<span>The correct answer should be two oxygen atoms. That's because it's properties are similar to carbon insofar that it can form four bonds, so if it forms bonds with 2 oxygen atoms then it will have all four bonds created since Oxygen forms double bonds. This would make SiO2 which is also known worldwide as silica.</span>
Correct question is;
Resistance of a given wire is obtained by measuring the current flowing in it and the voltage difference applied across it. If the percentage errors in the measurement of the current and the voltage difference are 3% each, then error in the value of resistance of the wire is?
Answer:
6%
Explanation:
From ohms law, we know that;
R = V/I
Where;
R is resistance
V is voltage
I is current
Now, the percentage error in the resistance is given by the formula;
ΔR/R = ΔV/V+ ΔI/I
We are told that the current and the voltage difference have a percentage error of 3% each. Thus;
ΔR/R = 3% + 3% = 6%
NH4^+ + OH^- ==> NH3 + H2O
<span>Set up an ICE chart. </span>
<span>initial: </span>
<span>NH4^+ = mL x M = 100 x 0.1 = 10 millimoles </span>
<span>NH3 = mL x M = 80 x 0.20 = 16 millimoles. </span>
<span>change: </span>
<span>we add 0.2 g NaOH which is 0.2/40 = 0.005 moles or 5 millimoles. </span>
<span>NH3 = +5 millimoles </span>
<span>NH4^+ = -5 millimoles </span>
<span>equilibrium: </span>
<span>NH3 = 16 + 5 = 21 millimoles. </span>
<span>NH4^+ = 10 - 5 = 5 millimoles. </span>
<span>You may substitute millimoles in place of concn (since millimoles/mL = molarity and the mL (180 mL) appears in both numerator and denominator) OR you can divide millimoles/180 mL to arrive at concn for both base and acid and substitute those numbers. Plug those into the HH equation and solve for pH. </span>
<span>For part d, just set up the HH equation and pH = pKa + log (base/acid). The question is asking you to calculate pH if base and acid were equal. So plug in the same number (any number you choose) for base and acid and calculate. Note that the log of 1 = 0</span>
Answer:
11460 years
Explanation:
0.693/t1/2 = 2.303/t log (No/N)
t1/2 = half life of the carbon
t = age of the fossil
No= amount of radioactive material originally present
N= amount of radioactive material present at time=t
No= mass of carbon + nitrogen = 5g
0.693/5730 = 2.303/t log (5/1.25)
1.21 ×10^-4 = 1.3866/t
t= 1.3866/1.21 ×10^-4
t= 11460 years