The effluent flow in concentration and particulate mass flow will be 0.198m³/sec.
<h3>How to calculate the effluent flow?</h3>
It should be noted that the total inflow will be equal to the total outflow. Therefore,
0.2 + 0.048 = 0.05 + We
Collect like terms
Qe = 0.2 + 0.048 - 0.05
Qe = 0.198m³/sec
The concentration will be:
= (360 × 1000)/0.05
= 7200mg/L.
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Answer:
Indeed, the two samples should contain about the same number of gas particles. However, the molar mass of
is larger than that of
(by a factor of about
.) Therefore, the mass of the
sample is significantly larger than that of the
sample.
Explanation:
The
and the
sample here are under the same pressure and temperature, and have the same volume. Indeed, if both gases are ideal, then by Avogadro's Law, the two samples would contain the same number of gas particles (
and
molecules, respectively.) That is:
.
Note that the mass of a gas
is different from the number of gas particles
in it. In particular, if all particles in this gas have a molar mass of
, then:
.
In other words,
.
.
The ratio between the mass of the
and that of the
sample would be:
.
Since
by Avogadro's Law:
.
Look up relative atomic mass data on a modern periodic table:
Therefore:
.
.
Verify whether
:
- Left-hand side:
. - Right-hand side:
.
Note that the mass of the
sample comes with only two significant figures. The two sides of this equations would indeed be equal if both values are rounded to two significant figures.
Answer:
The value of an integer x in the hydrate is 10.
Explanation:

Molarity of the solution = 0.0366 M
Volume of the solution = 5.00 L
Moles of hydrated sodium carbonate = n


Mass of hydrated sodium carbonate = n= 52.2 g
Molar mass of hydrated sodium carbonate = 106 g/mol+x18 g/mol



Solving for x, we get:
x = 9.95 ≈ 10
The value of an integer x in the hydrate is 10.
Answer:
It has 6 protons and its Carbon 14
Explanation: