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Neko [114]
3 years ago
13

Looking at the following equation, what coefficient should be placed in front of the H2? Fe + H2SO4 --> Fe2(SO4)3 + H2

Chemistry
1 answer:
Hatshy [7]3 years ago
6 0

Answer:

3 (three)

Explanation:

2 Fe + 3H2SO4 = Fe2(SO4)3 + 3 H2 (basically just balance both sides)

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Write the structure of butanoate ion.
drek231 [11]

Answer:

Explanation:

Molecular FormulaC4H7O2

Average mass87.098 Da

Monoisotopic mass87.045151 Da

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3 years ago
What is an example of a qualitative observation?
Lunna [17]

Answer:

20 cats.

Explanation:

7 0
3 years ago
What volume of 0.08892 M HNO3 is required to react completetly with 0.2352 g of potassium hydrogen phosphate?
galina1969 [7]

Answer:

0.0303 Liters

Explanation:

Given:

Mass of the potassium hydrogen phosphate = 0.2352

Molarity of the HNO₃ Solution = 0.08892 M

Now,

From the reaction it can be observed that 1 mol of potassium hydrogen phosphate reacts with 2 mol of HNO₃

The number of moles of 0.2352 g of potassium hydrogen phosphate

= Mass / Molar mass

also,

Molar mass of potassium hydrogen phosphate

= 2 × (39.09) + 1 + 30.97 + 4 × 16 = 174.15 g / mol

Number of moles = 0.2352 / 174.15 = 0.00135 moles

thus,

The number of moles of HNO₃ required for  0.00135 moles

= 2 ×  0.00135 mol of HNO₃

= 0.0027 mol of HNO₃

Now,

Molarity = Number of Moles / Volume

thus,

for 0.0027 mol of HNO₃, we have

0.08892 = 0.0027 / Volume

or

Volume =  0.0303 Liters

8 0
3 years ago
Read 2 more answers
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℃.

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The flow of electricity is an electric current. The path that an electric currents floes is a circuit.
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