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neonofarm [45]
3 years ago
12

a sample of ideal gas at room temperature occupies a volume of 12.0 L at a pressure of 392 torr. if the pressure changes to 1960

torr, with no change in temperature or moles of gas, what is the new volume, V2?
Chemistry
1 answer:
Setler79 [48]3 years ago
3 0
852 x 40.0= 4260 x v2 
<span>v2= 8 L </span>

<span>b) 852 x 40 = p2 x 69.0 </span>
<span>p2= 494 torr</span>
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What would most accurately measure the volume of water or other liquid?
Afina-wow [57]

Answer:

D. Graduated Cylinder

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how much energy would it take to heat a section of the copper tubing that weights about 660.0 gram, from 12.93 degree Celsius to
expeople1 [14]

Answer:

                      Q  =  2647 J

Explanation:

                    Specific heat capacity is the amount of energy required by one Kg of a substance to raise its temperature by 1 °C.

In thermodynamics the equation used is as follow,

                                                 Q  =  m Cp ΔT

Where;

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           ΔT  =  Change in Temperature  =  23.35 °C - 12.93 °C  =  10.42 °C

Putting values in eq. 1,

                            Q  =  660 g × 0.3850 J.g⁻¹.°C⁻¹ ×  10.42 °C

                            Q  =  2647 J

8 0
3 years ago
How many 325mg aspirin tablets can be made from 3628 grams of aspirins ? Need an answer by tonight it’s getting late .
krek1111 [17]

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3 years ago
Please help me with chemistry. Thanks
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Constant:
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Independent:
<span>volume of gas

Dependent:
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7 0
3 years ago
Read 2 more answers
A common laboratory reaction is the neutralization of an acid with a base. When 31.8 mL of 0.500 M HCl at 25.0°C is added to 68.
lord [1]

Answer:

The correct answer to the following question will be "90.6 kJ/mol".

Explanation:

The total reactant solution will be:

(31.8 \ mL+68.9 \ mL)\times 1.07\ g/mL = 107.74 \ g

The produced energy will be:

=4.18 \ J/(gK)\times 107.74 \ g\times (28.2-25.0)K

=450.35\times 3.2

=1441.12 \ J

The reaction will be:

⇒  HCl+NaOH \rightarrow NaCl+H_{2}O

Going to look at just the amounts of reactions with the same concentrations, we notice that they're really comparable.  

Therefore, the moles generated by NaCl will indeed be:

=  (\frac{31.8}{1000} \ L)\times (0.500 \ M \ HCl/L)\times \frac{1 \ mol \ NaCl}{1 \ mol \ HCl}

=  0.0318\times 0.500

=  0.0159 \ mole  \ of \ NaCl

Now,

=  \frac{1441.12 \ J}{0.0159 \ moles \ NaCl}

=  906364.7

=  90.6 \ KJ/mol \ NaCl

7 0
3 years ago
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