Bronchodilators <span>are prophylactic agents used to treat bronchoconstriction.</span>
Answer:
The answer is A.
Explanation:
C6H12O6 + 6O2 -> 6CO2 + 6H20
C6H12O6 + 6O2 -> 6CO2 + 6H20At the reactant side, there are 6 carbon atoms, 12 hydrogen atoms and 18 oxygen atoms.
atoms, 12 hydrogen atoms and 18 oxygen atoms.At product side, there are 6 carbon atoms, 12 hydrogen atoms and 18 oxygen atoms.
(In a balanced equation, the no. of atoms on the reactant side must be <em><u>e</u></em><em><u>q</u></em><em><u>u</u></em><em><u>a</u></em><em><u>l</u></em><em> </em>to the no. of atoms on the product side)
Answer:
C. The half-life of C-14 is about 40,000 years.
Explanation:
The only false statement from the options is that the half-life of C-14 is 40,000yrs.
The half-life of an isotope is the time it takes for half of a radioactive material to decay to half of its original amount. C-14 has an half-life of 5730yrs. This implies that during every 5730yrs, C-14 will reduce to half of its initial amount.
- All living organisms contain both stable C-12 and the unstable isotope of C-14
- The lower the C-14 compared to the C-12 ratio in an organism, the older it is.
First let's find out the oxidation number of Fe in K₄[Fe(CN)₆] compound.
The oxidation number of cation, K is +1. Hence, the total charge of the anion, [Fe(CN)₆] is -4. CN has charge has -1. There are 6 CN in anion. Let's assume the oxidation number of Fe is 'a'.
Sum of the oxidation numbers of each element = Charge of the compound
a + 6 x (-1) = -4
a -6 = -4
a = +2
Hence, oxidation number of Fe in [Fe(CN)₆]⁴⁻ is +2.
Now Fe has the atomic number as 26. Hence, number of electrons in Fe at ground state is 26.
Electron configuration = 1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁶ 4s² = [Ar] 3d⁶ 4s²
When making Fe²⁺, Fe releases 2 electrons. Hence, the number of electrons in Fe²⁺ is 26 - 2 = 24.
Hence, the electron configuration of Fe²⁺ = 1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁶
= [Ar] 3d⁶
Hence, the number of 3d electrons of Fe in K₄[Fe(CN)₆] compound is 6.
Answer:
1.09 moles of NaOH
Explanation:
First of all, to calculate moles, you need to find the molar mass of NaOH.
Let us first find the molar mass of NaOH then.
Na = 23.0 amu
O = 16.0
H = 1.0
They are 1 nitrogen atom, 1 oxygen atom, and one hydrogen atom.
So do this.
23.0(1) + 16.0(1) + 1.0(1) = 40 g/mol.
Now use dimensional analysis to show your work
43.5 g of NaOH * 1 mol of NaOH / 40 g/mol of NaOH
The grams cancel out.
43.5 / 40.0 = 1.0875
Use sig figs and round the answer to the nearest hundredths place.
1.0875 = 1.09
So the final answer is 1.09 moles of NaOH
Hope it helped!
