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arlik [135]
4 years ago
12

Please help :') *picture attached !*

Chemistry
1 answer:
DaniilM [7]4 years ago
6 0

Answer:

Pressure is 2.32 atm

Explanation:

Pressure . Volume = moles . Constant Ideal Gases . Temperature (K)

Pressure = (moles . Constant Ideal Gases . T° K) / Volume

T°K = T°C + 273 → 61°C + 273 = 334 K

Let's replace every data:

P = (0.214 mol . 0.082 L.atm/mol.K . 334K) / 2.53L

P = 2.32 atm

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Answer: operating machines they are easy to handle
8 0
3 years ago
What does he want? change into passive ​
Marina CMI [18]
There’s no picture if you meant to add a picture
3 0
3 years ago
Mothballs are composed primarily of the hydrocarbon naphthalene (C10H8). When 1.025 g of naphthalene is burned in a bomb calorim
Serggg [28]

Answer:

\Delta E_{rxn} for combustion of naphthalene is -5164 kJ/mol

Explanation:

\Delta E_{rxn}=C_{calorimeter}\times \Delta T_{calorimeter}

where, C refers heat capacity and \Delta T refers change in temperature.

Here, \Delta T_{calorimeter}=(32.33-24.25)^{0}\textrm{C}=8.08^{0}\textrm{C}

So, \Delta E_{rxn}=(5.11\frac{kJ}{^{0}\textrm{C}})\times (8.08^{0}\textrm{C})=41.3kJ

\Delta E_{rxn} is generally expressed in terms of per mole unit of reactant. Also, \Delta E_{rxn} should be negative as it is an exothermic reaction (temperature increases).

Molar mass of naphthalene is 128.17 g/mol

So, 1.025 g of naphthalene = \frac{1.025}{128.17}moles of naphthalene

                                              = 0.007997 moles of naphthalene

\Delta E_{rxn}=-\frac{41.3kJ}{0.007997mol}=-5164kJ/mol

5 0
3 years ago
A solid reactant is placed into a beaker of a warm water. The liquid vigorously bubbles as the solid dissolves into the solution
Nataly [62]

Answer:

Fewer bubbles will be produced because of fewer collisions of reactant molecules

Explanation:

As the solid dissolves into the solution after the liquid has been vigorously bubbled, if the temperature of the liquid is reduced a little, what will happen is that fewer bubbles will be produced as a result of lesser amount of collisions occurring between the reactant molecules

3 0
3 years ago
How many helium atoms are there in a helium blimp containing 536 kg of helium? n = atoms?
Mariulka [41]

Answer:- 8.07*10^2^8atoms

Solution:- It is a unit conversion where it asks to convert kg of He to atoms. We need to convert kg to grams, grams to moles and finally moles to atoms.

We know that, 1 kg = 1000 g

Molar mass of He is 4.0 grams per mole and 1 mole of an atom equals to Avogadro number of atoms.

The set up would be as given below:

536kg(\frac{1000g}{1kg})(\frac{1mole}{4g})(\frac{6.02*10^2^3atoms}{1mole})

= 8.07*10^2^8atoms

So, there are 8.07*10^2^8atoms of He in it's 536 kg.



5 0
4 years ago
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