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Vanyuwa [196]
3 years ago
14

What are all the different types of chemical change?

Chemistry
1 answer:
creativ13 [48]3 years ago
4 0

Answer:

the types of chemical reaction are combination, decomposition, single replacement, double replacement, combustion

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What is physical property?
ExtremeBDS [4]
A physical property is any property that is measurable, whose value describes a state of a physical system. The changes in the physical properties of a system can be used to describe its changes between momentary states. Physical properties are often referred to as observables. They are not modal properties. 
3 0
3 years ago
Boron has an atomic mass of 10.811. What does this mean?
icang [17]

Answer:

Knowing that boron has an atomic mass of 10,811 means that all boron isotopes on average weigh 10,811 u.

Explanation:

The atomic mass of an atom is the mass of the atom measured in u (unified atomic mass unit), although we can also express it as Da (Dalton's unit)

Atomic mass refers to the average mass that all isotopes of that element have.

When we speak of isotopes we are referring to the element itself but with a different number of neutrons, which makes it modify its mass number.

6 0
3 years ago
Perform the indicated conversion: 1.345 kcal =
Novay_Z [31]

Answer:

Use x calc

Explanation:

4 0
3 years ago
Fill in the chart to describe and give examples of physical changes.
frosja888 [35]

Answer:

here are some examples of physical change!!!

Explanation:

-An ice cube melting into water in your drink.

-Freezing water to make ice cubes.

-Boiling water evaporating.

-Hot shower water turning to steam.

-Steam from the shower condensing on a mirror.

8 0
3 years ago
Read 2 more answers
Assume that silver and gold form ideal, random mixtures. Calculate the mass of pure Ag needed to cause an entropy increase of 20
KengaRu [80]

Answer:

m_{Ag}=2,265.9g

Explanation:

Hello!

In this case, since the definition of entropy in a random mixture is:

\Delta S=-n_TR\Sigma[x_i*ln(x_i)]

For this silver-gold mixture we write:

\Delta S=-(n_{Au}+n_{Ag})R\Sigma[\frac{n_{Au}}{n_{Au}+n_{Ag}} *ln(\frac{n_{Au}}{n_{Au}+n_{Ag}} )+\frac{n_{Ag}}{n_{Au}+n_{Ag}} *ln(\frac{n_{Ag}}{n_{Au}+n_{Ag}} )]

By knowing the moles of gold:

n_{Au}=100g*\frac{1mol}{197g} =0.508mol

It is possible to write the aforementioned formula in terms of the variable x representing the moles of silver:

20\frac{J}{mol}=-(0.508+x)8.314\frac{J}{mol*K} \Sigma[\frac{0.508}{0.508+x} *ln(\frac{0.508}{0.508+x} )+\frac{x}{0.508+x} *ln(\frac{x}{0.508+x} )]

Which can be solved via Newton-Raphson or a solver software, in this case, I will provide you the answer:

x=n_{Ag}=21.0molAg

So the mass is:

m_{Ag}=21.0mol*\frac{107.9g}{1mol}\\ \\m_{Ag}=2,265.9g

Best regards!

3 0
3 years ago
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