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mr Goodwill [35]
3 years ago
6

A block has a mass of 30g and dimensions of 1.5cm by 4.8cm by 2.3 cm. Will the block float or sink in

Chemistry
1 answer:
zimovet [89]3 years ago
3 0

The block will sink

Explanation:

In general, an object with a higher density than water will sink  in it, and one with a lower density will float. Now, to know if the block will float or sink, the first step is to calculate the density of the block, and to do this, the volume and mass of the block are required.

Mass of the block: 30 grames

Volume of the block:

The volume of the block is not provided, but this can be found using the dimensions given.

V (Volume) = s (side) x s x s

V = 1.5 cm x 4.8 cm x 2.3 cm

V = 16.56 cm^{3}

Density of the block:

To find the density simply divide the mass into the volume

D (density) = M (Mass) ÷ V (Volume)

D = 30 grams ÷ 16.56 cm^{3}

D = 1. 811 grams/ cm^{3}

The density of the block is  1. 811 grams/ cm^{3}

Additionally, the density of the water is 1 gram/ cm^{3} or 1 gram/mL. According to this, the density of the block is higher than the density of the water, and therefore this is the block sinks.

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What is the percent yield for a process in which 10.4g of CH3OH reacts and 10.1 g of CO2 is formed
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Answer:

A. 70.7%

Explanation:

In the first step lets compute the molar mass of CH₃OH and CO

Molar Mass of CH₃OH =  1(12.01 g/mol) + 4(1.008 g/mol) +1(16.00 g/mol)

                                     = 32.042 g/mol

Molar Mass of CO₂      = 1(12.01 g/mol) + 2(16.00 g/mol)  

                                     = 44.01 g/mol

                                   

Mass of only one reactant i.e. CH₃OH is given so  it must be the limiting reactant. Next, the theoretical yield is calculated directly as follows:

Given mass of CH₃OH is 10.4 g. So we have:

                                     10.4g CH₃OH

Convert grams of CH₃OH to moles of CH₃OH utilizing molar mass of CH₃OH as:

                          1 mol CH₃OH / 32.042 g CH₃OH

Convert CH₃OH to moles of CO₂ using mole ratio as:

                             2 mol CO₂ / 2 mol CH₃OH

Convert moles of  CO₂ to grams of  CO₂ utilizing molar mass of  CO₂ as:

                           44.01 g/mol CO₂ / 1 mol CO₂

Now calculating theoretical yield using above steps:

[ 10.4 g CH₃OH ]  [1 mol CH₃OH / 32.042 g CH₃OH ]  [2 mol CO₂ / 2 mol CH₃OH]  [44.01 g/mol CO₂ / 1 mol CO₂]

Multiplication is performed here. We are left with 10.4 and 44.01 g CO₂ from numerator terms in the above equation and 32.042 from denominator terms after cancellation process of above terms. So this equation becomes:

= ( 10.4 ) ( 44.01 ) g CO₂ / 32.042

= 457.704/32/042

=  14.28 g CO₂

Theoretical yield =  14.28 g CO₂  

Finally compute the percent yield for a process in which 10.4g of CH₃OH reacts and 10.1 g of CO₂ is formed:

percent yield = (actual yield / theoretical yield) x 100

As we have calculated theoretical yield which is 14.28 g CO₂ and actual yield is 10.1 g CO₂ So,

percent yield = (10.1 g CO₂ / 14.28 g CO₂) x 100%

                       = 0.707 x 100%

                       = 70.7 %

Hence option A 70.7% yield is the correct answer.

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