Red blood cells. Need glucose oxygen to convert to carbon dioxide
Answer:
8.55 × 10²⁴ Ions
Explanation:
Ammonium Chloride is an ionic compound which contains a monatomic anion (Cl⁻ ; Chloride) and a polyatomic cation (NH₄⁺ ; Ammonium).
Hence, when added in water Ammonium Chloride ionizes as;
NH₄Cl → NH₄⁺ + Cl⁻
Hence, we can say that it produces two ions when dissolved in water.
Also,
We know that 1 mole of any substance contains exactly 6.022 × 10²³ particles which is also called as Avogadro's Number. So in order to calculate the number of ions contained by 7.1 moles of NH₄Cl, we will use following relation to first calculate the number of molecules as;
Moles = Number of Molecules ÷ 6.022 × 10²³ Molecules.mol⁻¹
Solving for Number of Molecules,
Number of Molecules = Moles × 6.022 × 10²³ Molecules.mol⁻¹
Putting values,
Number of Molecules = 7.1 mol × 6.022 × 10²³
Number of Molecules = 4.27 × 10²⁴ Molecules
So,
As,
1 Molecule of NH₄Cl contained = 2 Ions
So,
4.27 × 10²⁴ Molecules of NH₄Cl will contain = X ions
Solving for X,
X = 2 Ions × 4.27 × 10²⁴ Molecules / 1 Molecule
X = 8.55 × 10²⁴ Ions
The population would decrease, the more predators there are the more food needed for the species .
Answer:
Mass of original sample = 100 g
Explanation:
Half life of cesium-137 = 30.17 years
Where, k is rate constant
So,
The rate constant, k = 0.02297 year⁻¹
Time = 90.6 years
Using integrated rate law for first order kinetics as:
![[A_t]=[A_0]e^{-kt}](https://tex.z-dn.net/?f=%5BA_t%5D%3D%5BA_0%5De%5E%7B-kt%7D)
Where,
![[A_t]](https://tex.z-dn.net/?f=%5BA_t%5D)
is the concentration at time t
![[A_0]](https://tex.z-dn.net/?f=%5BA_0%5D)
is the initial concentration
Initial concentration = ?
Final concentration = 12.5 grams
Applying in the above equation, we get that:-
![12.5\ g=[A_0]e^{-0.02297\times 90.6}](https://tex.z-dn.net/?f=12.5%5C%20g%3D%5BA_0%5De%5E%7B-0.02297%5Ctimes%2090.6%7D)
![[A_0]=\frac{12.5}{e^{-0.02297\times 90.6}}\ g=100\ g](https://tex.z-dn.net/?f=%5BA_0%5D%3D%5Cfrac%7B12.5%7D%7Be%5E%7B-0.02297%5Ctimes%2090.6%7D%7D%5C%20g%3D100%5C%20g)
<u>Mass of original sample = 100 g</u>
This is the balanced eq
N2 + 3H2 -> 2NH3
first you need to find mole of N2 by using
mol = mass ÷ molar mass.
mol N2= 20g ÷ (14.01×2)g/mol
=0.7138mol
then look at the coefficient between H2 and NH3.
it is N2:NH3
1:2
0.7138:0.7138×2
0.7138:1.4276 moles
moles of NH3 = 1.4276 moles
