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3 years ago

Any atom can be considered stable if it has what ?

Chemistry

1 answer:

Zinaida [17]3 years ago

5 0

Any atom can be considered stable if it has an equal number of protons and electrons, giving it a net charge of zero.

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A student decomposed 3.67g of copper (ii) hydroxide into copper (ii) oxide. how many ml of 3m h2so4 is need to react with all th

vredina [299]

In the presence of heat, copper (II) hydroxide decomposes in to copper (II) oxide.
Cu(OH)₂ (s) ----> CuO (s) + H₂O (l)
upon decomposition, water is removed from Cu(OH)₂
the amount of Cu(OH)₂ decomposed - 3.67 g
number of moles of Cu(OH)₂ - 3.67 g / 97.5 g/mol = 0.038 mol
stoichiometry of Cu(OH)₂ to CuO is 1:1
therefore number of CuO moles formed are - 0.038 mol
CuO reacts with sulfuric acid to form CuSO₄
CuO + H₂SO₄ ---> CuSO₄ + H₂O
stoichiometry of CuO to H₂SO₄ is 1:1
therefore number of H₂SO₄ moles that should react is 0.038 mol
the molarity of H₂SO₄ is 3M
this means that in 1000 ml - 3 mol of H₂SO₄ present
so if 3 mol are present in 1000 ml
then volume for 0.038 mol = 1000/3 * 0.038
= 12.67 ml

6 0

4 years ago

Continental waters are classified into two categories Lentic and Lotic. A Lotic Ecosystem has flowing waters, such as creeks,

Flauer [41]

Wind because it’s abiotic which means non living

8 0

3 years ago

What's the balanced equation for magnesium and hydrochloric acid?

Roman55 [17]

Mg + 2 HCl ------> MgCl₂ +H₂O

6 0

4 years ago

An aluminum cup of 140 cm3 capacity is completely filled with glycerin at 20°C. How much glycerin will spill out of the cup if t

mel-nik [20]

Explanation:

The given data is as follows.

coefficient of volume expansion of glycerin ( $\beta$ ) = $5.1 \times 10^{-4} 1/^{o}C$

linear expansion coefficient of aluminum, $(\alpha_{A})$ = $23 \times 10^{-6} 1/^{o}C$

Volume = 100 $cm^{3}$

The increase in volume of the cup will be calculated as follows.

$\Delta V_{c} = V \times 3 \times \alpha \times \Delta T$

= $100 cm^{3} \times 3 \times 23 \times 10^{-6} \times (31 - 20)^{o}C$

= $75900 \times 10^{-6} cm^{3}$

= 0.0759 $cm^{3}$

Formula for increase in volume of glycerine is as follows.

$\Delta V_{g} = V \times \beta_{g} \times \Delta T$

= $100 cm^{3} \times 5.1 \times 10^{-4} 1/^{o}C \times (31 - 20)^{o}C$

= 0.5610 $cm^{3}$

Therefore, volume of glycerin spilled is calculated as follows.

$\Delta V = \Delta V_{g} - \Delta V_{c}$

= (0.5610 - 0.0759) $cm^{3}$

= 0.4851 $cm^{3}$

Thus, we can conclude that 0.4851 $cm^{3}$ glycerin will spill out of the cup.

6 0

3 years ago

The heat of vaporization for liquid zinc is 1.76 kj/g. How much heat is needed to boil 11.2 g of liquid zinc already at its boil

sladkih [1.3K]

<h3>Answer:</h3>

= 19.712 kJoules

<h3>Explanation:</h3>

Heat of vaporization refers to the amount of heat required to change a unit mass of a substance from liquid to gaseous state without change in temperature.

To calculate the amount of heat, we use,

Amount of heat = Mass × Heat of vaporization

Q = m×Lv

Given;

Mass of liquid Zinc = 11.2 g

Lv of liquid Zinc = 1.76 kJ/g

Therefore;

Q = 11.2 g × 1.76 kJ/g

= 19.712 kJ

Thus, the amount of heat needed to boil 11.2 g of zinc is 19.712 kilo-joules.

6 0

3 years ago