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olga_2 [115]
3 years ago
10

Determine the Concentration of the Unknown Strong Acid

Chemistry
1 answer:
Natasha2012 [34]3 years ago
3 0
Hey baby I know it’s late but I wanted to say how are u doing
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The table shows columns that brenda uses for her notes on the properties of elements. her notes state that some elements can rea
AleksandrR [38]

The element should be placed in the column for metalloids. Option A

<h3>What is the property of metals?</h3>

We know that the properties of metals can be used to identify an unknown substances as a metal. We know that metals are able to react with acids to liberate the metal salt and hydrogen gas. Metals do not react with bases because they can not accept electrons.

Metalloids are generally amphoteric in nature. They can react with both the acid and the basses to form compounds. The nonmetals do not react with acid or base.

Looking at the result, we are told that the entry in the second row have the entries  nothing, nothing, nothing. If this is the outcome, then it follows that the element should be placed in the column for metalloids.

Learn more about metalloids:brainly.com/question/2548493

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4 0
2 years ago
Write the balanced equation for the burning of nonane, c9h20, in air.
Ivanshal [37]
C9H20 + 14O2 --> 9CO2 + 10H2O
3 0
3 years ago
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daser333 [38]
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8 0
3 years ago
Question
madam [21]

Answer:

The specific heat of the metal is 2.09899 J/g℃.

Explanation:

Given,

For Metal sample,

mass = 13 grams

T = 73°C

For Water sample,

mass = 60 grams

T = 22°C.

When the metal sample and water sample are mixed,

The addition of metal increases the temperature of the water, as the metal is at higher temperature, and the  addition of water decreases the temperature of metal. Therefore, heat lost by metal is equal to the heat gained by water.

Since, heat lost by metal is equal to the heat gained by water,

Qlost = Qgain

However,

Q = (mass) (ΔT) (Cp)

(mass) (ΔT) (Cp) = (mass) (ΔT) (Cp)

After mixing both samples, their temperature changes to 27°C.

It implies that , water sample temperature changed from  22°C to 27°C and metal sample temperature changed from 73°C to 27°C.

Since, Specific heat of water = 4.184 J/g°C

Let Cp be the specific heat of the metal.

Substituting values,

(13)(73°C - 27°C)(Cp) = (60)(27°C - 22℃)(4.184)

By solving, we get Cp =

Therefore, specific heat of the metal sample is 2.09899 J/g℃.

5 0
3 years ago
I begin the reaction with 0.45 g of beryllium. If my actual yield of beryllium chloride (mm = 79.91 g/mol) was 3.5 grams, what w
trasher [3.6K]

The percentage of yield was 777.78%

<u>Explanation:</u>

We have the equation,

Be [s]  +  2 HCl [aq]  →  BeCl 2(aq]  + H 2(g]  ↑ Be (s]  + 2 HCl [aq]  →  BeCl 2(aq]  + H 2(g] ↑

To find the percent yield we have the formula

Percentage of Yield= what you actually get/ what you should theoretically get  x 100

                                   =3.5 g/0.45 g 100

                                    = 777.78 %

The percentage of yield was 777.78%

5 0
3 years ago
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