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4 years ago

How to Prevent Onions from Making You Cry

Chemistry

1 answer:

sergejj [24]4 years ago

6 0

Answer:

Deleting everything except the first sentences of each numbered step.

Explanation:

The first sentences actually tell you what to do while the rest just explains why they're telling you to do it. If you read that article, all you would want to know are the steps to prevent crying while cutting onions, not necessarily the science behind those reasons.

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Which of the following examples describes an increase in potential energy? Question 1 options: a basketball falling through a ne

MArishka [77]

Paint being carried up a ladder

4 0

3 years ago

Read 2 more answers

Calculate the average atomic mas of an unknown element that has two naturally-occurring isotopes.

professor190 [17]

The average atomic mass if the element above is calculated by the sum of the product of the isotope abundance and its atomic mass unit. It is expressed as:

Average atomic mass = Σ xi(Mi)
<span>Average atomic mass = (.7547 x 248.7) + (.2453 x 249.4) = 248.87
</span>
Hope this helps.

8 0

3 years ago

Read 2 more answers

How many moles of NH3 can be produced from 12.0 mol of H2 and excess N2? Express your answer numerically in moles. View Availabl

VladimirAG [237]

Answer:

A) 8.00 mol NH₃

B) 137 g NH₃

C) 2.30 g H₂

D) 1.53 x 10²⁰ molecules NH₃

Explanation:

Let us consider the balanced equation:

N₂(g) + 3 H₂(g) ⇄ 2 NH₃(g)

Part A

3 moles of H₂ form 2 moles of NH₃. So, for 12.0 moles of H₂:

$12.0molH_{2}.\frac{2molNH_{3}}{3molH_{2}} =8.00molNH_{3}$

Part B:

1 mole of N₂ forms 2 moles of NH₃. And each mole of NH₃ has a mass of 17.0 g (molar mass). So, for 4.04 moles of N₂:

$4.04molN_{2}.\frac{2molNH_{3}}{1molN_{2}} .\frac{17.0gNH_{3}}{1molNH_{3}} =137gNH_{3}$

Part C:

According to the <em>balanced equation</em> 6.00 g of H₂ form 34.0 g of NH₃. So, for 13.02g of NH₃:

$13.02gNH_{3}.\frac{6.00gH_{2}}{34.0gNH_{3}} =2.30gH_{2}$

Part D:

6.00 g of H₂ form 2 moles of NH₃. An each mole of NH₃ has 6.02 x 10²³ molecules of NH₃ (Avogadro number). So, for 7.62×10⁻⁴ g of H₂:

$7.62 \times 10^{-4} gH_{2}.\frac{2molNH_{3}}{6.00gH_{2}} .\frac{6.02\times 10^{23}moleculesNH_{3} }{1molNH_{3}}=1.53\times10^{20}moleculesNH_{3}$

3 0

4 years ago

Using the formula M1V1 = M2V2 , how many milliliters of a 2. 50 M hydrochloric acid solution is required to make 100. 0 mL of a

gladu [14]

Dilution of the solution can be calculated by the formula of the molarity and volume. The initial volume of 2.50 M solution was 30 mL.

<h3>What is the relationship between molar concentration and dilution?</h3>

Molar concentration or the dilution factor is in an inverse relationship and with an increase in the dilution, the molarity of the solution decreases.

Given,

Initial molarity = 2.50 M

initial volume = ?

Final molarity = 0.750 M

Final volume = 100.0 ml

Substituting values in the formula:

$\begin{aligned}\rm V_{1} &= \rm \dfrac{M_{2}V_{2}}{M_{1}}\\\\&= \dfrac{0.750 \times 100}{2.50}\\\\&= 30 \;\rm mL\end{aligned}$

Therefore, 30 mL was the initial volume of the solution before it was diluted.

Learn more about dilution here:

brainly.com/question/26896011

3 0

3 years ago

The unknown ionic compound gave a red flame test, NR with ammonium carbonate, and a red/orange halide test. What is the unknown

harkovskaia [24]

Answer:

This question is incomplete, however, the unknown compound can be inferred to be "Lithium Bromide"

Explanation:

The unknown compound firstly is said to be an ionic compound. An ionic/electrovalent compound is a compound in which it's constituent ions transfer/receive electron(s). They are mostly made of group 1 and group 7 elements. Examples include NaCl, NaF, LiF and KCl.

Also, the ion (metallic ion) that produces a red flame test colour in a flame test is the <u>Lithium ion (Li⁺).</u> Also, when dissolved in water or hexane, the only halogen that produces a red/orange colour is bromine. Hence, the unknown ionic compound can be inferred to be Lithium Bromide.

5 0

3 years ago