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Anastaziya [24]
3 years ago
15

All are examples of charged particles except _________. a. neutrons b. protons c. electrons

Chemistry
2 answers:
trapecia [35]3 years ago
5 0
A. Neutrons is not chaged
NeX [460]3 years ago
4 0
According to the basic facts, the only one non-charged subatomic particle from the scale is definitely neutrons because protons is always positively charged whereas elecrtons is <span>negatively charged. So, the right answer is A.</span>
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Vhat are some of the benefits and drawbacks of recycling?​
Eva8 [605]

Answer:

benefits: can help reduce waste in land fills, and promotes reusability (turned into different products)

drawbacks: can be much more expensive to salvage than to just throw it in a landfill, and most of the time products cannot be recycled effectively due to its material.

5 0
3 years ago
Convert 3.30 g of copper (II) hydroxide Cu(OH)2 to molecules.
ratelena [41]

Answer:

0.18× 10²³ molecules

Explanation:

Given data:

Mass of copper hydroxide = 3.30 g

Number of molecules = ?

Solution:

Number of moles = mass/molar mass

Number of moles = 3.30 g/97.56 g/mol

Number of moles = 0.03 mol

Avogadro number:

The given problem will solve by using Avogadro number.

It is the number of atoms , ions and molecules in one gram atom of element, one gram molecules of compound and one gram ions of a substance.  The number 6.022 × 10²³ is called Avogadro number.

1 mole = 6.022 × 10²³ molecules

0.03 mol × 6.022 × 10²³ molecules / 1mol

0.18× 10²³ molecules

6 0
3 years ago
Determine the molarity of a solution with a volume of 435. mL and 0.550 mol of solute dissolved.
lora16 [44]

Answer:

M = 1.26

Explanation:

Molarity = mole of solution/liters of solution
435mL/1000 = .435L
Plugging in the numbers into the formula, we get:
Molarity = .550 mol/.435L = 1.26 M

8 0
2 years ago
<img src="https://tex.z-dn.net/?f=H_2PO_4%5E-%28aq%29%20%5Crightarrow%20H%5E%2B%28aq%29%20%2B%20HPO_4%5E%7B2-%7D%28aq%29" id="Te
klasskru [66]

Answer:

The pH of the buffer solution = 8.05

Explanation:

Using the Henderson - Hasselbalch equation;

pH = pKa₂ + log ( [HPO₄²-]/[H₂PO4⁻]

where pKa₂ = -log (Ka₂) = -log ( 6.1 * 10⁻⁸) = 7.21

Concentration of OH⁻ added = 0.069 M (i.e. 0.069 mol/L)

[H₂PO4⁻] after addition of OH⁻ = 0.165 - 0.069 = 0.096 M

[HPO₄²-] after addition of OH⁻ = 0.594 + 0.069 = 0.663 M

Therefore,

pH = 7.21 + log (0.663 / 0.096)

pH = 7.21 + 0.84

pH = 8.05

4 0
3 years ago
What are the products of the complete combustion of a hydrocarbon in excess oxygen
dangina [55]
The products for the complete combustion of a hydrocarbon in excess air is carbon dioxide and water. Any hydrocarbon when reacted with oxygen will always yield the said products. Incomplete combustion, on the other hand, yields carbon monoxide and water.
4 0
4 years ago
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