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Len [333]
2 years ago
10

What is the probability of a short plant in the cross between a Tt and a tt plant?

Chemistry
1 answer:
kondaur [170]2 years ago
7 0
75% would be your answer since the lower letter represents a shorter plant
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Check all that apply...helppppp
mote1985 [20]

Answer:

dfgh

Explanation:

3 0
3 years ago
The estimated heat of vaporization of diethyl ether using the Chen's rule is A. 29.7 KJ/mol B. 33.5 KJ/mol C. 26.4 KJ/mol D. 36.
Brums [2.3K]

Answer:

C. 26.4 kJ/mol

Explanation:

The Chen's rule for the calculation of heat of vaporization is shown below:

\Delta H_v=RT_b\left [ \frac{3.974\left ( \frac{T_b}{T_c} \right )-3.958+1.555lnP_c}{1.07-\left ( \frac{T_b}{T_c} \right )} \right ]

Where,

\Delta H_v is the Heat of vaoprization (J/mol)

T_b is the normal boiling point of the gas (K)

T_c is the Critical temperature of the gas (K)

P_c is the Critical pressure of the gas (bar)

R is the gas constant (8.314 J/Kmol)

For diethyl ether:

T_b=307.4\ K

T_c=466.7\ K

P_c=36.4\ bar

Applying the above equation to find heat of vaporization as:

\Delta H_v=8.314\times307.4 \left [ \frac{3.974\left ( \frac{307.4}{466.7} \right )-3.958+1.555ln36.4}{1.07-\left ( \frac{307.4}{466.7} \right )} \right ]

\Delta H_v=26400 J/mol

The conversion of J into kJ is shown below:

1 J = 10⁻³ kJ

Thus,

\Delta H_v=26.4 kJ/mol

<u>Option C is correct</u>

6 0
3 years ago
Activity C: Hydrolysis of Carbohydrates As discussed earlier, disaccharides are composed of two monosaccharides linked together.
adelina 88 [10]

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.

4 0
3 years ago
Plz someone help, really struggling
frez [133]

Answer:

22.9 Liters CO(g) needed

Explanation:

2CO(g)     +   O₂(g)    =>    2CO₂(g)

? Liters          32.65g

                 = 32.65g/32g/mol

                 =   1.02 moles O₂

Rxn ratio for CO to O₂ = 2 mole CO(g) to 1 mole O₂(g)

∴moles CO(g) needed = 2 x 1.02 moles CO(g) = 2.04 moles CO(g)

Conditions of standard equation* is STP (0°C & 1atm) => 1 mole any gas occupies 22.4 Liters.

∴Volume of CO(g) = 1.02mole x 22.4Liters/mole = 22.9 Liters CO(g) needed

___________________

*Standard Equation => molecular rxn balanced to smallest whole number ratio coefficients is assumed to be at STP conditions (0°C & 1atm).

6 0
2 years ago
A gas starts at a volume of 23 L and a pressure of 1.23 atm. What is the new pressure if you
Anna71 [15]

Answer:

<h2>1.89 atm</h2>

Explanation:

The new volume can be found by using the formula for Boyle's law which is

P_1V_1 = P_2V_2

Since we are finding the new volume

V_2 =  \frac{P_1V_1}{P_2}  \\

From the question we have

V_2 =  \frac{1.23 \times 23}{15}  =  \frac{28.29}{15}  \\  = 1.886

We have the final answer as

<h3>1.89 atm</h3>

Hope this helps you

7 0
2 years ago
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