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In-s [12.5K]
2 years ago
13

Mike and Mariah are in the same science class. Mike wants to know how the water cycle transports energy and matter. Which of the

following examples in Mariah's notebook should she share as the best example of how the water cycle transports both energy and matter?
(A) Water transforms from a solid to a liquid to a gas.
(B) Water transported through the water cycle is never created or lost.
(C) Water vapor precipitates back into the ocean or the land after heating.
(D) Water vapor rises into the air where colder temperatures cause it to condense and form clouds.
Chemistry
2 answers:
Jobisdone [24]2 years ago
5 0
The answer would be D water vapor...
lord [1]2 years ago
3 0

Answer:(D) Water vapor rises into the air where colder temperatures cause it to condense and form clouds.

Explanation: I just know it's defenitely not A or B. Hope my answer helps

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What is the pH of a substance that has a hydrogen ion concentration of<br> 8.8x10^(-5) M? *
stealth61 [152]

Answer:

The pH of the substance is 4,06.

Explanation:

The pH indicates the acidity or basicity of a substance. PH values between 0 and less than 7 indicate acidic solutions, 7 neutral and greater than 7 to 14 basic. It is calculated as:

pH = -log (H +)

pH= -log (8.8x10^-5)

<em>pH=4,06</em>

5 0
3 years ago
How many moles are in 8.63 x 103 atoms of Li?
Ira Lisetskai [31]
<h3>Answer:</h3>

1.43 × 10⁻²⁰ mol Li

<h3>General Formulas and Concepts:</h3>

<u>Math</u>

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

  1. Brackets
  2. Parenthesis
  3. Exponents
  4. Multiplication
  5. Division
  6. Addition
  7. Subtraction
  • Left to Right

<u>Chemistry</u>

<u>Atomic Structure</u>

  • Using Dimensional Analysis
  • Avogadro's Number - 6.022 × 10²³ atoms, molecules, formula units, etc.
<h3>Explanation:</h3>

<u>Step 1: Define</u>

8.63 × 10³ atoms Li

<u>Step 2: Identify Conversions</u>

Avogadro's Number

<u>Step 3: Convert</u>

  1. Set up:                              \displaystyle 8.63 \cdot 10^3 \ atoms \ Li(\frac{1 \ mol \ Li}{6.022 \cdot 10^{23} \ atoms \ Li})
  2. Multiply/Divide:                \displaystyle 1.43355 \cdot 10^{-20} \ moles \ Li

<u>Step 4: Check</u>

<em>Follow sig fig rules and round. We are given 3 sig figs.</em>

1.43355 × 10⁻²⁰ mol Li ≈ 1.43 × 10⁻²⁰ mol Li

4 0
2 years ago
What is the molar mass of a gas if 0.102 g of the gas occupies 0.070 L at STP? a. 32.6g / mol b. 28.2 g/mol c. 40.0g / mol d. 47
alexgriva [62]

Answer:

A. 32.6 g/mol

Explanation:

First convert the volume of gas to moles using the ratio 1 mol / 22.4 L at STP.

0.070 L • (1 mol / 22.4 L) = 0.00313 mol

Now divide the grams of gas by the moles of gas:

0.102 g / 0.00313 mol = 32.6 g/mol

8 0
2 years ago
.500 mol of br2 and .500 mol of cl2 are placed in a .500L flask and allowed to reach equilibrium. at equilibrium the flask was f
Tasya [4]

Answer : The value of K_c for the given reaction is, 0.36

Explanation :

Equilibrium constant : It is defined as the equilibrium constant. It is defined as the ratio of concentration of products to the concentration of reactants.

The equilibrium expression for the reaction is determined by multiplying the concentrations of products and divided by the concentrations of the reactants and each concentration is raised to the power that is equal to the coefficient in the balanced reaction.

As we know that the concentrations of pure solids and liquids are constant that is they do not change. Thus, they are not included in the equilibrium expression.

The given equilibrium reaction is,

Br_2(aq)+Cl_2(aq)\rightleftharpoons 2BrCl(aq)

The expression of K_c will be,

K_c=\frac{[BrCl]^2}{[Br_2][Cl_2]}

First we have to calculate the concentration of Br_2,Cl_2\text{ and }BrCl.

\text{Concentration of }Br_2=\frac{Moles}{Volume}=\frac{0.500mol}{0.500L}=1M

\text{Concentration of }Cl_2=\frac{Moles}{Volume}=\frac{0.500mol}{0.500L}=1M

\text{Concentration of }BrCl=\frac{Moles}{Volume}=\frac{0.300mol}{0.500L}=0.6M

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

K_c=\frac{[BrCl]^2}{[Br_2][Cl_2]}

K_c=\frac{(0.6)^2}{(1)\times (1)}

K_c=0.36

Therefore, the value of K_c for the given reaction is, 0.36

6 0
3 years ago
Read 2 more answers
If you produce 7 mols of KBr, how many mols of MgBr2 did you need?
djyliett [7]

Answer:

hmm

Explanation:

hmm

7 0
2 years ago
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