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Luden [163]
3 years ago
14

A(n) 816 ml sample of hydrogen was collected over water at 21.0◦c on a day when the barometric pressure was 755 torr. what volum

e would the dry hydrogen occupy under standard conditions? the vapor pressure of water at 21.0◦c is 19.0 torr. answer in units of ml.
Chemistry
1 answer:
hammer [34]3 years ago
6 0
The correct answer is A

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Which part of the human nervous system links the brain to the rest of the body
lara31 [8.8K]

Answer: The Spinal Cord

Explanation: The spinal cord runs along the dorsal side of the body and links the brain to the rest of the body.

8 0
3 years ago
What is the IUPAC name of this compound? CI-CH2-CH2 - CH2 - CH 2 -Cl​
MissTica

Answer:

methyl chloride i guess

8 0
3 years ago
What is the mass of 5 moles of Nitrogen gas
love history [14]
Molar mass N₂ = 28.0 g/mol

1 mole ------------- 28.0 g
5 moles ------------ ?

mass = 5 . 28.0 / 1

mass = 140 g

hope this helps!
5 0
3 years ago
Two solutions, initially at 24.60 °C, are mixed in a coffee cup calorimeter (Ccal = 15.5 J/°C). When a 100.0 mL volume of 0.100
-Dominant- [34]

Answer:

∆H°rxn in kJ/mol AgCl = - 59.61 kJ/mol

Explanation:

To solve this question we need to calculate the heat absorbed by the cup calorimeter and by the water in the solutions. We will also need to calculate the amount in moles produced by the reaction since we want to know the ∆H°rxn in kJ/mol AgCl .

mol AgNO₃ = 100 mL x 1L/1000 mL x 0.100 mol/L = 0.01 mol

mol NaCl = 100 mL x 1L/1000 mL x 0.200 mol/L = 0.02 mol

Therefore our limiting reagent is the 0.01 mol AgNO₃ and 0.01 mol AgCl will be produced according to the stoichiometry of the reaction:

AgNO₃ + NaCl ⇒ AgCl + NaNO₃

Heat absorbed by the water:

qw = m(H₂O) x c x ΔT where  m (H₂O) = 200 g ( the density of final solution is  1  g/ml)

c = specific heat of water = 4.18 J/gºC

ΔT = change in temperature =  (25.30 - 24.60 ) ºC = 0.7ºC

qw = 200 g x 4.18 J/gºC x 0.7 ºC = 585.20 J

Heat Absorbed by the calorimeter :

q cal = C cal x  ΔT  = 15.5 J/ºC x 0.7ºC = 10.85 J

Total Heat released by the combustion = qw + qcal = 585.20 J +10.85 J

=  596.05 J

We have to change the sign to this quantity since it is an exotermic reaction  ( ΔT is positive ) and have the ∆Hrxn

∆H rxn  = -596.05 J  

but this  is not what we are being asked since this heat was released by the formation of  0.0100 mol of AgCl so finally  

∆H°rxn = -596.05 J /0.01 mol  = -59,605 J/mol x 1 kJ/1000J = -59.61 kJ/mol

5 0
3 years ago
Given an initial cyclopropane concentration of 0.00560 m, calculate the concentration of cyclopropane that remains after 1.50 ho
maria [59]

<span>We can solve this problem by assuming that the decay of cyclopropane follows a 1st order rate of reaction. So that the equation for decay follows the expression:</span>

A = Ao e^(- k t) 

Where,

A = amount remaining at time t = unknown (what to solve for) <span>
Ao = amount at time zero = 0.00560 M </span><span>
<span>k = rate constant
t = time = 1.50 hours or 5400 s </span></span>

The rate constant should be given in the problem which I think you forgot to include. For the sake of calculation, I will assume a rate constant which I found in other sources:

k = 5.29× 10^–4 s–1                     (plug in the correct k value)

<span>Plugging in the values in the 1st equation:</span>

A = 0.00560 M * e^(-5.29 × 10^–4 s–1 * 5400 s )

A = 3.218 <span>× 10^–4 M           (simplify as necessary)</span>

8 0
3 years ago
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