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pishuonlain [190]

3 years ago

12

Describe Pasteur’s results and how they helped disprove spontaneous generation. (This response must be multiple complete sentenc

es.)

1 answer:

mina [271]3 years ago

3 0

Answer::

Explanation:

You might be interested in

How many valance electron dose an atom of oxgen have? A 6 B 4 C 2 D 8

Maurinko [17]

Answer:

6

Explanation:

oxygen has 12 protons and neutrons in total

8 0

3 years ago

The total surface area for the Asia consisting of forest, cultivated land, grass land, and desert is approximately 4.46 x 107 km

tester [92]

Explanation:

We calculate the amount of carbon fixed as follows.

$455 \times 10^{3} kg/km^{2} \times 4.46 \times 10^{7} km^{2}$

= $2029.3 \times 10^{10}$ kg

$\Delta H_{f}_{C_{6}H_{12}O_{6}}$ = -1273.1 kJ/mol

$\Delta H_{f}_{CO_{2}}$ = -393.5 kJ/mol

$\Delta H_{f}_{H_{2}O}$ = -285.8 kJ/mol

Hence, total $\Delta H$ reaction will be as follows.

$\Delta H_{rxn} = (\Delta H_{f}_{C_{6}H_{12}O_{6}}) - 6(\Delta H_{f}_{CO_{2}}) - 6(\Delta H_{f}_{H_{2}O})$

= $-1273.1 kJ/mol - (6 \times 393.5 kJ/mol) - (6 \times 285.8 kJ/mol)$

= 2802.7 kJ/mol

Therefore, calculate the number of moles of fixed carbon as follows.

$n_{c, fixed} = \frac{m_{c, fixed}}{\text{Molar mass of C in kg/mol}}$

= $\frac{4.55 \times 10^{5}kg km^{-2} year^{-1}}{12.01 \times 10^{-3} kg/mol}$

= $3.786 \times 10^{7} mol km^{-2} year^{-1}$

Thus, we can conclude that the annual enthalpy change resulting from the photosynthetic carbon fixation over the land surface is $3.786 \times 10^{7} mol km^{-2} year^{-1}$ .

7 0

3 years ago

For the function F(x) = 1, which of these could be a value of F(x) when x is<br> close to zero?

Sonbull [250]

Answer:

A. -10,000

Explanation:

When x approaches zero, the graph tells us that F(x) either becomes extremely large or extremely small. The best answer is therefore A: -10,000. Best, because we aren't given the exact value of x, but clearly the other results require that x be greater than 1 or less than -1, which is far from zero.

7 0

2 years ago

When 85 L of gas at 1.376 atm has its pressure changed to 0.154 atm, what is the new volume?

igor_vitrenko [27]

Answer:

<h2>759.48 L</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

$V_2 = \frac{85 \times 1.376}{0.154} = \frac{116.96}{0.154} \\ = 759.480519...$

We have the final answer as

<h3>759.48 L</h3>

Hope this helps you

4 0

3 years ago

Given the equilibrium constants for the following two reactions in aqueous solution at 25 ∘C HNO2(aq)H2SO3(aq)⇌⇌H+(aq) + NO2−(aq

krok68 [10]

Answer : The value of $K_c$ for the final reaction is, 184.09

Explanation :

The equilibrium reactions in aqueous solution are :

(1) $HNO_2(aq)\rightleftharpoons H^+(aq)+NO_2^-(aq)$ $K_{c_1}=4.5\times 10^{-4}$

(2) $H_2SO_3(aq)\rightleftharpoons 2H^+(aq)+SO_3^{2-}(aq)$ $K_{c_2}=1.1\times 10^{-9}$

The final equilibrium reaction is :

$2HNO_2(aq)+SO_3^{2-}(aq)\rightleftharpoons H_2SO_3(aq)+2NO_2^-(aq)$ $K_{c}=?$

Now we have to calculate the value of $K_c$ for the final reaction.

Now equation 1 is multiply by 2 and reverse the equation 2, we get the value of final equilibrium reaction and the expression of final equilibrium constant is:

$K_c=\frac{(K_{c_1})^2}{K_{c_2}}$

Now put all the given values in this expression, we get :

$K_c=\frac{(4.5\times 10^{-4})^2}{1.1\times 10^{-9}}=184.09$

Therefore, the value of $K_c$ for the final reaction is, 184.09

4 0

3 years ago