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3 years ago

What is the specific heat of silicon if the temperature of a 4.11 g sample of silicon

Chemistry

1 answer:

monitta3 years ago

4 0

Answer:

The right solution is "0.7107 J/g°C".

Explanation:

The given values are:

Mass,

m = 4.11 g

Temperature,

ΔT = 3.8°C

Heat,

Q = 11.1 J

The specific heat of silicon will be:

⇒ $C=\frac{Q}{m \Delta T}$

On substituting the values, we get

⇒ $=\frac{11.1}{4.11\times 3.8}$

⇒ $=\frac{11.1}{15.618}$

⇒ $=0.7107 \ J/g^{\circ}C$

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Which substance is not a structural isomer of hexyne?

stich3 [128]

Answer:

c) 3,3-dimethylpent-1-yne

5 0

3 years ago

Natural gas is stored in a spherical tank at a temperature of 13°C. At a given initial time, the pressure in the tank is 117 kPa

drek231 [11]

Answer:

1. the absolute pressure in the tank before filling = 217 kPa

2. the absolute pressure in the tank after filling = 312 kPa

3. the ratio of the mass after filling M2 to that before filling M1 = 1.44

The correct relation is option c ( $\frac{M_{2} }{M_{1} } = \frac{P_{2} T_{1} }{P_{1} T_{2} }$ )

Explanation:

To find -

1. What is the absolute pressure in the tank before filling?

2. What is the absolute pressure in the tank after filling?

3. What is the ratio of the mass after filling M2 to that before filling M1 for this situation?

As we know that ,

Absolute pressure = Atmospheric pressure + Gage pressure

So,

Before filling the tank :

Given - Atmospheric pressure = 100 kPa , Gage pressure = 117 kPa

⇒Absolute pressure ( p1 ) = 100 + 117 = 217 kPa

Now,

After filling the tank :

Given - Atmospheric pressure = 100 kPa , Gage pressure = 212 kPa

⇒Absolute pressure (p2) = 100 + 212= 312 kPa

Now,

As given, volume is the same before and after filling,

i.e. $V_{1}$ = $V_{2}$

As we know that, P ∝ M

⇒ $\frac{p_{1} }{p_{2} } = \frac{m_{1} }{m_{2} }$

⇒ $\frac{m_{2} }{m_{1} } = \frac{p_{2} }{p_{1} }$

⇒ $\frac{m_{2} }{m_{1} } = \frac{312 }{217 } = 1.4378$ ≈ 1.44

Now, as we know that PV = nRT

As V is constant

⇒ P ∝ MT

⇒ $\frac{P}{T}$ ∝ M

⇒ $\frac{M_{2} }{M_{1} } = \frac{P_{2} T_{1} }{P_{1} T_{2} }$

So, The correct relation is c option.

6 0

3 years ago

How would you prepare 500 mL of 0.360 M solution of CaCl2 from<br> solid CaCl2?

LenKa [72]

We need to measure 20.0 grams of CaCl₂ to prepare 500 mL of 0.360 M solution.

First, we need to determine the required moles of CaCl₂. We have 500 mL (0.500 L) of a 0.360 M solution (0.360 moles of CaCl₂ per liter of solution).

$0.500 L \times \frac{0.360mol}{L} = 0.180 mol$

Then, we will convert 0.180 moles to grams using the molar mass of CaCl₂ (110.98 g/mol).

$0.180 mol \times \frac{110.98g}{mol} = 20.0 g$

To prepare the solution, we weigh 20.0 g of CaCl₂ and add it to a beaker with enough distilled water to dissolve it. We stir it, heat it if necessary, and when we have a solution, we transfer it to a 500 mL flask and complete it to the mark with distilled water.