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marusya05 [52]
3 years ago
10

How many significant figures does the number 125,000 have?

Chemistry
2 answers:
larisa [96]3 years ago
6 0

Answer:

3

Explanation:

1, 2, and 5.

Please mark me brainliest :D

ANEK [815]3 years ago
5 0

Answer:

6

Explanation:

you count the zero because they are in the right side

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Sergio [31]

Answer:

C. The Protons

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3 years ago
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A piece of iron can change in different ways. How is iron bending different from iron rusting
fenix001 [56]
Iron bending is just a physical change because the iron has only changed shape but there is no new substance. Iron rusting is a chemical change because there is a new substance which is rust.
7 0
4 years ago
HELP!
ki77a [65]

Answer:

(1)There are 1.5 moles of water in a 27 gram sample of water. The molar mass of water is 18.02 gmol g m o l .(2)

AnswersChemistryGCSEArticle

What is the mass (g) of 0.25mols of NaCl?

What you need for these equations are a calculator, periodic table and the following equation:

Mass (g) = Mr x Moles (important equation to remember)

In this case we already know the moles as it's in the question, 0.25 moles.

to find the Mr, you need to look at your periodic table. Find the relative atomic mass of Na and Cl and add the two numbers together.

Na = 22.99

Cl = 35.45

NaCl = 58.4

Now just put all of the numbers into the equation.

0.25 x 58.4 = 14.6g

4 0
3 years ago
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Metals posses varying reduction potentials. Iron can reduce Cu+2 (aq) to copper metal. Silver can reduce Au+ (aq) to gold metal.
natta225 [31]

<u>Answer:</u> The true statement is iron can reduce Au^+(aq) to gold metal

<u>Explanation:</u>

Single displacement reaction is defined as the reaction in which more reactive element displaces a less reactive element from its chemical reaction.

The reactivity of metal is determined by a series known as reactivity series. The metals lying above in the series are more reactive than the metals which lie below in the series.

A+BC\rightarrow AC+B

Metal A is more reactive than metal B.

We are given:

Iron can reduce copper, silver can reduce gold, sodium can reduce iron and copper can reduce silver metal.

The increasing order of reactivity thus follows:

Au

where, sodium is most reactive and gold is least reactive

For the given options:

<u>Option 1:</u>  Copper cannot easily reduce sodium ion to sodium metal because it is less reactive.

Cu(s)+Na^+(aq.)\rightarrow \text{ No reaction}

<u>Option 2:</u>  Iron cant easily reduce gold ion to gold metal because it is more reactive.

Fe(s)+3Au^+(aq.)\rightarrow Fe^{3+}(aq.)+3Au(s)

<u>Option 3:</u>  Silver cannot easily reduce iron ion to iron metal because it is less reactive.

Ag(s)+Fe^{3+}(aq.)\rightarrow \text{ No reaction}

Hence, the true statement is iron can reduce Au^+(aq) to gold metal

7 0
3 years ago
What is the molarity of 225 grams of Cu(NO2)2 in a total volume of 2.59 L?
Kay [80]

Answer:

The molarity is 0.56\frac{moles}{L}

Explanation:

In a mixture, the chemical present in the greatest amount is called a solvent, while the other components are called solutes. Then, the molarity or molar concentration is the number of moles of solute per liter of solution.

In other words, molarity is the number of moles of solute that are dissolved in a given volume.

The Molarity of a solution is determined by:

Molarity (M)=\frac{number of moles of solute}{Volume}

Molarity is expressed in units (\frac{moles}{liter}).

Then you must know the number of moles of Cu(NO₂)₂. For that it is necessary to know the molar mass. Being:

  • Cu: 63.54 g/mol
  • N: 14 g/mol
  • O: 16 g/mol

the molar mass of Cu(NO₂)₂ is:

Cu(NO₂)₂= 63.54 g/mol + 2*(14 g/mol + 2* 16 g/mol)= 155.54 g/mol

Now the following rule of three applies: if 155.54 g are in 1 mole of the compound, 225 g in how many moles are they?

moles=\frac{225 g*1 mole}{155.54 g}

moles= 1.45

So you know:

  • number of moles of solute= 1.45 moles
  • volume=2.59 L

Replacing in the definition of molarity:

Molarity=\frac{1.45 moles}{2.59 L}

Molarity= 0.56\frac{moles}{L}

<u><em>The molarity is 0.56</em></u>\frac{moles}{L}<u><em></em></u>

5 0
3 years ago
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