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White raven [17]

2 years ago

5

HELP WITH SCIENCE!! 7TH GRADE! DUE TODAY!

1 answer:

ehidna [41]2 years ago

3 0

I am not positive this is what you want but here lol:

The Earth's surface is about 71 percent water spreading over many oceans, rivers, and lakes. This also includes the droplets of water on the morning grass, the undrained water from the rain, puddles that never evaporated, etc. As this water is heated up by the Sun, its' molecules start to move faster as it turns into gas and evaporates. As it evaporates it rises, however, the cooler temperatures in the high altitude of Earth's atmosphere make it condense into large clouds. When so much water has evaporated and the cloud very dense, gravity brings down the water from the clouds onto the ground in the form of rain. This completes the water cycle as the water will either turn into runoff, stay on the surface of the Earth, or seep into the ground. Then the water evaporates again, and the whole process starts over.

*Hope this helped!! Have a good night/ day wherever you are! : )*

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A 0.187 M weak acid solution has a pH of 3.99. Find Ka for the acid. Express your answer using two significant figures.

Anna35 [415]

Answer:

5.56 × 10⁻⁸

Explanation:

Step 1: Given data

Concentration of the weak acid (Ca): 0.187 M

pH of the solution: 3.99

Step 2: Calculate the concentration of H⁺

We will use the following expression.

pH = -log [H⁺]

[H⁺] = antilog -pH = antilog -3.99 = 1.02 × 10⁻⁴ M

Step 3: Calculate the acid dissociation constant (Ka)

We will use the following expression.

$Ka = \frac{[H^{+}]^{2} }{Ca} = \frac{(1.02 \times 10^{-4})^{2} }{0.187} = 5.56 \times 10^{-8}$

3 0

2 years ago

In methane combustion, the following reaction pair is important: At 1500 K, the equilibrium constant Kp has a value of 0.003691

andre [41]

Explanation:

Let us assume that the value of $K_{r}$ = $2.82 \times 10^{5} \times e^({\frac{-9835}{T}}) m^{6}/mol^{2}s$

Also at 1500 K, $K_{r}$ = $2.82 \times 10^{5} \times e^({\frac{-9835}{1500}}) m^{6}/mol^{2}s$

$K_{r} = 400.613 m^{6}/mol^{2}s$

Relation between $K_{p}$ and $K_{c}$ is as follows.

$K_{p} = K_{c}RT$

Putting the given values into the above formula as follows.

$K_{p} = K_{c}RT$

$0.003691 = K_{c} \times 8.314 \times 1500$

$K_{c} = 2.9 \times 10^{-7}$

Also, $K_{c} = \frac{K_{f}}{K_{r}}$

or, $K_{f} = K_{c} \times K_{r}$

= $2.9 \times 10^{-7} \times 400.613$

= $1.1617 \times 10^{-4} m^{6}/mol^{2}s$

Thus, we can conclude that the value of $K_{f}$ is $1.1617 \times 10^{-4} m^{6}/mol^{2}s$ .

6 0

2 years ago

Which technology was essential for the development of the cell theory?

vodka [1.7K]

Answer: Microscope

Explanation: This is a kid answering :)

8 0

2 years ago

Which of the following is not an example of a molecule?

9966 [12]

D. F

Molecules are a group of bonded atoms but Fluorine stands on its own

4 0

3 years ago

Read 2 more answers

Interstellar space has an average temperature of about 10 K , and an average density of hydrogen atoms of about one hydrogen ato

Bas_tet [7]

Answer:

$2.2508\times 10^{19} m$ meter the mean free path of hydrogen atoms in interstellar space.Explanation:

The mean free path equation is given as:

$\lambda =\frac{1}{\sqrt{2}\pi d^2 n}$

Where"

d = diameter of hydrogen atom in meters

n = number of molecules per unit volume

We are given: d = 100 pm = $100\times 10^{-12}= 10^{-10} m$

$n = 1 m^{-3}$

$\lambda =\frac{1}{\sqrt{2}\pi (10^{-10} m)^2\times 1 m^{-3}}$

$\lambda =2.2508\times 10^{19} m$

$2.2508\times 10^{19} m$ meter the mean free path of hydrogen atoms in interstellar space.

7 0

2 years ago