The answer is A: Between 50 and 5,000 amino acids
<em>answer:</em><em> </em><em>option </em><em>d </em><em>(</em><em>2</em><em>×</em><em>m</em><em>o</em><em>l</em><em>a</em><em>r</em><em> </em><em>mass </em><em>of </em><em>H </em><em>+</em><em>2</em><em>×</em><em>m</em><em>o</em><em>l</em><em>a</em><em>r</em><em> </em><em>mass </em><em>of </em><em>O</em>
<em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em>
(B. 3) 172 All nonzero digits are significant.
(A. 4) 450.0 x 10^3 Trailing zeroes after the decimal point are significant.
(A. 4) 3427 All nonzero digits are significant.
(B. 3) 0.0000455 Leading zeroes are not significant.
(B. 3) 0.00456 Leading zeroes are not significant.
(C. 5) 2205.2 Zeroes between nonzero digits are significant.
(C. 5) 107.20 Trailing zeroes after the decimal point are significant.
(B. 3) 0.0473 Leading zeroes are not significant.
Answer:
The unknown temperature is 304.7K
Explanation:
V1 = 100mL = 100*10^-3L
P1 = 99.10kPa = 99.10*10³Pa
V2 = 74.2mL = 74.2*10^-3L
P2 = 133.7kPa = 133.7*10³Pa
T2 = 305K
T1 = ?
From combined gas equation,
(P1 * V1) / T1 = (P2 * V2) / T2
Solving for T1,
T1 = (P1 * V1 * T2) / (P2 * V2)
T1 = (99.10*10³ * 100*10^-3 * 305) / (133.7*10³ * 74.2*10^-3)
T1 = 3022550 / 9920.54
T1 = 304.67K
T1 = 304.7K
Answer:
a) Unsaturated
b) Supersaturated
c) Unsaturated
Explanation:
A saturated solution contains the <u>maximum amount of a solute that will dissolve in a given solvent at a specific temperature</u>.
An unsaturated solution contains <u>less solute than it has the capacity to dissolve. </u>
A supersaturated solution, <u>contains more solute than is present in a saturated solution</u>. Supersaturated solutions are not very stable. In time, some of the solute will come out of a supersaturated solution as crystals.
According to these definitions and considering that the solubility of KCl in 100 mL of H₂O at <u>20 °C is 34 g</u>, and at <u>50 °C is 43 g</u> we can label the solutions:
a) 30 g in 100 mL of H₂O at 20 °C ⇒ unsaturated
b) 65 g in 100 mL of H₂O at 50 °C ⇒ supersaturated
c) 42 g in 100 mL of H₂O at 50 °C and slowly cooling to 20 °C to give a clear solution <u>with no precipitate</u> ⇒ unsaturated (if it were saturated it would have had precipitate)
