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3 years ago

To convert from °F to °C: T(°C) = T(°F - 32) × 5/9

Chemistry

1 answer:

DaniilM [7]3 years ago

4 0

Answer:

Your notation is a bit confusing, let me write it more clearly.

Explanation:

( Temperature in °F − 32) × 5/9 = Temperature in °C

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A gas that occupies 50.0 liters has its volume increased to 68 liters when the pressure was changed to 3.0 ATM. What was the ori

Yuliya22 [10]

Answer:

4.1 atm = 3,116 mmHg = 415.4 kPa

Explanation:

According to Boyle's law, as volume is increased the pressure of the gas is decreased. That can be expressed as:

P₁ x V₁= P₂ x V₂

Where P₁ and V₁ are the initial pressure and volume respectively, and P₂ and V₂ are final pressure and volume, respectively.

From the problem, we have:

V₁= 50.0 L

V₂= 68.0 L

P₂= 3.0 atm

Thus, we calculate the initial pressure as follows:

P₁= (P₂ x V₂)/V₁= (3.0 atm x 68.0 L)/(50.0 L)= 4.08 atm ≅ 4.1 atm

To transform to mmHg, we know that 1 atm= 760 mmHg:

4.1 atm x 760 mmHg/1 atm = 3,116 mmHg

To transform to kPa we use: 1 atm= 101.325 kPa

4.1 atm x 101.325 kPa = 415.4 kPa

5 0

3 years ago

After sugar dissolves in water, the sugar is no

katrin [286]

Answer:

Because it devolved

Explanation:

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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 $0.449J/g^oC$ ).

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ANSWER: B salt does not evaporate with water

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