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3 years ago

You have a 15.0 gram sample of gold at 20.0°C. How much heat does it take to raise the temperature to 100.0°C?

Chemistry

1 answer:

Nadusha1986 [10]3 years ago

7 0

Answer:

=154.8 J

Explanation:

The rise in temperature is contributed by the change in temperature.

Change in enthalpy = MC∅, where M is the mass of the substance, C is the specific heat capacity and ∅ is the change in temperature.

Change in temperature = 100.0°C-20.0°C=80°C

ΔH=MC∅

The specific heat capacity of gold= 0.129 J/g°C

ΔH= 15.0g×0.129J/g°C×80°C

=154.8 J

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Which gas law (choices are Charles' Law, Gay-Lussac's Law, or Boyle's Law) explains each scenario:________.

posledela

Answer:

A. Boyle's Law

B. Charles' Law

C. Gay-Lussac's Law

Explanation:

An air bag inflates due to the decomposition of sodium azide or NaN₃ to completely fill the bag with nitrogen gas which is an example of Boyle's law, which states that the pressure of a given mass of gas is inversely proportional to its volume, hence due to the estricted volume of the airbag, the pressure of the nitrogen gas in the bag increses protecting the occupants of a cr from injuries in a crash

Helium balloon decrease in sice in a freezer is an example of Charlles law which states that the volume of a given mass of gas is nverslely proportionl to its temperature at constant pressure

A can of spray paint will explode if tossed into a fire is an example of Gay-Lussac's Law which states that the pressure of a given mass of gas is directly proportional to its temperature hence the increased pressure causes the can ti explode

4 0

2 years ago

Which of the following things would be the MOST irresistible for a dog?

kolbaska11 [484]

Answer:

B.a fire hydrant

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5 0

2 years ago

Read 2 more answers

Picture of gas laws question below:

Vera_Pavlovna [14]

Volume of the tank is 5.5 litres.

Explanation:

mass of the CO2 is given 8.6 grams

Pressure of the gas is 89 Kilopascal which is 0.8762 atm

Temperature of the gas is 29 degrees ( 0 degrees +273.5= K) so (29+273)

R = gas constant 0.0821 liter atmosphere per kelvin)

FROM THE IDEAL GAS LAW

PV=nRT ( P Pressure, V Volume, n is number of moles of gas, R gas constant, Temperature in Kelvin)

no of moles = mass/atomic mass

= 8.6/44

= 0.195 moles

now putting the values in equation

V=nRT/P

= 0.195*0.0821*302/ 0.8762

= 5.5 litres.

As the carbon dioxide gas occupies the volume os the tank hence volume of tank is 5.5 litres.

4 0

2 years ago

Consider 100.0 g samples of two different compounds consisting only of carbon and oxygen. One compound contains 27.2 g of carbon

Pani-rosa [81]

Answer: The ratio of carbon in both the compounds is 1 : 2

Explanation:

Law of multiple proportions states that when two elements combine to form two or more compounds in more than one proportion. The mass of one element that combine with a given mass of the other element are present in the ratios of small whole number. For Example: $Cu_2O\text{ and }CuO$

For Sample 1:

Total mass of sample = 100 g

Mass of carbon = 27.2 g

Mass of oxygen = (100 - 27.7) = 72.8 g

To formulate the formula of the compound, we need to follow some steps:

Step 1: Converting the given masses into moles.

Moles of Carbon = $\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{27.2g}{12g/mole}=2.26moles$

Moles of Oxygen = $\frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{72.8g}{16g/mole}=4.55moles$

Step 2: Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 2.26 moles.

For Carbon = $\frac{2.26}{2.26}=1$

For Oxygen = $\frac{4.55}{2.26}=2.01\approx 2$

Step 3: Taking the mole ratio as their subscripts.

The ratio of C : O = 1 : 2

Hence, the formula for sample 1 is $CO_2$

For Sample 2:

Total mass of sample = 100 g

Mass of carbon = 42.9 g

Mass of oxygen = (100 - 42.9) = 57.1 g

To formulate the formula of the compound, we need to follow some steps:

Step 1: Converting the given masses into moles.

Moles of Carbon = $\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{42.9g}{12g/mole}=3.57moles$

Moles of Oxygen = $\frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{57.1g}{16g/mole}=3.57moles$

Step 2: Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 3.57 moles.

For Carbon = $\frac{3.57}{3.57}=1$

For Oxygen = $\frac{3.57}{3.57}=1$

Step 3: Taking the mole ratio as their subscripts.

The ratio of C : O = 1 : 1

Hence, the formula for sample 1 is $CO$

In the given samples, we need to fix the ratio of oxygen atoms.

So, in sample one, the atom ratio of oxygen and carbon is 2 : 1.

Thus, for 1 atom of oxygen, the atoms of carbon required will be = $\frac{1}{2}\times 1=\frac{1}{2}$

Now, taking the ratio of carbon atoms in both the samples, we get:

$C_1:C_2=\frac{1}{2}:1=1:2$

Hence, the ratio of carbon in both the compounds is 1 : 2

8 0

2 years ago

How many grams of material is lost in the aqueous phase if two extractions are carried out on 100 mL of a 5% (m/v) aqueous solut

san4es73 [151]

Answer:

4.94g of material

Explanation:

Partition coefficient (Kp) of a substance is defined as the ratio between concentration of organic solution and aqueous solution, that is:

Kp = 8 = Concentration in Ethyl acetate / Concentration in water

100mL of a 5% solution contains 5g of material in 100mL of water. Thus:

8 = X / 100mL / (5g-X) / 100mL

Where X is the amount of material in grams that comes to the organic phase.

8 = X / 100mL / (5g-X) / 100mL

8 = 100X / (500-100X)

4000 - 800X = 100X

4000 = 900X

4.44g = X

Thus, in the first extraction you will lost 4.44g of material from the aqueous phase.

And will remain 5g-4.44g = 0.56g.

In the second extraction:

8 = X / 100mL / (0.56g-X) / 100mL

8 = 100X / (56-100X)

448 - 800X = 100X

448 = 900X

0.50g = X

In the second extraction, you will extract 0.50g of material

Thus, after the two extraction you will lost:

4.44g + 0.50g = 4.94g of material

6 0

2 years ago