Answer:
B. 0.7%
Explanation:
Given data:
Actual boiling point = 82.5°C
Experimental boiling point = 83.1°C
Percent error = ?
Solution:
Formula:
Percent error = ( actual value - experimental value / actual value )×100
by putting values,
Percent error = (82.5°C - 83.1 °C /82.5°C) × 100
Percent error = 0.007 × 100
Percent error = 0.7 %
Negative sign shows that experimental value is greater than accepted value. It can not written in result.
p =
<em>p equals m over V</em>
hope this helps!!
Answer:
n = 3
Explanation:
Given the formula for the transition energy of an atom with 1 electron:

For the H transition n=5 to n=2:

Then we solve for nf with Z=2 (Helium)


Is near 3, actually the energy of the transitions are:
H (5⇒2) = -2.85 eV = 434 nm (Dark blue)
He (4⇒3) = -2.64 eV = 469 nm (Light blue)
I thought it was cool to see the actual colors. Included them.
Answer:
0.031moles
Explanation:
ok so it would help if I knew the marks but 1 mole = 1000L so if the volume was 31L then if you convert you will get that answer
5.732 grams of AgCl is formed when 0.200 L of 0.200 M AGNO3 reacts with an excess of CaCl2.
Explanation:
The balanced equation:
2 AgNO3(aq) + CaCl2(aq) -----> 2 AgCl(s) + Ca(NO3)2(aq)
data given:
volume of AgNO3 = 0.2 L
molarity of AgNO3 = 0.200 M
atomic weight of AgCl= 143.32 gram/mole
from the formula, number of moles can be calculated
Molarity = 
number of moles of AgNO3 = 0.04
From the reaction:
2 moles of AgNO3 reacts to form 2 moles of AgCl
0.04 moles of AgNO3 reacts to form x mole of AgCl
= 
= 0.04 moles of AgCl is formed
mass of AgCl formed is calculated by multiplying number of moles with atomic mass of AgCl
mass of AgCl = 0.04 x 143.32
= 5.732 grams of AgCl is formed.