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alekssr [168]
3 years ago
11

If barium was to react with oxygen to form ionic compounds, how many oxygen atoms would be needed? Explain.

Chemistry
1 answer:
Dafna11 [192]3 years ago
8 0

Answer:

1 oxygen atoms

Explanation:

When barium combines with oxygen to form an ionic compound, it will only require just an oxygen atom.

An ionic compound is formed due to the electrostatic attraction between oppositely charged ions

A metal loses electrons to be gained by the non - metal and in the process they attract each other with a force.

 Barium has two valence electrons, to be stable, it loses the two;

 Oxygen has six valence electrons, to be stable it will gain two valence electrons to form an octet.

 So;

     One barium atom will successfully bind with One oxygen atom to form

               BaO

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disa [49]
Zero (0) molecules of glucose are produced.
4 0
3 years ago
A gas containing 80% CH4 and 20% He is sent through a quart/ diffusion tube (see Figure P8.8) to recover the helium, Twenty perc
elena55 [62]

For a gas containing 80% CH4 and 20% He is sent through a quart diffusion tube, the composition is mathematically given as

%He=12.5%

%CH4=87.5%

<h3>What is the composition of the waste gas if 100 kg moles of gas are processed per minute?</h3>

Generally, the equation for the Material balance  is mathematically given as

F=R+W

Therefore

100=0.20*1000+W

W=80kmol/min

In conclusion, waste gas compose

2.0/100*100=50/100*20+%*80

Hence

%He=12.5%

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3 0
2 years ago
What is an intensive property? *
Dennis_Churaev [7]

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This is the answer

8 0
3 years ago
33.56 g of fructose (C6H,206) and 18.88 g of water are mixed to obtain a 40.00 ml solution a. What is this solution's density? b
Darina [25.2K]

Explanation:

Mass of fructose = 33.56 g

Mass of water =  18.88  g

Total mass of the solution =  Mass of fructose + Mass of water = M

M = 33.56 g + 18.88  g =52.44 g

Volume of the solution = V = 40.00 mL

Density =\frac{Mass}{Volume}

a) Density of the solution:

\frac{M}{V}=\frac{52.44 g}{40.00 mL}=1.311 g/mL

b) Molar mass of fructose = 180.16 g/mol

Moles of fructose = n_1=\frac{ 33.56 g}{180.16 g/mol}=0.1863 mol

Molar mass of water = 18.02 g/mol

Moles of water= n_2=\frac{ 18.88 g}{18.02 g/mol}=1.0477 mol

Mole fraction of fructose in this solution:\chi_1

\chi_1=\frac{n_1}{n_1+n_2}=\frac{0.1863 mol}{0.1863 mol+1.0477 mol}

\chi_1=0.1510

Mole fraction of water = \chi_2=1-\chi_1=0.8490

c) Average molar mass of of the solution:

=\chi_1\times 180.16 g/mol+\chi_2\times 18.02 g/mol

=0.1510\times 180.16 g/mol+0.8490\times 18.02 g/mol=42.50 g/mol

d) Mass of 1 mole of solution = 42.50 g/mol

Density of the solution = 1.311 g/mL

d) Specific molar volume of the solution:

\frac{\text{Average molar mass}}{\text{Density of the mass}}

=\frac{42.50 g/mol}{1.311 g/mL}=32.42 mL/mol

5 0
3 years ago
Two objects are brought into contact Object 1 has mass 0.76 kg, specific heat capacity 0.87) g'c and initial temperature 52.2 'C
taurus [48]

Answer:

T_F=77.4\°C

Explanation:

Hello there!

In this case, according to the given information, it turns out possible to set up the following energy equation for both objects 1 and 2:

Q_1=-Q_2

In terms of mass, specific heat and temperature change is:

m_1C_1(T_F-T_1)=-m_2C_2(T_F-T_2)

Now, solve for the final temperature, as follows:

T_F=\frac{m_1C_1T_1+m_2C_2T_2}{m_1C_1+m_2C_2}

Then, plug in the masses, specific heat and temperatures to obtain:

T_F=\frac{760g*0.87\frac{J}{g\°C} *52.2\°C+70.7g*3.071\frac{J}{g\°C}*154\°C}{760g*0.87\frac{J}{g\°C} +70.7g*3.071\frac{J}{g\°C}} \\\\T_F=77.4\°C

Yet, the values do not seem to have been given correctly in the problem, so it'll be convenient for you to recheck them.

Regards!

4 0
3 years ago
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