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Mademuasel [1]
3 years ago
9

What is the difference between property and change?

Chemistry
2 answers:
miss Akunina [59]3 years ago
8 0
If your talking about physical property and chemical change then , a physical property is something that can be observed without changing and a chemical change , changes into a new substance.
 
maria [59]3 years ago
5 0
A physical property is an aspect of matter that can be observed or measured without changing it. Examples of physical properties include color, molecular weight and volume. A chemical property may only be observed by changing the chemical identity of a substance.
You might be interested in
Is Helium a compound,element, or.mixture
dusya [7]

Answer:

element

Explanation:

we know that helium is a pure substance although helium atoms are sometimes mixed with their isotopes it is still the same element. since there is no other element combined with helium this makes it an element.

5 0
3 years ago
Prepare indicators by using natural materials like extract of red cabbage leaves, crushed beetroot,
AysviL [449]

Answer:

Y

ou have learnt in your previous classes that the sour and bitter

tastes of food are due to acids and bases, respectively, present in them.

If someone in the family is suffering from a problem of acidity after

overeating, which of the following would you suggest as a remedy– lemon

juice, vinegar or baking soda solution?

n Which property did you think of while choosing the remedy?

Surely you must have used your knowledge about the ability of

acids and bases to nullify each other’s effect.

n Recall how we tested sour and bitter substances without tasting

them.

You already know that acids are sour in taste and change the colour

of blue litmus to red, whereas, bases are bitter and change the colour of

the red litmus to blue. Litmus is a natural indicator, turmeric is another

such indicator. Have you noticed that a stain of curry on a white cloth

becomes reddish-brown when soap, which is basic in nature, is scrubbed

on it? It turns yellow again when the cloth is washed with plenty of

water. You can also use synthetic indicators such as methyl orange and

phenolphthalein to test for acids and bases.

In this Chapter, we will study the reactions of acids and bases, how

acids and bases cancel out each other’s effects and many more interesting

things that we use and see in our day-to-day life. Do You Know?

Litmus solution is a purple dye, which is extracted from lichen, a plant belonging to

the division Thallophyta, and is commonly used as an indicator. When the litmus

solution is neither acidic nor basic, its colour is purple. There are many other natural

materials like red cabbage leaves, turmeric, coloured petals of some flowers such as

Hydrangea, Petunia and Geranium, which indicate the presence of acid or base in a

solution. These are called acid-base indicators or sometimes simply indicators.

2021–22

18 Science

2.1 UNDERSTANDING THE CHEMICAL PROPERTIES OF

ACIDS AND BASES

2.1.1 Acids and Bases in the Laboratory

Activity 2.1

These indicators tell us whether a substance is acidic or basic by

change in colour. There are some substances whose odour changes in

acidic or basic media. These are called olfactory indicators. Let us try

out some of these indicators.

Explanation:

8 0
3 years ago
At 14,000 ft elevation the air pressure drops to 0.59 atm. Assume you take a 1L sample of air at this altitude and compare it to
Ray Of Light [21]

Answer:

There are 0.1125 g of O₂ less in 1 L of air at 14,000 ft than in 1 L of air at sea level.

Explanation:

To solve this problem we use the ideal gas law:

PV=nRT

Where P is pressure (in atm), V is volume (in L), n is the number of moles, T is temperature (in K), and R is a constant (0.082 atm·L·mol⁻¹·K⁻¹)

Now we calculate the number of moles of air in 1 L at sea level (this means with P=1atm):

1 atm * 1 L = n₁ * 0.082 atm·L·mol⁻¹·K⁻¹ * 298 K

n₁=0.04092 moles

Now we calculate n₂, the number of moles of air in L at an 14,000 ft elevation, this means with P = 0.59 atm:

0.59 atm * 1 L = n₂ * 0.082 atm·L·mol⁻¹·K⁻¹ * 298 K

n₂=0.02414 moles

In order to calculate the difference in O₂, we substract n₂ from n₁:

0.04092 mol - 0.02414 mol = 0.01678 mol

Keep in mind that these 0.01678 moles are of air, which means that we have to look up in literature the content of O₂ in air (20.95%), and then use the molecular weight to calculate the grams of O₂ in 20.95% of 0.01678 moles:

0.01678mol*\frac{20.95}{100} *32\frac{g}{mol} =0.1125 gO_{2}

4 0
4 years ago
What type of circuit is it when there are 2 lights in a circuit but only one is working?
IceJOKER [234]
Series circuit is the answer

5 0
3 years ago
A 1.40 L sample of O2 at 645 Torr and 25 °C, and a 0.751 L sample of N2 at 1.13 atm and 25 °C, are both transferred to the same
anyanavicka [17]

Answer:

  • P(O₂) = 0.595 atm
  • P(N₂) = 0.424 atm
  • Total Pressure = 1.019 atm

Explanation:

To solve this problem we use PV=nRT for both gases in their containers, in order to <u>calculate the moles of each one</u>:

  • O₂:

645 Torr ⇒ 645 /760 = 0.85 atm

25°C ⇒ 25 + 273.16 = 298.16 K

0.85 atm * 1.40 L = n * 0.082 atm·L·mol⁻¹·K⁻¹ *298.16 K

n = 0.0487 mol O₂

  • N₂:

1.13 atm * 0.751 L = n * 0.082 atm·L·mol⁻¹·K⁻¹ *298.16 K

n = 0.0347 mol N₂

Now we can <u>calculate the partial pressure for each gas in the new container</u>, because the number of moles did not change:

  • O₂:

P(O₂) * 2.00 L = 0.0487 mol O₂ * 0.082 atm·L·mol⁻¹·K⁻¹ *298.16 K

P(O₂) = 0.595 atm

  • N₂:

P(N₂) * 2.00 L = 0.0347 mol N₂ * 0.082 atm·L·mol⁻¹·K⁻¹ *298.16 K

P(N₂) = 0.424 atm

Finally we add the partial pressures of all gases to <u>calculate the total pressure</u>:

  • Pt = 0.595 atm+ 0.424 atm = 1.019 atm
6 0
4 years ago
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