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2 years ago

How does ocean currents affect climate i need it in 8 sentences.

1 answer:

5 0

Ocean currents move both warm and cool water across many miles. Warm water can be carried to a cooler place like Alaska. It can bring warmer temperatures as well as warmer water to Alaska. Just like the warm water, cold water can do this as well. If cold water moves from Alaska to The Gulf of Mexico, it can cause the water and air to cool. As air travels over the ocean water vapor can condense into it. The current’s flow depends on what variables are in play. There are many different currents that connect together to heat up and cool off as they move.

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The first step in industrial nitric acid production is the catalyzed oxidation of ammonia. Without a catalyst, a different react

murzikaleks [220]

<u>Answer:</u> The value of $K_{eq}$ is $4.84\times 10^{-5}$

<u>Explanation:</u>

We are given:

Initial moles of ammonia = 0.0280 moles

Initial moles of oxygen gas = 0.0120 moles

Volume of the container = 1.00 L

Concentration of a substance is calculated by:

$\text{Concentration}=\frac{\text{Number of moles}}{\text{Volume}}$

So, concentration of ammonia = $\frac{0.0280}{1.00}=0.0280M$

Concentration of oxygen gas = $\frac{0.0120}{1.00}=0.0120M$

The given chemical equation follows:

$4NH_3(g)+3O_2(g)\rightleftharpoons 2N_2(g)+6H_2O(g)$

<u>Initial:</u> 0.0280 0.0120

<u>At eqllm:</u> 0.0280-4x 0.0120-3x 2x 6x

We are given:

Equilibrium concentration of nitrogen gas = $3.00\times 10^{-3}M=0.003$

Evaluating the value of 'x', we get:

$\Rightarrow 2x=0.003\\\\\Rightarrow x=0.0015M$

Now, equilibrium concentration of ammonia = $0.0280-4x=[0.0280-(4\times 0.0015)]=0.022M$

Equilibrium concentration of oxygen gas = $0.0120-3x=[0.0120-(3\times 0.0015)]=0.0075M$

Equilibrium concentration of water = $6x=(6\times 0.0015)]=0.009M$

The expression of $K_{eq}$ for the above reaction follows:

$K_{eq}=\frac{[H_2O]^6\times [N_2]^2}{[NH_3]^4\times [O_2]^3}$

Putting values in above expression, we get:

$K_{eq}=\frac{(0.009)^6\times (0.003)^2}{(0.022)^4\times (0.0075)^3}\\\\K_{eq}=4.84\times 10^{-5}$

Hence, the value of $K_{eq}$ is $4.84\times 10^{-5}$

4 0

3 years ago

Would someone please help?

sveta [45]

The season that is starting is winter.
The answer to 21 is (4)

3 0

3 years ago

An iron loses electrons. State the type of charge it will end up with

erastova [34]

Answer:

Most atoms do not have eight electrons in their valence electron shell. Some atoms have only a few electrons in their outer shell, while some atoms lack only one or two electrons to have an octet. In cases where an atom has three or fewer valence electrons, the atom may lose those valence electrons quite easily until what remains is a lower shell that contains an octet. Atoms that lose electrons acquire a positive charge as a result because they are left with fewer negatively charged electrons to balance the positive charges of the protons in the nucleus. Positively charged ions are called cations. Most metals become cations when they make ionic

4 0

2 years ago

If we use 18 moles of H2 how many moles of H2O do we make? Use this chemical equation: 6H2 + O2 → 3H2O

Ilya [14]

<h3>Answer:</h3>

9 mol H₂O

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

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right

<u>Chemistry</u>

<u>Atomic Structure</u>

Reading a Periodic Table
Moles
Avogadro's Number - 6.022 × 10²³ atoms, molecules, formula units, etc.

<u>Stoichiometry</u>

Using Dimensional Analysis
Analyzing reactions RxN

<h3>Explanation:</h3>

<u>Step 1: Define</u>

[RxN - Balanced] 6H₂ + O₂ → 3H₂O

[Given] 18 mol H₂

[Solve] mol H₂O

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

[RxN] 6 mol H₂ → 3 mol H₂O

<u>Step 3: Stoich</u>

[DA] Set up conversion: $\displaystyle 18 \ mol \ H_2(\frac{3 \ mol \ H_2O}{6 \ mol \ H_2})$
[DA] Simplify: $\displaystyle 18 \ mol \ H_2(\frac{1 \ mol \ H_2O}{2 \ mol \ H_2})$
[DA] Divide [Cancel out units]: $\displaystyle 9 \ mol \ H_2O$