Answer:
<h2>The answer is 334 g</h2>
Explanation:
The mass of a substance when given the density and volume can be found by using the formula
<h3>mass = Density × volume</h3>
From the question
volume of ethanol = 423 cm³
density = 0.789 g/cm³
So we have
mass = 0.789 × 423 = 333.747
We have the final answer as
<h3>334 g</h3>
Hope this helps you
Answer:
Explanation:
Although the context is not clear, let's look at the oxidation and reduction processes that will take place in a Fe/Sn system.
The problem states that anode is a bar of thin. Anode is where the process of oxidation takes place. According to the abbreviation 'OILRIG', oxidation is loss, reduction is gain. Since oxidation occurs at anode, this is where loss of electrons takes place. That said, tin loses electrons to become tin cation:
Similarly, iron is cathode. Cathode is where reduction takes place. Reduction is gain of electrons, this means iron cations gain electrons and produce iron metal:
The net equation is then:
However, this is not the case, as this is not a spontaneous reaction, as iron metal is more reactive than tin metal, and this is how the coating takes place. This implies that actually anode is iron and cathode is tin:
Actual anode half-equation:
Actual cathode half-equation:
Actual net reaction:
Answer:
The molar mass of lysine using the ideal gas equation for this problem is 146.25 g/mole.
Explanation:
The ideal gas equation PV = nRT, was derived from the ABC laws (Avogadros, Boyles and Charles laws). We need to obtain the value for the number of moles n.
The parameters of this equation are:
P = 1.918 atm
V = 750.0mL = 0.75L
n = ?
R = 0.0821
T = 25 degree celcius = 25 + 273 = 298 degree kelvin.
From this formular, n = (PV)/(RT)
n = (1.918 X 0.75)/(0.0821 X 298 )
n = 0.0588
n, no of mole = mass/molar mass
0.0588 = 8.6/MM
MM = 8.6/0.0588
MM = 146.25g/mole.
Answer:
Potassium selenide
Explanation:
Potassium selenide (K2Se)
Answer:
Part A: 36 MBq; Part B: 18 MBq
Explanation:
The half-life is the time it takes for half the substance to disappear.
The activity decreases by half every half-life
A =Ao(½)^n, where n is the number of half-lives.
Part A
3.0 da = 1 half-life
A = Ao(½) = ½ × 72 MBq = 36 MBq
Part B
6.0 da = 2 half-lives
A = Ao(½)^2 = ¼ × 72 MBq = 18 MBq