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snow_tiger [21]
3 years ago
13

Where is the type of force a person stands on the floor and his weight is 752 in in the pressure exerted from his feet on the fl

oor it is 9002PA what is the area of his feet in square meters
Chemistry
1 answer:
ladessa [460]3 years ago
3 0

Answer:

0.084m^2

Explanation:

Data obtained from the question include:

W (weight) = F (force) = 752N

P (pressure) = 9002Pa

Recall:

101325Pa = 101325N/m2

Therefore, 9002Pa = 9002N/m2

Pressure exerted on an object is simply force per unit area. It is represented mathematically as:

Pressure = Force /Area

With the above formula, we can easily find the area as follow:

Pressure = Force /Area

P = F/A

9002 = 752/A

Cross multiply to express in linear form

9002 x A = 752

Divide both side by 9002

A = 752/9002

A = 0.084m^2

Therefore, the area of his feet is 0.084m^2

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Test tubes, spot and/or well plate will be used to combine substances in bottles and form a reaction is referred to a true statement.

<h3>What is a Chemical reaction?</h3>

This is the reaction which involves substances known as reactants combining new substances which are referred to as products. This is usually done in the laboratory and equipment and tools such as test tubes, thermometer etc are used.

Test tubes, spot and/or well plate are often used to combine substances in bottles and form a reaction which must be handled with care to ensure it doesn't spill.

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Answer:

Explanation:

1. find the molar mass (amu) of each element and add them to get the whole molar mass.

2. divide the 1 element molar mass with the whole molar mass

3. multiple by 100 and that gives you the % composition.

<h2><u><em>56-57: NaCl</em></u></h2>

1. Na(22.99amu) + Cl (35.453amu)=58.443

2(Na):   \frac{22.99}{58.443} = .393

2(Cl): \frac{35.453}{58.443}= .607

3(Na): .393 * 100=39.3%

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<h2><u>58-60 </u>K_{2} CO_{3}<u /></h2>

1. K: (39.098)(2)=78.196

_ C: (12.011)(1)= 12.011

_O: (15.99)(3) = 47.997

78.196+12.011+47.997= 138.204

2:K: \frac{78.196}{138.204}= .566 <u>Step </u>3: (.566)(100)= 56.6%

2: C: \frac{12.011}{138.204}= .087 <u>Step 3</u>: (.087)(100)= 8.7%

2: O: \frac{47.997}{138.204}= .347 <u>Step 3</u>: (.347)(100) = 34.7%

<h2>61-62 Fe_{3} O_{4}</h2>

1. Fe (55.845)(3)= 167.535

_ O (15.999)(4) = 63.996

167.535+63.996=231.531

2: Fe: \frac{167.535}{231.531}= .724 Step 3: (.724)(100)= 72.4%

2: O : \frac{63.996}{231.531}= .276 Step 3: (.276)(100) = 27.6%

<h2>63-65 C_{3}H_{5}(OH)_{3}</h2>

1.

C(12.011*3)=36.033

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