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DerKrebs [107]
3 years ago
11

A cylinder measuring 9.7 cm wide and 12. cm high is filled with gas. The piston is pushed down with a steady force measured to b

e 18. N. piston cylinder Calculate the pressure of the gas inside the cylinder. Write your answer in units of kilopascals. Be sure your answer has the correct number of significant digits. kPa Check Explanation 5,174 30
Chemistry
1 answer:
bezimeni [28]3 years ago
6 0

Answer:

2.40 kPa

Explanation:

Given data:

radius or width of the cylinder d= 9.7 cm

radius r= 4.85 cm

height h= 12 cm

force applied from the piston F= 18 N

Area of the cross section of cylinder A=πr^2

= \pi (0.0485)^2= 0.007386 m^2

now, pressure P= Force / Area

P= \frac{18}{0.007386}

= 2437.04 \frac{N}{m^2}

= 2.437 kPa

correct to two significant figure = 2.40 kPa

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Aluminum metal and manganese (IV) oxide react to produce aluminum oxide and manganese metal. What mass of manganese (IV) oxide r
gayaneshka [121]

Answer: 1.77 kg of manganese (IV) oxide reacts to produce 1.12kg of manganese metal.

Explanation:

The balanced chemical equation is:

4Al+3MnO_2\rightarrow 2Al_2O_3+3Mn

To calculate the moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text {Molar mass}}

moles of manganese = \frac{1.12\times 1000g}{55g/mol}=20.4mol

According to stoichiometry :

3 moles of Mn is produced by = 3 moles of MnO_2

Thus 20.4 moles of Mn is produced by =\frac{3}{3}\times 20.4=20.4moles  of MnO_2

Mass of MnO_2=moles\times {\text {Molar mass}}=20.4moles\times 86.9g/mol=1773g=1.773kg    (1kg=1000g)

Thus 1.77 kg of manganese (IV) oxide reacts to produce 1.12kg of manganese metal.

6 0
3 years ago
What is the concentration of kl solution if 20.68g of solute was added to enough water to form 100ml solution?
rewona [7]

Answer:

1.25 M

Explanation:

Step 1: Given data

Mass of KI (solute): 20.68 g

Volume of the solution: 100 mL (0.100 L)

Step 2: Calculate the moles of solute

The molar mass of KI is 166.00 g/mol.

20.68 g × 1 mol/166.00 g = 0.1246 mol

Step 3: Calculate the molar concentration of KI

Molarity is equal to the moles of solute divided by the liters of solution.

M = 0.1246 mol/0.100 L= 1.25 M

5 0
3 years ago
A chemical reaction is shown below:
BabaBlast [244]

Answer:

Mass = 8.46 g

Explanation:

Given data:

Mass of water produced = ?

Mass of glucose = 20 g

Mass of oxygen = 15 g

Solution:

Chemical equation:

C₆H₁₂O₆ + 6O₂     →   6H₂O + 6CO₂

Number of moles of glucose:

Number of moles = mass/molar mass

Number of moles = 20 g/ 180.16 g/mol

Number of moles = 0.11 mol

Number of moles of oxygen:

Number of moles = mass/molar mass

Number of moles = 15 g/ 32 g/mol

Number of moles = 0.47 mol

now we will compare the moles of water with oxygen and glucose.

               C₆H₁₂O₆           :            H₂O

                   1                   :              6

                 0.11                :           6/1×0.11 = 0.66

                   O₂               :            H₂O

                   6                   :              6

                 0.47                :           0.47

Less number of moles of water are produced by oxygen thus it will limit the yield of water and act as limiting reactant.

Mass of water produced:

Mass = number of moles × molar mass

Mass = 0.47 mol  ×18 g/mol

Mass = 8.46 g

8 0
3 years ago
Copper has two naturally occurring isotopes. Cu−63 has a mass of 62.939 amu and relative abundance of 69.17%.
fiasKO [112]
The answer is 64.907 amu.

The atomic mass of an element is the average of the atomic masses of its isotopes. The relative abundance of isotopes must be taken into consideration, therefore:
atomic mass of copper = atomic mass of isotope 1 * abundance 1 + atomic mass of isotope 2 * abundance 2

We know:
atomic mass of copper = 63.546 amu
The atomic mass of isotope 1 is: 62.939 amu
The abundance of isotope 1 is: 69.17% = 0.6917
The atomic mass of isotope 1 is: x
The abundance of isotope 2: 100% - 69.17% = 30.83% = 0.3083

Thus:
63.546 amu = 62.939 amu * 0.6917 + x * 0.3083
63.546 <span>amu = 43.535 amu + 0.3083x
</span>⇒ 63.546 amu - 43.535 amu = 0.3083x
⇒ 20.011 amu = 0.3083x
   ⇒ x = 20.011 amu ÷ 0.3083 = 64.907 amu
7 0
3 years ago
If a frog initially contained 2 grams of carbon-14 and the half-life of carbon-14 is 5,730 years, how much carbon-14 remains in
andreyandreev [35.5K]
Half life is the time taken for a radioactive isotope to decay by half its original mass. In this case the half life of carbon-14 is 5.730 years. 
Using the formula;
New mass = original mass × (1/2)^n; where n is the number of half lives (in this case n=1 ) 
New mass = 2 g × (1/2)^1     
                  = 1 g
Therefore; the mass of carbon-14 that remains will be 1 g 
5 0
3 years ago
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