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2 years ago

What can occur in physical change ?

1 answer:

5 0

Within a physical change, an element can change forms, such as going from solid to a liquid through melting. Color change can also occur during a physical change. Physical changes are very different from chemical changes. In a chemical change the element itself changes into something else within a reaction, such as combustion (burning).

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If the container is closed and the ethanol is allowed to reach equilibrium with its vapor, how many grams of liquid ethanol rema

SVETLANKA909090 [29]

Explanation:

Let us assume that the given data is as follows.

V = 3.10 L, T = $19^{o}C$ = (19 + 273)K = 292 K

P = 40 torr (1 atm = 760 torr)

So, P = $\frac{40 torr}{760 torr} \times 1 atm$

= 0.053 atm

n = ?

According to the ideal gas equation, PV = nRT.

Putting the given values into the above equation to calculate the value of n as follows.

PV = nRT

$0.053 atm \times 3.10 L = n \times 0.0821 L atm/mol K \times 292 K$

0.1643 = $n \times 23.97$

n = $6.85 \times 10^{-3}$

It is known that molar mass of ethanol is 46 g/mol. Hence, calculate its mass as follows.

No. of moles = $\frac{mass}{\text{molar mass}}$

$6.85 \times 10^{-3} = \frac{mass}{46 g/mol}$

mass = $315.1 \times 10^{-3}$ g

= 0.315 g

Thus, we can conclude that the mass of liquid ethanol is 0.315 g.

4 0

2 years ago

How can you balance this chemical equation?HSiCl3+H2O~H10Si10O15+HCL

qwelly [4]

10HSiCl3 + 15H2O = H10Si10O15 + 30HCl

4 0

3 years ago

Give five properties of water.

Gelneren [198K]

Polarity, cohesion, adhesion, surface tension, high specific heat, and evaporating cooling

3 0

2 years ago

Read 2 more answers

For which of the following elements would the most common ion be expected to have a larger radius than that of its corresponding

Andre45 [30]

Answer: The ion that is expected to have a larger radius than the corresponding atom is chlorine.

Explanation:

There are two types of ions:

Cations: They are formed when an atom looses its valence electrons. They are positive ions.
Anions: They are formed when an atom gain electrons in its outermost shell. They are negative ions.

For positive ions, the removal of electron increases the nuclear charge for an outermost electron because the outermost electrons are more strongly attracted by the nucleus. So, the effective nuclear charge increases for cations and thus, the size of the cation will be smaller than that of the corresponding atom.

For negative ions, the addition of electron decreases the nuclear charge for an outermost electron because the outermost electrons are less strongly attracted by the nucleus. So, the effective nuclear charge decreases for anions and thus, the size of the anion will be larger than that of the corresponding atom.

For the given options:

Option a: Chlorine

Chlorine gains 1 electron and form $Cl^-$ ion

Option b: Sodium

Sodium looses 1 electron and form $Na^+$ ion

Option c: Copper

Copper looses 2 electrons and form $Cu^{2+}$ ion

Option d: Strontium

Strontium looses 2 electrons and form $Sr^{2+}$ ion

Hence, the ion that is expected to have a larger radius than the corresponding atom is chlorine.

8 0

2 years ago

In each case, calculate the appropriate ratio to show that the information given is consistent with the law of multiple proporti

Alchen [17]

Answer:

(a) 3:2; (b) 2:1

Explanation:

The Law of Multiple Proportions states that when two elements A and B combine to form two or more compounds, the masses of B that combine with a given mass of A are in the ratios of small whole numbers.

That is, if one compound has a ratio r₁ and the other has a ratio r₂, the ratio of the ratios r is in small whole numbers.

(a) Ammonia and hydrazine.

In ammonia, the mass ratio of H:N is r₁ = 0.2158/1

In hydrazine, the mass ratio of H:N is r₂ = 0.1439/1

The ratio of the ratios is:

$r = \dfrac{r_{1}}{r_{2}} = \dfrac{ 0.2158}{0.1439} = \dfrac{1.500 }{1} = \dfrac{2.999}{2} \approx \mathbf{\dfrac{3}{2}}\\\\\text{The relative amounts of H in the two compounds are in the ratio }\boxed{\mathbf{\dfrac{3}{2}}}$

(b) Nitrogen oxides

In nitrogen monoxide, the mass ratio of O:N is r₁ = 1.142/1

In dinitrogen monoxide, the mass ratio of O:N is r₂ = 0.571/1

The ratio of the ratios is:

$r = \dfrac{r_{1}}{r_{2}} = \dfrac{ 1.142}{0.571} = \dfrac{2.000 }{1} \approx \mathbf{\dfrac{2}{1}}\\\\\text{The relative amounts of O in the two compounds are in the ratio }\boxed{\mathbf{\dfrac{2}{1}}}$

8 0

3 years ago