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Leviafan [203]
3 years ago
7

How many moles of each reactant are needed to produce 3.60 x 10^2g ch3oh

Chemistry
1 answer:
zysi [14]3 years ago
3 0
Methanol is prepared by reacting Carbon monoxide and Hydrogen gas,

                                   CO  +  2 H₂    →    CH₃OH

Calculating Moles of CO:
                                       According to equation,

             32 g (1 mole) of CH₃OH is produced by  =  1 Mole of CO
So,
             3.60 × 10² g of CH₃OH is produced by  =  X Moles of CO

Solving for X,
                       X  =  (3.60 × 10² g × 1 Mole) ÷ 32 g

                       X  =  11.25 Moles of CO

Calculating Moles of H₂:
                                       According to equation,

             32 g (1 mole) of CH₃OH is produced by  =  2 Mole of H₂
So,
             3.60 × 10² g of CH₃OH is produced by  =  X Moles of H₂

Solving for X,
                       X  =  (3.60 × 10² g × 2 Mole) ÷ 32 g

                       X  =  22.5 Moles of H₂

Result:
            3.60 × 10² g of CH₃OH
is produced by reacting 11.25 Moles of CO and 22.5 Moles of H₂.
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The length breadth and thickness of a brick is 18cm 8 cm and 5cm respictively find the area of the widest part of rhe brick​
Sunny_sXe [5.5K]

Answer:

144cm²

Explanation:

Given dimensions:

     Length x breadth x thickness

        18cm  x   8cm    x    5cm

The widest part of this figure will be the face containing the length and the breadth.

The breadth is the width of the figure;

Area of the widest part  = length x breadth = 18cm x 8cm  = 144cm

The area of the widest part of the figure is 144cm²

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3 years ago
How many grams of Br are in 335g of CaBr2
ASHA 777 [7]
The answer is 267.93 g

Molar mass of CaBr2 is the sum of atomic masses of Ca and Br:
Mr(CaBr2) = Ar(Ca) + 2Ar(Br)
Ar(Ca) = 40 g/mol
Ar(Br) = 79.9 g/mol
Mr(CaBr2) = 40 + 2 * 79.9 = 199.8 g/mol

The percentage of Br in CaBr2 is:
2Ar(Br) / Mr(CaBr2) * 100 = 2 * 79.9 / 199.8 * 100 = 79.98%

Now make a proportion:
x g in 79.98%
335 g in 100%
x : 79.98% = 335 g : 100%
x = 79.98% * 335 g : 100%
x = 267.93 g
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A. High intermolecular forces of attraction. If there are high intermolecular forces, the molecules will need large energies to escape into the liquid. The substance will nave a high melting point.

The other options are <em>incorrect </em>because they are <em>weak force</em>s. They would cause <em>low melting points</em>.

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