Answer:
Sucrose: glucose and fructose
Explanation:
<em>What monosaccharides will result from the hydrolysis of sucrose?</em>
<em>Sucrose</em> is a <em>disaccharide</em> composed of 2 different <em>monosaccharides</em>: glucose and fructose joining by a 1 ⇒ 2 bond. These monosaccharides will be released upon the hydrolysis of sucrose.
<em>What monosaccharide will result from the hydrolysis of starch?</em>
<em>Starch</em> is a <em>polysaccharide</em> composed of numerous glucose monomers joined by glycosidic bonds (1 ⇒ 4 and 1 ⇒ 6). These monosaccharides will be released upon the hydrolysis of starch.
Answer:
B = (2.953 × 10⁻⁹⁵) N.m⁹
Explanation:
At equilibrium, where the distance between the two ions (ro) is the sum of their ionic radii, the force between the two ions is zero.
That is,
Fa + Fr = 0
Fa = - Fr
Fa = (|q₁q₂|)/(4πε₀r²)
Fr = -B/(r^n) but n = 9
Fr = -B/r⁹
(|q₁q₂|)/(4πε₀r²) = (B/r⁹)
|q₁| = |q₂| = (1.6 × 10⁻¹⁹) C
(1/4πε₀) = k = (8.99 × 10⁹) Nm²/C²
r = 0.097 + 0.181 = 0.278 nm = (2.78 × 10⁻¹⁰) m
(k|q₁q₂|)/(r²) = (B/r⁹)
(k × |q₁q₂|) = (B/r⁷)
B = (k × |q₁q₂| × r⁷)
B = [8.99 × 10⁹ × 1.6 × 10⁻¹⁹ × 1.6 × 10⁻¹⁹ × (2.78 × 10⁻¹⁰)⁷]
B = (2.953 × 10⁻⁹⁵) N.m⁹
Answer:

Explanation:
Hello there!
In this case, according to the given STP (standard pressure and temperature), it is possible for us to realize that the equation to use here is the Avogadro's law as a directly proportional relationship between moles and volume:

In such a way, given the initial volume and both initial and final moles, we can easily compute the final volume as shown below:

Best regards!
Answer:
17.0 g of hask2 lahhwle 11c u2b8ss
<span>graduated cylinder is the answer</span>