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sweet-ann [11.9K]

3 years ago

5

the coast of California on the western side of the state has a voice climate wild Eastern side of the desert.what could cause th

is diffrence

2 answers:

diamong [38]3 years ago

7 0

fires could cause a big difference in the weather of California

Elanso [62]3 years ago

7 0

I think the answer is B, Mountains prevent moist air from blowing over into the desert.

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Which of the following is true about matter

SashulF [63]

Answer: b- it is always a substance

Explanation: matter can be any liquid, gas, or solid that takes up space so the answer cannot be C or D. For answer number A to be true, you would have to remove the word more, because matter just has mass and takes up space it does not have "more mass"

8 0

3 years ago

Give the name of one or more polysaccharides that matches each of the following descriptions:

adell [148]

Answer:

A Cellulose not digested by humans.

b. the storage form of carbohydrates in plants is starch

C amylose contains 1-4 glycosidic bond

D Glycogen and starch are highly branched polysaccharides.

Explanation:

8 0

2 years ago

What happens to population growth in a logistic growth pattern as it reaches carrying capacity?

Leno4ka [110]

I believe the answer would be C. slows down, hope this helps:)

4 0

2 years ago

Aspirin (acetylsalicylic acid, C9H8O4) is a weak monoprotic acid. To determine its acid-dissociation constant, a student dissolv

Triss [41]

Answer:

1. $3.57\times 10^{-4}$ was the $K_a$ value calculated by the student.

2. $5.93\times 10^{-4}$ was the $K_b$ of ethylamine value calculated by the student.

Explanation:

1.

The $pH$ value of Aspirin solution = 2.62

$pH=-\log[H^+]$

$[H^+]=10^{-2.62}=0.00240 M$

Moles of s asprin = $\frac{2.00 g}{180 g/mol}=0.01111 mol$

Volume of the solution = 0.600 L

The initial concentration of Aspirin = c = $\frac{0.01111 mol}{0.600 L}=0.0185 M$

$HAs\rightleftharpoons As^-+H^+$

initially

c 0 0

At equilibrium

(c-x) x x

The expression of dissociation constant :

$K_a=\frac{[As^-][H^+]}{[HAs]}$ :

$K_a=\frac{x\times x }{(c-x)}$

$=\frac{0.00240 M\times 0.00240 M}{(0.0185-0.00240 )}$

$K_a=3.57\times 10^{-4}$

$3.57\times 10^{-4}$ was the $K_a$ value calculated by the student.

2.

The $pH$ value of ethylamine = 11.87

$pH+pOH=14$

$pOH=14-11.87=2.13$

$pOH=-\log[OH^-]$

$[OH^-]=10^{-2.13}=0.00741 M$

The initial concentration of ethylamine = c = 0.100 M

$C_2H_5NH_2+H_2O\rightleftharpoons C_2H_5NH_3^{+}+OH^-$

initially

c 0 0

At equilibrium

(c-x) x x

The expression of dissociation constant :

$K_b=\frac{[C_2H_5NH_3^{+}][OH^-]}{[C_2H_5NH_2]}$ :

$K_b=\frac{x\times x}{(c-x)}$

$=\frac{0.00741\times 0.00741}{(0.100-0.00741)}$

$K_b=5.93\times 10^{-4}$

$5.93\times 10^{-4}$ was the $K_b$ of ethylamine value calculated by the student.

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