Answer is: Filtering a mixture will separate it into substances.
Pure substance is made of only one type of atom (element) or only one type of molecule, it has definite and constant composition with distinct chemical properties. Pure substances can be separated chemically, not physically, that is difference between pure substances and mixtures.
Filtration is use for separation of heterogenous mixture, for example, sand and water.
Answer:
Sodium dihydrogenphosphate = NaH₂PO₄
Sodium monohydrogenphosphate = Na₂HPO₄
Explanation:
A buffer solution is a solution is a solution that resists changes to its oH when a little quantity of strong acid or strong base is added to it.
They are solutions of weak acids or weak bases and their salts known as conjugate base or conjugate acids respectively for the weak acids and weak bases.
For example, a solution of the weak acid ethanoic acid and its salt or conjugate base, sodium ethanoate serves as a buffer solution.
In biochemical experiments, where the pH of the reaction medium is kept as constant and as close as possible to that of the internal environment, buffer solutions are widely used. One of the commonly used buffers is the phosphate buffer. The phosphate buffer consists of the acid salts sodium dihydrogenphosphate and sodium monohydrogenphosphate. Sodium dihydrogenphosphate serves as the weak acid while sodium monohydrogenphosphate serves as the conjugate base.
The formulas of these two compounds are given below:
Sodium dihydrogenphosphate = NaH₂PO₄
Sodium monohydrogenphosphate = Na₂HPO₄
Answer:
1223.38 mmHg
Explanation:
Using ideal gas equation as:
where,
P is the pressure
V is the volume
n is the number of moles
T is the temperature
R is Gas constant having value = 
Also,
Moles = mass (m) / Molar mass (M)
Density (d) = Mass (m) / Volume (V)
So, the ideal gas equation can be written as:
Given that:-
d = 1.80 g/L
Temperature = 32 °C
The conversion of T( °C) to T(K) is shown below:
T(K) = T( °C) + 273.15
So,
T = (32 + 273.15) K = 305.15 K
Molar mass of nitrogen gas = 28 g/mol
Applying the equation as:
P × 28 g/mol = 1.80 g/L × 62.3637 L.mmHg/K.mol × 305.15 K
⇒P = 1223.38 mmHg
<u>1223.38 mmHg must be the pressure of the nitrogen gas.</u>
Answer is in picture below.
Use 100 grams of the compound:
ω(Cl) = 85.5% ÷ 100%.
ω(Cl) = 0.855; mass percentage of the chlorine in the compound.
m(Cl) = 0.855 · 100 g.
m(Cl) = 85.5 g; mass of chlorine.
m(C) = 100 g - 85.5 g.
m(C) = 14.5 g; mass of carbon.
n(Cl) = m(Cl) ÷ M(Cl).
n(Cl) = 85.5 g ÷ 35.45 g/mol.
n(Cl) = 2.41 mol; amount of chlorine.
n(C) = 14.5 g ÷ 12 g/mol.
n(C) = 1.21 mol; amount of carbon.
n(Cl) : n(C) = 2.41 mol : 1.21 mol = 2 : 1.
This compound is dichlorocarbene CCl₂.
