The options for given question are as follow,
1) Methane molecules show hydrogen bonding.
<span>2) Ammonia molecules show hydrogen bonding. </span>
<span>3) Methane has stronger hydrogen bonding than ammonia. </span>
<span>4) Both the compounds do not show hydrogen bonding. </span>
<span>5) Both the compounds have strong hydrogen bonding.
</span>
Answer:
Correct answer is Option-2 (Ammonia molecules show hydrogen bonding).
Explanation:
Hydrogen bond interactions are formed when a partial positive hydrogen atom attached to most electronegative atom of one molecule interacts with the partial negative most electronegative element of another molecule. So, in Ammonia hydrogen gets partial positive charge as nitrogen is highly electronegative. While the C-H bond in Methane is non-polar and fails to form hydrogen bond interactions.
False. At equilibrium, the rate of forward reaction is equal to the rate of backward reaction. The net concentration of both products and reactants won't change, but the reactions still take place.
<h3>
Answer:</h3>
2 L Ne
<h3>
General Formulas and Concepts:</h3>
<u>Math</u>
<u>Pre-Algebra</u>
Order of Operations: BPEMDAS
- Brackets
- Parenthesis
- Exponents
- Multiplication
- Division
- Addition
- Subtraction
<u>Chemistry</u>
<u>Atomic Structure</u>
- Using Dimensional Analysis
- STP (Standard Conditions for Temperature and Pressure) = 22.4 L per mole at 1 atm, 273 K
<h3>
Explanation:</h3>
<u>Step 1: Define</u>
0.07 mol Ne (g)
<u>Step 2: Identify Conversions</u>
STP - 22.4 L per mole
<u>Step 3: Convert</u>
- Set up:
- Multiply:
<u>Step 4: Check</u>
<em>Follow sig fig rules and round. We are given 1 sig fig.</em>
1.568 L Ne ≈ 2 L Ne
Polar covalent, it is what happens when there is unequal bond between two atoms, where one has a slightly negative charge and one has a slightly positive charge.
Answer:
Y = 62.5%
Explanation:
Hello there!
In this case, for the given chemical reaction whereby carbon dioxide is produced in excess oxygen, it is firstly necessary to calculate the theoretical yield of the former throughout the reacted 10 grams of carbon monoxide:
Finally, given the actual yield of the CO2-product, we can calculate the percent yield as shown below:
Best regards!