Answer:
Proteins and nucleic acids
Explanation:
Nitrogen compounds in animals that are no longer of use, or are in access are excreted from the animals body, and are thus called nitrogenous waste. These nitrogenous waste can be excreted in three different ways.
1. Ammonia
2. Urea
3. Uric acid
Answer: 12.78ml
Explanation:
Given that:
Volume of KOH Vb = ?
Concentration of KOH Cb = 0.149 m
Volume of HBr Va = 17.0 ml
Concentration of HBr Ca = 0.112 m
The equation is as follows
HBr(aq) + KOH(aq) --> KBr(aq) + H2O(l)
and the mole ratio of HBr to KOH is 1:1 (Na, Number of moles of HBr is 1; while Nb, number of moles of KOH is 1)
Then, to get the volume of a 0.149 m potassium hydroxide solution Vb, apply the formula (Ca x Va)/(Cb x Vb) = Na/Nb
(0.112 x 17.0)/(0.149 x Vb) = 1/1
(1.904)/(0.149Vb) = 1/1
cross multiply
1.904 x 1 = 0.149Vb x 1
1.904 = 0.149Vb
divide both sides by 0.149
1.904/0.149 = 0.149Vb/0.149
12.78ml = Vb
Thus, 12.78 ml of potassium hydroxide solution is required.
Answer:
d.) granulated
Because the granulated sugar has a greater surface area.
Answer:
0.0468 g.
Explanation:
- The decay of radioactive elements obeys first-order kinetics.
- For a first-order reaction: k = ln2/(t1/2) = 0.693/(t1/2).
Where, k is the rate constant of the reaction.
t1/2 is the half-life time of the reaction (t1/2 = 1620 years).
∴ k = ln2/(t1/2) = 0.693/(1620 years) = 4.28 x 10⁻⁴ year⁻¹.
- For first-order reaction: <em>kt = lna/(a-x).</em>
where, k is the rate constant of the reaction (k = 4.28 x 10⁻⁴ year⁻¹).
t is the time of the reaction (t = t1/2 x 8 = 1620 years x 8 = 12960 year).
a is the initial concentration (a = 12.0 g).
(a-x) is the remaining concentration.
∴ kt = lna/(a-x)
(4.28 x 10⁻⁴ year⁻¹)(12960 year) = ln(12)/(a-x).
5.54688 = ln(12)/(a-x).
Taking e for the both sides:
256.34 = (12)/(a-x).
<em>∴ (a-x) = 12/256.34 = 0.0468 g.</em>
Answer: snake grass hopper and monkey
Explanation: i took the test and this is what i got