In this question, you are given the average cofactor mass per cell (41.5pg) and the total cells count(105 cells). You are asked how much cofactor that will be found from those cells(microgram= 10^6 picogram). Then the calculation would be:
Cofactor mass= cofactor per cell * cell count= 41.5pg/cell * 105 cells= 4357.5pg= 4.36 x 10^3pg
Then convert the picogram(pg) into microgram: 4.36 x 10^3pg/ (10^6pg/microgram)= 4.36x10^-3 microgram or 0.00436 microgram
if 105 cells mean 10^5 cells, the answer should be 4.15 microgram
The oxidation state of hydrogen gas is 0 and oxidation state of hydrogen cation is +1.
There’s an increase in oxidation number therefore it’s an oxidation reaction.
Oxidation reactions give out electrons. The masses and charges on both sides should be balanced
Half reaction is
H2 —> 2H+ +2e
Answer : The percent difference between the ideal and real gas is, 4.06 %.
Explanation : Given,
Ideal pressure (true value) = 49.3 atm
Real pressure (measured value) = 47.3 atm
The formula used to calculate percent difference is :
Percent difference = 
Percent difference = 
Percent difference = 4.06 %
Therefore, the percent difference between the ideal and real gas is, 4.06 %.
Answer:
0.019 moles of M2CO3
Explanation:
M2CO3(aq) + BaCl2 (aq) --> 2MCl (aq) + BaCO3(s)
From the equation above;
1 mol of M2CO3 reacts to produce 1 mol of BaCO3
Mass of BaCO3 formed = 3.7g
Molar mass of BaCO3 = 197.34g/mol
Number of moles = Mass / Molar mass = 3.7 / 197.34 = 0.0187 ≈ 0.019mol
Since 1 mol of M2CO3 reacts with 1 mol of BaCO3,
1 = 1
x = 0.019
x = 0.019 moles of M2CO3
<h3>
Answer:</h3>
19.3 g/cm³
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Explanation:</h3>
Density of a substance refers to the mass of the substance per unit volume.
Therefore, Density = Mass ÷ Volume
In this case, we are given;
Mass of the gold bar = 193.0 g
Dimensions of the Gold bar = 5.00 mm by 10.0 cm by 2.0 cm
We are required to get the density of the gold bar
Step 1: Volume of the gold bar
Volume is given by, Length × width × height
Volume = 0.50 cm × 10.0 cm × 2.0 cm
= 10 cm³
Step 2: Density of the gold bar
Density = Mass ÷ volume
Density of the gold bar = 193.0 g ÷ 10 cm³
= 19.3 g/cm³
Thus, the density of the gold bar is 19.3 g/cm³
