Answer:
pOH = 1.3, pH = 12.7
Explanation:
Since NaOH is a strong base, it will completely ionize; further, since it completely ionizes, our hydroxide concentration (a product of the ionization) will be the same as the given concentration of NaOH.
NaOH -> Na⁺ + OH⁻, [OH⁻] = 5.0 x 10^-2 M
pOH is the negative log of the hydroxide concentration, so plug our hydroxide concentration in:
pOH = -log[OH⁻] = -log[5.0 x 10^-2 M] = 1.3
Since pH + pOH = 14, we can plug in pOH and solve for pH:
pH + 1.3 = 14
pH = 14 - 1.3 = 12.7
Thus, our pOH = 1.3 and pH = 12.7.
Okay, so you need to start by finding the molar mass (grams in one mole) of nitrogen monoxide (NO).
N=14g
O=16g
So we know that NO has a molar mass of 30 grams, then just divide your given mass of 22.5 grams by the molar mass of 30 grams to find the number of molecules in your sample. The answer should be .75 molecules. Hope this helps!
In order to determine the concentration of ammonium ions in
the solution prepared by mixing solutions of ammonium sulfate, (NH4)2SO4, and ammonium
nitrate, first calculate the amount of ammonium ions for each solution.<span>
<span>For ammonium sulfate sol'n: 0.360 L x 0.250 mol(NH4)2SO4/ L x 2 mol NH4+ /1 mol(NH4)2SO4 =
0.18 mol NH4+
<span>For ammonium nitrate sol'n: 0.675 x 1.2 mol NH4NO3/L x 1 mol NH4+ /1 molNH4NO3
= 0.81 mol NH4+
Thus, the amount of NH4+ ions is (0.18 + 0.81) mol or 0.99
mol NH4+. To get the concentration, multiply this to the volume of solution
which is assumed to be additive, such that:</span></span></span>
M NH4+ in sol’n = 0.99 mol NH4+/1.035 L = 0.9565 mol NH4+/ L
sol’n
Plant cells<span> are </span>eukaryotic cells<span>. Prokaryotic </span>cells<span> do not contain a membrane bound nucleus, mitochondria or other membrane bound </span>cell<span> structures (organelles), the DNA of prokaryotic </span>cells<span> are located in the cytoplasm of the </span>cell<span>. ... </span>Plant cells<span> are </span>eukaryotic<span> because they have a nuclear membrane.
so therefore, A rose thorn is a eukaryotic plant cell.</span>
