Answer:
m = 0.3249 g
Explanation:
First, I'm assuming you have a reaction of mercury(II) oxyde descomposition. If this is the case, then the equation to use is the following:
HgO ---------> Hg + O2
Balancing the equation:
2HgO ----------> 2Hg + O2
This means that 2 moles of HgO reacts to produce 1 mole of O2, so, we first calculate the moles of O2, then, the moles of HgO and finally the mass:
We have the volume of O2, the pressure and temperature, so let's use the ideal gas equation:
PV = nRT
Solving for n:
n = PV/RT
R: 0.082 L atm / K mole
T = 70 + 273 = 343 K
V = 83 / 1000 = 0.083 L
Calculating n:
n = 1 * 0.083 / 0.082 * 343
n = 0.003 moles
as stated before, 2 moles of HgO reacts with 1 mole of oxygen so:
2 moles HgO = 1 moles O2
moles HgO = moles O2 / 2
moles HgO = 0.003 / 2 = 0.0015 moles
Finally, to calculate the mass:
m = n * MM
the molar mass of HgO is 216.59 g/mol, so replacing:
m = 0.0015 * 216.59
m = 0.3249 g
<u>Answer:</u> The volume of NaOH solution required to reach the half-equivalence point is 0.09 mL
<u>Explanation:</u>
The chemical equation for the dissociation of butanoic acid follows:
![CH_3CH_2CH_2COOH\rightleftharpoons CH_3CH_2CH_2COO^-+H^+](https://tex.z-dn.net/?f=CH_3CH_2CH_2COOH%5Crightleftharpoons%20CH_3CH_2CH_2COO%5E-%2BH%5E%2B)
The expression of
for above equation follows:
![K_a=\frac{[CH_3CH_2CH_2COO^-][H^+]}{[CH_3CH_2CH_2COOH]}](https://tex.z-dn.net/?f=K_a%3D%5Cfrac%7B%5BCH_3CH_2CH_2COO%5E-%5D%5BH%5E%2B%5D%7D%7B%5BCH_3CH_2CH_2COOH%5D%7D)
We are given:
![[CH_3CH_2CH_2COOH]=0.888M\\K_a=1.54\times 10^{-5}](https://tex.z-dn.net/?f=%5BCH_3CH_2CH_2COOH%5D%3D0.888M%5C%5CK_a%3D1.54%5Ctimes%2010%5E%7B-5%7D)
![[CH_3CH_2CH_2COO^-]=[H^+]](https://tex.z-dn.net/?f=%5BCH_3CH_2CH_2COO%5E-%5D%3D%5BH%5E%2B%5D)
Putting values in above expression, we get:
![1.54\times 10^{-5}=\frac{[H^+]^2}{0.888}](https://tex.z-dn.net/?f=1.54%5Ctimes%2010%5E%7B-5%7D%3D%5Cfrac%7B%5BH%5E%2B%5D%5E2%7D%7B0.888%7D)
![[H^+]=-0.0037,0.0037](https://tex.z-dn.net/?f=%5BH%5E%2B%5D%3D-0.0037%2C0.0037)
Neglecting the negative value because concentration cannot be negative
To calculate the volume of base, we use the equation given by neutralization reaction:
![n_1M_1V_1=n_2M_2V_2](https://tex.z-dn.net/?f=n_1M_1V_1%3Dn_2M_2V_2)
where,
are the n-factor, molarity and volume of acid which is butanoic acid
are the n-factor, molarity and volume of base which is NaOH.
We are given:
![n_1=1\\M_1=\frac{0.0037}{2}M\text{ (half equivalence)}\\V_1=20.00mL\\n_2=1\\M_2=0.425M\\V_2=?mL](https://tex.z-dn.net/?f=n_1%3D1%5C%5CM_1%3D%5Cfrac%7B0.0037%7D%7B2%7DM%5Ctext%7B%20%28half%20equivalence%29%7D%5C%5CV_1%3D20.00mL%5C%5Cn_2%3D1%5C%5CM_2%3D0.425M%5C%5CV_2%3D%3FmL)
Putting values in above equation, we get:
![1\times \frac{0.0037}{2}\times 20.00=1\times 0.425\times V_2\\\\V_2=\frac{1\times 0.0037\times 20}{1\times 0.425\times 2}=0.087mL](https://tex.z-dn.net/?f=1%5Ctimes%20%5Cfrac%7B0.0037%7D%7B2%7D%5Ctimes%2020.00%3D1%5Ctimes%200.425%5Ctimes%20V_2%5C%5C%5C%5CV_2%3D%5Cfrac%7B1%5Ctimes%200.0037%5Ctimes%2020%7D%7B1%5Ctimes%200.425%5Ctimes%202%7D%3D0.087mL)
Hence, the volume of NaOH solution required to reach the half-equivalence point is 0.09 mL
Answer:
B. a type of carbohydrate
Explanation:
Sugar is a type of carbohydrate.
Hope it helped brainiest plz
1 μL = 10^-6 L by definition
So...
82 μL = 82x10^-6 L = 8.2x10^-5 L
Answer:
Explanation:
<u><em>What is the potential energy of a 2-kg book sitting on a shelf 2 meters above the ground?</em></u>
<em></em>
PE = MGH, that is, potential energy = mass x gravity x height.
2*2*9.8=
39.2 joules