<span>440 g
First, determine the volume of each sheet. And it's easier if each dimension is using the same unit of measure. So each sheet is 28 cm by 22 cm by 0.30 cm. Multiply them together
28 cm * 22 cm * 0.30 cm= 184.8 cm^3
Since we have 2 identical sheets, double the total volume
184.8 cm^3 * 2 = 369.6 cm^3
Now multiple the volume by the density
369.6 cm^3 * 1.2 g/cm^3 = 443.52 g
Round the results to 2 significant digits since all of the given figures are only 2 significant digits long.
443.52 g = 440 g</span>
Answer:
The atomic mass of second isotope is 7.016
Explanation:
Given data:
Average Atomic mass of lithium = 6.941 amu
Atomic mass of first isotope = 6.015 amu
Relative abundance of first isotope = 7.49%
Abundance of second isotope = ?
Atomic mass of other isotope = ?
Solution:
Total abundance = 100%
100 - 7.49 = 92.51%
percentage abundance of second isotope = 92.51%
Now we will calculate the mass if second isotope.
Average atomic mass of lithium = (abundance of 1st isotope × its atomic mass) +(abundance of 2nd isotope × its atomic mass) / 100
6.941 = (6.015×7.49)+(x×92.51) /100
6.941 = 45.05235 + (x92.51) / 100
6.941×100 = 45.05235 + (x92.51)
694.1 - 45.05235 = (x92.51)
649.04765 = x
92.51
x = 485.583 /92.51
x = 7.016
The atomic mass of second isotope is 7.016
Answer:
Zero
Explanation:
FrBr is an ionic compound
.
Fr is in Group 1. Br is in Group 17.
The charges on the ions are +1 and -1, respectively.
The compound consists of Fr⁺Br⁻ ions.
However, there are equal numbers of + and - charges, so
The overall charge of the compound is zero.
Answer:
110ml
Explanation:
<em>Using the dilution equation, C1V1 = C2V2</em>
<em>Where C1 is the initial concentration of solution</em>
<em>C2 is final concentration of solution</em>
<em>V1 is intital volume of solution</em>
<em>V2 is final volume of solution.</em>
From the question , C1=6M, C2=0.5M, V1=10ml, V2=?
volume of water added = final volume -initial volume
= 120-10
=110ml
