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ANTONII [103]
3 years ago
6

SCIENCE QUESTIONS about chemical and physical changes (image linked with a question)

Chemistry
1 answer:
konstantin123 [22]3 years ago
8 0
Chemical is the answer why don’t you hook it up bro
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Identify ways to reduce the risk of liquid bumping while heating.
postnew [5]

The most common way of preventing bumping is by adding one or two boiling chips to the reaction vessel. However, these alone may not prevent bumping and for this reason it is advisable to boil liquids in a boiling tube, a boiling flask, or an Erlenmeyer flask.

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2 years ago
Please help me..<br>what is carbon 1/12​
tatuchka [14]

Answer:

It is expressed as a multiple of one-twelfth the mass of the carbon-12 atom, 1.992646547 × 10−23 gram, which is assigned an atomic mass of 12 units. ... In this scale 1 atomic mass unit (amu) corresponds to 1.660539040 × 10−24 gram.

6 0
3 years ago
Which statement describes the composition of potassium chlorate, KClO3?
Aliun [14]
The statement that best describes the composition of potassium chlorate, KCIO3 is "<span> The proportion by mass of elements combined in potassium chlorate is fixed."</span>
5 0
3 years ago
Read 2 more answers
The expression below was formed by combining different gas laws. V is proportional to StartFraction n T over P EndFraction. Whic
Ronch [10]

Answer:

The Ideal gas law

Explanation:

From the given question, we have:

V \alpha \frac{nT}{P}

where each variable has its usual meaning.

Thus,

V = \frac{nRT}{P}

where R is the ideal gas constant

cross multiply to have;

PV = nRT

This implies that the volume of the gas is directly proportional to the number of moles of the gas.

Therefore, the law can be used to determine the relationship between the volume and number of moles is the ideal gas law.

3 0
2 years ago
The equilibrium constant is given for two of the reactions below. Determine the value of the missing equilibrium constant. 2A(g)
monitta

Answer:

The value of the missing equilibrium constant ( of the first equation) is 1.72

Explanation:

First equation: 2A + B ↔ A2B   Kc = TO BE DETERMINED

 ⇒ The equilibrium expression for this equation is written as: [A2B]/[A]²[B]

Second equation: A2B + B ↔ A2B2   Kc= 16.4

⇒ The equilibrium expression is written as: [A2B2]/[A2B][B]

Third equation:  2A + 2B ↔ A2B2     Kc = 28.2

⇒ The equilibrium expression is written as: [A2B2]/ [A]²[B]²

If we add the first to the second equation

2A + B + B ↔ A2B2   the equilibrium constant Kc will be X(16.4)

But the sum of these 2 equations, is the same as the third equation ( 2A + 2B ↔ A2B2)   with Kc = 28.2

So this means: 28.2 = X(16.4)

or X = 28.2/16.4

X = 1.72

with X = Kc of the first equation

The value of the missing equilibrium constant ( of the first equation) is 1.72

7 0
3 years ago
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