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3 years ago

Which climate is most likely found in a high pressure belt

1 answer:

5 0

Answer:subtropical highs. ... Near the poles the pressure is high and it is known as the polar high. These pressure belts are not permanent in nature

Explanation: The horse latitudes are subtropical regions known for calm winds and little precipitation. ... Unable to sail and resupply due to lack of wind, crews often ran out of drinking water. To conserve scarce water, sailors on these ships would sometimes throw the horses they were transporting overboard.

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exactly 1mol of n2o4 is placed in an empty 1 l container. if at equilibrium n2o4 is dissociated 20%, what is the value of equili

egoroff_w [7]

Answer:

K = 0.2

Explanation:

Based on the chemical dissociation of N₂O₄:

N₂O₄ ⇄ 2NO₂

The equilibrium constant, K, of the reaction is:

K = [NO₂]² / [N₂O₄]

Now, if 20% of N₂O₄ is dissociated, 80% remains as N₂O₄ = 0.8mol/L = 0.8M

as 20% is dissociated, 0.2moles of N₂O₄ were dissociated and 0.2*2 = 0.4mol/L of NO₂ are produced.

Replacing in K:

K = [0.4M]² / [0.8M]

5 0

3 years ago

Some radioactive nuclides have very short half-lives, for example, I-31 has a half-life of approximately 8 days. Pu-234, by comp

lorasvet [3.4K]

Answer:

Here's what I find.

Explanation:

Iodine-131

Iodine-131 is both a beta emitter and a gamma emitter.

$_{53}^{131}\text{I}\longrightarrow \, _{54}^{131}\text{Xe} +\, _{-1}^{0}\text{e} +\, _{0}^{0}\gamma$

About 90 % of the energy is β-radiation and 10 % is γ-radiation. Both forms are highly energetic.

The main danger is from ingestion. The iodine concentrates in thyroid gland, where the β-radiation destroys cells up to 2 mm from the tissues that absorbed it.

Both the β- and γ-radiation cause cell mutations that can later become cancerous. Small doses, such as those absorbed from the nuclear disasters in the Ukraine and Japan, can cause cancers years after the original iodine has disappeared.

Plutonium-239

Plutonium-239 is an alpha emitter.

$_{94}^{239}\text{U} \longrightarrow \, _{92}^{235}\text{Xe} + \, _{2}^{4}\text{He}$

Alpha particles cannot penetrate the skin, so external exposure isn't much of a health risk.

However, they are extremely dangerous when they are inhaled and get inside cells. They travel first to the blood or lymph system and later to the bone marrow and liver, where they cause up to 1000 times more chromosomal damage than beta or gamma rays.

It takes about 20 years for plutonium to be eliminated from the liver around 50 years for from the skeleton, so it has a long time to cause damage.

6 0

3 years ago

At 25°C, the equilibrium constant Kc for the reaction in thesolvent CCl4 2BrCl <----> Br2 + Cl2 is 0.141. If the initial c

ivolga24 [154]

<u>Answer:</u> The equilibrium concentration of bromine gas is 0.00135 M

<u>Explanation:</u>

We are given:

Initial concentration of chlorine gas = 0.0300 M

Initial concentration of bromine monochlorine = 0.0200 M

For the given chemical equation:

$2BrCl\rightleftharpoons Br_2+Cl_2$

<u>Initial:</u> 0.02 0.03

<u>At eqllm:</u> 0.02-2x x 0.03+x

The expression of $K_c$ for above equation follows:

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

We are given:

$K_c=0.141$

Putting values in above equation, we get:

$0.141=\frac{x\times (0.03+x)}{(0.02-2x)^2}\\\\x=-0.96,+0.00135$

Neglecting the value of x = -0.96 because, concentration cannot be negative

So, equilibrium concentration of bromine gas = x = 0.00135 M

Hence, the equilibrium concentration of bromine gas is 0.00135 M

8 0

3 years ago

Given the unbalanced equation: PbCl₂ + Na₂CrO₂ ---> PbCrO₂ + NaCl. When properly balanced, the sum of the balancing coefficie

Veronika [31]

Answer:

A. 2

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3 0

2 years ago

How many moles of S would I have if I had 11 grams? (Stoichiometry) HELP

zepelin [54]

<h3>Answer:</h3>

0.34 mol S

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

<u>Math</u>

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right<u> </u>

<u>Chemistry</u>

<u>Atomic Structure</u>

Reading a Periodic Table

<u>Stoichiometry</u>

Using Dimensional Analysis

<h3>Explanation:</h3>

<u>Step 1: Define</u>

11 g S

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

[PT] Molar Mass of S - 32.07 g/mol

<u>Step 3: Convert</u>

Set up: $\displaystyle 11 \ g \ S(\frac{1 \ mol \ S}{32.07 \ g \ S})$
Multiply/Divide: $\displaystyle 0.343 \ mol \ S$

<u>Step 4: Check</u>

<em>Follow sig fig rules and round. We are given 2 sig figs.</em>

0.343 mol S ≈ 0.34 mol S

4 0

2 years ago