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olga2289 [7]
4 years ago
9

Gas molecules are constantly in motion. True or False?

Chemistry
1 answer:
SpyIntel [72]4 years ago
4 0
True gas moves throughout the atmosphere
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When aluminum is placed in concentrated hydrochloric acid, hydrogen gas is produced.
KatRina [158]

Answer : The mass of aluminium (Al) is 0.413 g

Solution : Given,

Volume of H_{2}(g) at STP = 513 ml = 0.513 L           ( 1 L = 1000 ml )

Molar mass of aluminium = 26.98 g/mole

First we have to calculate the moles of H_{2}(g).

At STP,    

1 mole occupies 22.4 L volume

now, 0.513 L gives \frac{1mole\times0.513L}{22.4L} moles of H_{2}(g)

The moles of H_{2}(g) = 0.0229 moles

The Net balanced chemical reaction is,

2Al(s)+6HCl(aq)\rightarrow 2AlCl_3(aq)+3H_2(g)

From the balanced chemical reaction, we conclude that

2 moles of Aluminium (Al) produces 3 moles of hydrogen gas

Now the number of moles of aluminium required in 0.0229 moles of hydrogen gas = \frac{2moles\times 0.0229moles}{3moles} = 0.0153 moles

Now we have to calculate the mass of aluminium.

Mass of aluminium = number of moles × Molar mass = 0.0153 moles × 26.98 g/mole = 0.413 g

The mass of aluminium required is 0.413 g.

7 0
3 years ago
Consider the following reaction occurring in a 1.0 L container:
jeka94

Answer:

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3 0
3 years ago
Given you have 1.93 mols of NaCl, find the molarity of a 2.5 L solution
Inessa [10]

Answer:

0.77 M

Explanation:

Molarity is the concentration of a solution per liter

C= concentration

n= number of moles

V= volume of solution

The formula we will use is C= n/V

C= n/V

C= (1.93 mol NaCl)/(2.5 L of solution)

C= 0.772 mol/L

Since this concentration is given in moles per liters of a solution, this concentration is also the molarity.

C= 0.772 mol/L

*Include two significant digits in final answer*

M= 0.77 M

5 0
3 years ago
In an atomic model that includes a nucleus, negative charge is
Kruka [31]
Seen in the electrons, which are found in the atomic cloud surrounding the nucleus.
8 0
3 years ago
You have 500.0 ml of a buffer solution containing 0.30 m acetic acid (ch3cooh) and 0.20 m sodium acetate (ch3coona). what will t
Nataly [62]
First, we should get moles acetic acid = molarity * volume

                                                                =0.3 M * 0.5 L

                                                                =  0.15 mol

then, we should get moles acetate = molarity * volume

                                                           = 0.2 M * 0.5L

                                                           = 0.1 mol

then, we have to get moles of OH- which added:

moles OH- = molarity  * volume

                   = 1 M    * 0.02L

                  = 0.02 mol

when the reaction equation is:


                 CH3COOH  +  OH-  → CH3COO-   +  H2O


moles acetic acid after adding OH- = (0.15-0.02) 
                                             
                                                            =  0.13M                                       

moles acetate after adding OH- =  (0.1 + 0.02)

                                                      =   0.12 M

Total volume = 0.5 L + 0.02 L= 0.52 L

∴[acetic acid] = moles acetic acid after adding OH- / total volume

                        = 0.13mol / 0.52L

                       = 0.25 M

and [acetate ) = 0.12 mol / 0.52L
 
                        = 0.23 M

by using H-H equation we can get PH:

PH = Pka + ㏒[salt/acid]

when we have Ka = 1.8 x 10^-5

∴Pka = -㏒Ka 

        = -㏒ 1.8 x 10^-5

       = 4.7

So by substitution:

∴ PH = 4.7 + ㏒[acetate/acetic acid]

         = 4.7 + ㏒(0.23/0.25)

        = 4.66
6 0
3 years ago
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