 g of caffeine ------------------ 1 can
10 g of caffeine ----------------- x
x = 10 x 1/0.0436 = 234.7 cans
An atom that undergoes radioactive decay and has a large nucleus most likely contains....<span>- More protons than electrons.</span>
<span>C. 11.2 L
There are several different ways to solve this problem. You can look up the density of CO2 at STP and work from there with the molar mass of CO2, but the easiest is to assume that CO2 is an ideal gas and use the ideal gas properties. The key property is that a mole of an idea gas occupies 22.413962 liters. And since you have 0.5 moles, the gas you have will occupy half the volume which is
22.413962 * 0.5 = 11.20698 liters. And of the available choices, option "C. 11.2 L" is the closest match.
Note: The figure of 22.413962 l/mole is using the pre 1982 definition of STP which is a temperature of 273.15 K and a pressure of 1 atmosphere (1.01325 x 10^5 pascals). Since 1982, the definition of STP has changed to a temperature of 273.15 K and a pressure of exactly 10^5 pascals. Because of this lower pressure, one mole of an ideal gas will have the higher volume of 22.710947 liters instead of the older value of 22.413962 liters.</span>
Answer:
Yes the two of the answer is True
Answer:
IUPAC Rules for Alkane Nomenclature
Find and name the longest continuous carbon chain.
Identify and name groups attached to this chain.
Number the chain consecutively, starting at the end nearest a substituent group.
Designate the location of each substituent group by an appropriate number and name.
Explanation:
