<span>The
kingdom, protista’s characteristics are that the organism (not a plant,
animal or fungus) are:
unicellular however some are multicellular like algae, are heterotrophic or
autotrophic, others lives in water while some live in moist areas or human body,
have a nucleus, cellular respiration is primarily aerobic, some are pathogenic
(e.g. causing Malaria) and reproduction is mitosis or meiosis. This kingdom
includes: Sacordinians – pseudopods (e.g. Amoeba, Foraminiferans<span>.)</span>, Zooflagellates – flagellates
(e.g. Trypanosoma gambiense),
Ciliaphorans – ciliates (e.g. paramecium) and Sporozoans (e.g. Plasmodium).</span>
Answer:
23.92 g
Explanation:
Molar mass of H2SO4 = (2×1)+32+(16×4)= 2+32+48= 82g/mol
H2SO4 + 2NaOH ---> Na2SO4 + 2H2O
I mole of H2SO4 = 2 moles of NaOH
24.5/82 = 24.5/82 × 2
= 0.598 moles of NaOH will neutralize
Mass= mole× molar mass
Molar mass of NaOH= 23+16+1 = 40g/mol
Mass= 0.598 × 40 = 23.92g of NaOH
Answer:
189.71 secs
Explanation:
We know that decomposition is a first order reaction;
So;
ln[A] = ln[A]o - kt
But;
[A]o = 1.00 M
[A] = 0.250 M
t =135 s
Hence;
ln[A] - ln[A]o = kt
k = ln[A] - ln[A]o/t
k = ln(1) - ln(0.250)/135
k =0 - (-1.386)/135
k = 1.386/135
k= 0.01
So time taken now will be;
ln[A] - ln[A]o = kt
t = ln[A] - ln[A]o/k
t = ln (3) - ln(0.450)/0.01
t = 1.0986 - (-0.7985)/0.01
t = 1.0986 + 0.7985/0.01
t = 189.71 secs
Answer:
non-polar covalent bonds
this is when the electronegativity difference between the two non-metal atoms is very little to cause a partial charge (delta positive and delta negative; δ+ and δ-) on the atoms
Answer:
we will use the Clausius-Clapeyron equation to estimate the vapour pressures of the boiling ethanol at sea level pressure of 760mmHg:
ln (P2/P1) = -)
where
P1 and P2 are the vapour pressures at temperatures T1 and T2
Δ
vapH = the enthalpy of vaporization of the ETHANOL
R = the Universal Gas Constant
In this problem,
P
1
=
100 mmHg
; T
1
=
34.7 °C
=
307.07 K
P
2
=
760mmHg
T
2
=T⁻²=?
Δ
vap
H
=
38.6 kJ/mol
R
=
0.008314 kJ⋅K
-1
mol
-1
ln
(
760/10)=(0.00325 - T⁻²) (38.6kJ⋅mol-1
/0.008314
)
0.0004368=(0.00325 - T⁻²)
T⁻²=0.002813
T² = 355.47K