Ask question

Ask question

All categories

Ksenya-84 [330]

3 years ago

5

Water decomposes into hydrogen and oxygen when an electric current is applied. How many grams of water must decompose to produce

2 grams of hydrogen and 16 grams of oxygen?

1 answer:

weqwewe [10]3 years ago

6 0

I believe this would be the method to carry out and find the solutions.

You might be interested in

Which scenario is most similar to the type of collision that gas particles have according to kinetic molecular theory?

Ugo [173]

I’m not for sure but I think it’s A.

6 0

4 years ago

Read 2 more answers

Given: 36.7 grams of CaF2 is added to 300 mL water. Find molarity?

BigorU [14]

<h3>Answer:</h3>

2 M

<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>Chemistry</u>

<u>Unit 0</u>

Reading a Periodic Table
Using Dimensional Analysis

<u>Aqueous Solutions</u>

Molarity = moles of solute / liters of solution

<h3>Explanation:</h3>

<u>Step 1: Define</u>

36.7 g CaF₂

300 mL H₂O

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

Molar Mass of Ca - 40.08 g/mol

Molar Mass of F - 19.00 g/mol

Molar Mass of CaF₂ - 40.08 + 2(19.00) = 78.08 g/mol

1000 mL = 1 L

<u>Step 3: Convert</u>

<em>Solute</em>

Set up: $\displaystyle 36.7 \ g \ CaF_2(\frac{1 \ mol \ CaF_2}{78.08 \ g \ CaF_2})$
Multiply: $\displaystyle 0.470031 \ mol \ CaF_2$

<em>Solution</em>

Set up: $\displaystyle 300 \ mL \ H_2O(\frac{1 \ L \ H_2O}{1000 \ mL \ H_2O})$
Multiply: $\displaystyle 0.3 \ L \ H_2O$

<u>Step 4: Find Molarity</u>

Substitute [M]: $\displaystyle x \ M = \frac{0.470031 \ mol \ CaF_2}{.3 \ L \ H_2O}$
Divide: $\displaystyle x = 1.56677 \ M$

<u>Step 5: Check</u>

<em>Follow sig fig rules and round.</em> <em>We are given 1 sig fig as our lowest.</em>

1.56677 M ≈ 2 M

8 0

3 years ago

what criteria are likely to dictate certain pairs of elements to undergo ionic bonding and others covalent bonding?

Paha777 [63]

Answer:

If one element is a metal and the other element is a non metal

Explanation:

Ionic bonds are formed between metals and non metals. The nonmetal atom receives electron(s) from a metal atom to form an ionic substance. This means that ionic substances are basically an ion pair.

Covalent bonds are mostly formed between two nonmetals. These elements share electrons with each other in order to form the compound.

7 0

3 years ago

PLEASE HELP FAST! When 23.3 g of O2 reacts with 18.3 g C10H8, what is the limiting reactant. The equation is C10H8 + 12O2 -->

Murljashka [212]

O₂ is the limiting reagent.

<h2>What is a limiting reactant?</h2>

A limiting reactant is described as the one that will be consumed first in a chemical reaction.

<h3>Calculation</h3>

C₁₀H₈ + 12O₂ → 10CO₂ + 4H₂O

Given, Mass of O₂ = 23.3 g

Mass of C₁₀H₈ = 18.3 g

Molar mass of O₂ = 32 g

No. of moles of O₂ = 32/23.3 = 1.4 moles

Molar mass of C₁₀H₈ = 128 g

No. of moles of C₁₀H₈ = 128/18.3 = 6.9 = 7 moles

According to the equation,

1 mole of C₁₀H₈ yields 10 moles of CO₂. So, 7 moles of C₁₀H₈, give 70 moles of CO₂.

12 moles of O₂ yields 10 moles of CO₂. So, 1.4 moles of O₂ give 1.16 moles of CO₂.

We do know, however, that based on our balanced equation, if we were to totally react all 1.4 moles of O2, we could only produce a maximum of 1.16 moles of CO2. Even while there is enough C10H8 to generate more, the amount of O2 we have is a limiting factor because it will be consumed first.

So, our limiting reactant is O₂.

Learn more about limiting reagents here:

brainly.com/question/11848702

#SPJ2

7 0

2 years ago

Read 2 more answers

What happened when bromine reaction with aluminum

Anit [1.1K]

Answer:

formation reaction.

Explanation:

Since the aluminum bromide produced in this reaction is formed from elements in their standard states, this is an example of a formation reaction.

8 0

3 years ago

Read 2 more answers