Answer:
Ammonia, urea, uric acid
Explanation:
The given compounds are:-
Urea which has a molecular formula of 
and has 2 atoms of nitrogen per molecule.
Ammonia which has a molecular formula of 
and has 1 atom of nitrogen per molecule.
Uric acid which has a molecular formula of 
and has 4 atoms of nitrogen per molecule.
Thus,
The order from least to most nitrogen is:-
Ammonia, urea, uric acid
Answer:
A
Explanation:
1g=1000mg
1g= 1000 000 micro gram
1g= 1000 000 000 nm = 10^9
You are given the neutralization of acetic acid with sodium hydroxide. Also, you are given the k for acetic acid, which is 1.8 x 10⁻⁵. You are asked to find the<span> approximate value of the equilibrium constant, kn, for the neutralization. We will have a reaction of both acetic acid and sodium hydroxide.
CH</span>₃COOH + NaOH → CH₃COONa + H₂O
which comes from
CH₃COOH → CH₃COO⁻ + H⁺
H⁺ + OH⁻ → H₂O<span>
</span>The k for water is always 1.0 x 10¹⁴. The Ksp for the reaction will be
<span>
Ksp = [</span>CH₃COOH][H₂O]
Ksp = (1.8 x 10⁻⁵)(1.0 x 10¹⁴)
<span>Ksp = 1.8 x 10</span>⁹
Answer:
The hot tea should transfer <em>25.63 kJ</em> the surroundings to cool the tea.
Explanation:
- To solve this problem, we can use the relation:
<em>Q = m.c.ΔT,</em>
where, Q is the amount of heat has to be transferred from the tea to the surroundings to cool the tea (Q = ??? J).
m is the mass of the hot tea (m = dV = (1.0 g/mL)(250 mL) = 250 g), suppose the density of water is the density of tea.
c is the specific heat of the hot tea (c = 4.10 J/°C.g).
ΔT is the difference between the initial and final temperature (ΔT = final T - initial T = 350 K - 375 K = -25°C).
<em>∴ Q = m.c.ΔT</em> = (250 g)(4.10 J/°C.g)(-25°C)) = <em>- 25630 J = - 25.63 kJ.</em>
<em>So, the hot tea should transfer 25.63 kJ the surroundings to cool the tea.</em>
yes substances Do react by mass
