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ahrayia [7]
3 years ago
11

How many moles of oxygen are required for the combustion of 35.5 g of Magnesium Mg+O2-->MgO Please balance

Chemistry
1 answer:
Verdich [7]3 years ago
6 0

Answer:

0.74 moles of oxygen are required for combustion

Explanation:

Given data:

Number of moles of oxygen required = ?

Mass of Mg = 35.5 g

Solution:

Chemical equation:

2Mg + O₂         →       2MgO

Number of moles of Mg:

Number of moles = mass/molar mass

Number of moles = 35.5 g/ 24 g/mol

Number of moles = 1.48 mol

Now we will compare the moles of Mg with oxygen.

                  Mg           :         O₂

                   2             :          1

                 1.48           :        1/2×1.48 = 0.74 mol

0.74 moles of oxygen are required for combustion.

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Answer:

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7 0
3 years ago
I need help with a questino
kipiarov [429]

Explanation:

What will the question be ?

6 0
3 years ago
Read 2 more answers
43 mg = [?]g <br>A. 0.043 g <br>B. 4.3 g <br>C. 4300 g <br>D. 43,000 g​
Ymorist [56]

Answer:

Option A (0.043 g) is the correct answer.

Explanation:

Given:

= 43 mg

As we know,

1 \ mg = \frac{1}{1000} \ g

then,

⇒ 43 \ mg = \frac{43}{1000}  \ g

              = 0.043 \ g

Thus, the above is the correct alternative.

4 0
3 years ago
What is the value of k, at 25 degrees Celsius
Dimas [21]
<h3>Answer:</h3>

298.15 K

<h3>Explanation:</h3>

W e are supposed to calculate the Value of K at 25°C

Assuming the value of K represent K, the question wants us to convert  degree Celsius to Kelvin.

  • To convert degrees Celsius to kelvin scale, we use the relationship;
  • Kelvin (K) = Degrees Celsius + 273.15 ; 273.5 is a constant
  • That is, to convert temperature from °C to Kelvin we add a constant of 273.15 to the °C given.

In this case;

Temperature is 273.15 °c

Thus, to Kelvin scale temperature will be;

= 25°C + 273.15

= 298.15 K

Therefore, the value of K, at 25°C is 298.15 K

7 0
3 years ago
opper(I) ions in aqueous solution react with NH 3 ( aq ) according to Cu + ( aq ) + 2 NH 3 ( aq ) ⟶ Cu ( NH 3 ) + 2 ( aq ) K f =
Y_Kistochka [10]

Answer:

55.373g/l

Explanation:

The dissolved amount of sparingly soluble salts is interlinked with a unitlesss quantity called as solubility product. It is a fixed quantity that only increases with the rise in temperatures and is used to predict the salting out of compounds. If the value of ionic product (Q) is larger than (Ksp), precipitation of compound occurs.

Given:

The solubility product of CuBr is 6.3×10−9.

The concentration of NH3 is 0.10 M.

Formula and Calculations:

The dissolution reaction (I) of CuBr is shown below.

The reaction showing dissolution of CuBr in NH3 is shown below.

The above reaction can be obtained by adding reaction (I) and (II) as shown below.

The equilibrium constants will get multiplied.

Suppose the solubility of CuBr is “s”.

It is given that concentration of NH3 is 0.10 M.

The equilibrium constant expression for the above reaction is as follows,

Here,

The concentration of pure solids is 1 M. Thus, the concentration of CuBr is 1 M.

As calculated, the value of Ksp is 396.9.

Substitute all the required values in above formula.  

On further solving above equation,

Therefore, the solubility of CuBr in ammonia is 0.386 M.

The formula to calculate solubility

Solubuility (g/l)= Molarity(M) x Molarmass

Chemistry homework question answer, step 2, image 10

The molar mass of CuBr is 143.45 g/mol.

The formula to calculate solubility in g/L is given below.

The molar mass of CuBr is 143.45 g/mol.

therefore,

solubility = 0.386M x 143.45g/mol

where (M = mol/l)

solubility = 55.375g/l

5 0
3 years ago
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