<h2>
Hey There!</h2><h2>
_____________________________________</h2><h2>
Answer:</h2><h2>
_____________________________________</h2><h2>
TRIPLE POINT:</h2>
Triple point is that temperature, at which the substance is found in all three states i.e. solid, liquid and gas. Every substance has different triple point, because Triple point is just above the freezing point and each substance has different freezing and boiling point.
<h2>
_____________________________________</h2><h2>WATER:</h2>
Water has a Melting point or Freezing Point (MP or FP) of
C or 273K and Boiling point (BP)
C or 373 K. Since the triple point is just above the MP or FP thus Triple Point of water is
C or 273.16K.
<h2>
_____________________________________</h2><h2>Best Regards,</h2><h2>'Borz'</h2><h2 />
In comparing two isotopes of the same element, the atomic number, stays the same.
Allow me to explain: The protons in an isotope ALWAYS stays the same, but the neutrons change. But, what's the atomic number made up of?
Protons, which means it won't change.
However, the atomic mass is made up of neutrons + protons, so only the atomic mass would change - not the atomic number.
Answer:
40 g NaOH
Step-by-step explanation:x
To make 500 mL of a 2 mol·L⁻¹ solution, the technician must measure the correct mass of NaOH.
Step 1. Calculate the <em>moles of NaOH
</em>
c = n/V Multiply both sides by V
n = Vc
c = 2 mol·L⁻¹
V =0.5 L
n = 0.5 × 2
n = 1 mol NaOH
Step 2. Calculate the <em>mass of NaOH
</em>
The molar mass of NaOH is 40.00 g/mol.
m = 1 × 40.00/1
m = <em>40 g NaOH
</em>
The technician must measure <em>40 g of NaOH</em>.
Answer:
0.005404 M
Explanation:

Since you added an excess of sodium carbonate you warrantied that all the
in the sample reacted with it. So we can say that the insoluble lead (II) carbonate
contains all the
ions in the original sample.
The moles of
are:

One mol of
is required to form one mol of
. So, the stoichiometric relationship between them is 1:1.
Knowing this, 0.00054 is also the number of moles of
in the original sample.
So, the concentration of
in the original sample is:
