Answer:
The most abundant isotope is 1.007 amu.
Explanation:
Given data:
Average atomic mass = 1.008 amu
Mass of first isotope = 1.007 amu
Mass of 2nd isotope = 2.014 amu
Most abundant isotope = ?
Solution:
First of all we will set the fraction for both isotopes
X for the isotopes having mass 2.014 amu
1-x for isotopes having mass 1.007 amu
The average atomic mass is 1.008 amu
we will use the following equation,
2.014x + 1.007 (1-x) = 1.008
2.014x + 1.007 - 1.007 x = 1.008
2.014x - 1.007x = 1.008 - 1.007
1.007 x = 0.001
x= 0.001/ 1.007
x= 0.0009
0.0009 × 100 = 0.09 %
0.09 % is abundance of isotope having mass 2.014 amu because we solve the fraction x.
now we will calculate the abundance of second isotope.
(1-x)
1-0.0009 = 0.9991
0.9991 × 100= 99.91%
Answer:
Yes
Explanation:
Because if you multiply it with waffles you get salad
Answer:
1.23 g/mL
-18.2%
Explanation:
We need to find the average, which is just the sum of the numbers divided by the number of numbers. Here, the sum will be 1.24 + 1.21 + 1.23 = 3.68 g/mL. There are 3 numbers, so divide 3.68 by 3: 3.68 / 3 ≈ 1.2266...
However, we need to round this and take into account significant figures. Each trial gave a number with 3 significant figures, so we round our number off to three: 1.22666... ≈ 1.23. So, circle the first number under the X column.
We now need to find the percent error, which is RE (%). To calculate this, we take the measured value (1.23 in this case) and subtract the exact value (1.50 here) from it, and then divide that by the exact value:
(1.23 - 1.50) / 1.50 ≈ -0.1822...
Again, we need to round to 3 significant figures, which would make it:
-0.1822... ≈ -0.182
Thus, circle the last number under the RE (%) column.
The answer is C. Here is the work:
<span>Moles=molarity * volume
2.5 * 0.5 = 1.25mols of NaCI
Mm of NaCI= 58.442468/mol
Multiply the moles by the Mm and you get the grams
1.25 * 58.442468 =73.1g
</span>
Answer:
gold is metal which is a very good conductor for electric current and most chips are made out of it.
