Answer:
1.) To convert between grams and moles, you would use the substance's molar mass. To go from grams to moles, divide the grams by the molar mass. 600 g58.443 g/mol = 10.27 mol of NaCl. It has been found that 1 mol of any gas at STP (Standard Temperature and Pressure = 0 °C and 1 atm) occupies 22.4 L
Answer:
58.94 mL
Explanation:
V1 = 48.3 mL V2 = v mL
T1 = 22 degree celsius OR 295 k T2 = 87 degree celsius OR 360 k
We will use the gas equation:
PV = nRT
Since the Pressure (p) , number of moles (n) and the universal gas constant(R) are all constants in this given scenario,
we can say that
V / T = k , (where k is a constant)
Since this is the first case,
V1 / T1 = k --------------------(1)
For case 2:
Since we have the same constants, the equation will be the same
V / T = k (where k is the same constant from before)
V2 / T2 = k (Since this is the second case) ------------------(2)
From (1) and (2):
V1 / T1 = V2 / T2
Now, replacing the variables with the given values
48.3 / 295 = v / 360
v = 48.3*360 / 295
v = 58.94 mL
Therefore, the final volume of the gas is 58.94 mL
They are called isotopes. One example is Carbon-14, which has 2 extra neutrons.
Chemical tests show that it is an alkaline earth metal. (a) The family of alkaline earth metals lies in group II of the periodic table indicating that each member of this family has 2 valence electrons. This therefore implies that the new element must also have only 2 electrons in its outermost orbit.
Answer: The equilibrium will shift to the left.
Explanation:
If sodium acetate is added to acetic acid, the dissociation of the acetic acid is suppressed. The equilibrium position shifts to the left and the hydrogen ion concentration decreases due to decreased ionization of the acid. This common ion solution becomes less acidic than pure acetic acid.
This is so because of common ion effect. The acetate ion already present from the dissociation of acetic acid is also present in the sodium acetate. The presence of a common ion usually shifts the equilibrium position towards the left.
NaCH3CO2 (s) → Na + (aq) + CH3CO2 −(aq)
CH3CO2H(aq) ⇌ H +(aq) + CH3CO 2− (aq)