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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.)
Chemistry
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
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