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3 years ago

Write the word and balanced chemical equations for the reaction between:

Chemistry

1 answer:

nadezda [96]3 years ago

3 0

Answer:

3 HNO₃ + Fe(OH)₃ → H₂O + Fe(NO₃)₃

Explanation:

An acid reacts with a base producing water and a salt. Having this in mind the reaction of nitric acid (HNO₃) and Iron (III) hydroxide (Fe(OH)₃) is:

HNO₃ + Fe(OH)₃ → H₂O + Fe(NO₃)₃

<em>The H⁺ of the acid reacts with the OH⁻ to produce H₂O. The other ions (Fe³⁺ and NO₃⁻) produce the salt</em>

<em />

There are 3 nitrates in products. To balance the nitrates:

<em>And this is the balanced reaction</em>

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Differences and similarities between liquid and gas

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<span>Various gases and liquids have different densities and combustion points.</span>

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3 years ago

92dm3 of a gas dissolved in a solvent. How many moles is this?

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Assuming its at r. t.p and pressure
no. of moles = 96/24=4moles
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3 years ago

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Dahasolnce [82]

Answer:

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Explanation:

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2 years ago

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3 years ago

n unknown metal is either aluminum, iron or lead. If 150. g of this metal at 150.0 °C was placed in a calorimeter that contains

Nitella [24]

Answer : The metal used was iron (the specific heat capacity is $0.449J/g^oC$ ).

Explanation :

In this problem we assumed that heat given by the hot body is equal to the heat taken by the cold body.

$q_1=-q_2$

$m_1\times c_1\times (T_f-T_1)=-m_2\times c_2\times (T_f-T_2)$

where,

$c_1$ = specific heat of unknown metal = ?

$c_2$ = specific heat of water = $4.184J/g^oC$

$m_1$ = mass of unknown metal = 150 g

$m_2$ = mass of water = 200 g

$T_f$ = final temperature of water = $34.3^oC$

$T_1$ = initial temperature of unknown metal = $150.0^oC$

$T_2$ = initial temperature of water = $25.0^oC$

Now put all the given values in the above formula, we get

$150g\times c_1\times (34.3-150.0)^oC=-200g\times 4.184J/g^oC\times (34.3-25.0)^oC$

$c_1=0.449J/g^oC$

Form the value of specific heat of unknown metal, we conclude that the metal used in this was iron (Fe).

Therefore, the metal used was iron (the specific heat capacity is $0.449J/g^oC$ ).

6 0

3 years ago