Answer:
Explanation:
q= mc theta
where,
Q = heat gained
m = mass of the substance = 670g
c = heat capacity of water= 4.1 J/g°C
theta =Change in temperature=(
66-25.7)
Now put all the given values in the above formula, we get the amount of heat needed.
q= mctheta
q=670*4.1*(66-25.7)
=670*4.1*40.3
=110704.1
Answer:
After measuring the solute, Carl should first dissolve the solid in a small amount of DI water before diluting to the total volume.
Explanation:
To ensure that all the solute dissolves in the solution, first dissolve the solid in less than the total volume of solution needed.
Answer:
4 g after 58.2 years
0.0156 After 291 years
Explanation:
Given data:
Half-life of strontium-90 = 29.1 years
Initially present: 16g
mass present after 58.2 years =?
Mass present after 291 years =?
Solution:
Formula:
how much mass remains =1/ 2n (original mass) ……… (1)
Where “n” is the number of half lives
to find n
For 58.2 years
n = 58.2 years /29.1 years
n= 2
or 291 years
n = 291 years /29.1 years
n= 10
Put values in equation (1)
Mass after 58.2 years
mass remains =1/ 22 (16g)
mass remains =1/ 4 (16g)
mass remains = 4g
Mass after 58.2 years
mass remains =1/ 210 (16g)
mass remains =1/ 1024 (16g)
mass remains = 0.0156g
Answer:
See explanation below for answers
Explanation:
We know that the balance is tared, so the innitial weight would be zero. Now, let's answer this by parts.
a) mass of displaced water.
In this case all we need to do is to substract the 0.70 with the 0.13 g. so:
mW = 0.70 - 0.13
mW = 0.57 g of water
b) Volume of water.
In this case, we have the density of water, so we use the formula for density and solve for volume:
d = m/V
V = m/d
Replacing:
Vw = 0.57/0.9982
Vw = 0.5710 mL of water
c) volume of the metal weight
In this case the volume would be the volume displaced of water, which would be 0.5710 mL
d) the mass of the metal weight.
In this case, it would be the mass when the metal weight hits the bottom which is 0.70 g
e) density.
using the above formula of density we calculate the density of the metal
d = 0.70 / 0.5710
d = 1.2259 g/mL
Explanation:
The given data is as follows.
= 253.4 nm =
(as 1 nm =
)
= 5,
= ?
Relation between energy and wavelength is as follows.
E = 
= 
=
J
= 
Hence, energy released is
.
Also, we known that change in energy will be as follows.

where, Z = atomic number of the given element


0.02 + 0.04 = 
= 
= 4
Thus, we can conclude that the principal quantum number of the lower-energy state corresponding to this emission is n = 4.