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beks73 [17]
3 years ago
6

Kinetic energy CANNOT be converted into potential energy True False

Chemistry
2 answers:
Alex777 [14]3 years ago
3 0

Answer:

false

Explanation:

mr Goodwill [35]3 years ago
3 0

Answer:

False

Explanation:

Because if you shoot a arrow and go back and get it and use it again it is potential

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The molar heat of fusion of platinum (Pt) is 4.700 kcal/mol. How much heat must be added to 85.5 g of solid platinum at its melt
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The heat required to completely melt the given substance, platinum, we just have to convert first the given mass in mole and multiply the answer to its molar heat of fusion.. 
                              Hf  = mass x (1/molar mass) x molar heat of fusion
                            Hf = (85.5 g) x (1 mole/195.08 g) x 4.70 kcal/mol
                                   Hf = 2.06 kcal
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A sample compound with a molar mass of 34.00g/mol is found to consist of 0.44g H and 6.92g O. Calculate both empirical and molec
balu736 [363]

Answer:

E.F= OH

M.F=O_{2} H_{2}

Explanation:

Empirical Formula

Step 1: Calculate mols of each element

0.44gH(1 mol H/ 1.008g H)= 0.4365 mol H

*note leave extra sig figs for calculations

6.92gO(1 mol O/ 16 g O)=0.4325 mol O

Step 2: Identify which is the smallest mol

*in this case 0.4365 mol H> 0.4325 mol O so we will use 0.4325 mol O

Step 3: Divide above calculations by the smallest mol

0.4325 mol O/0.4325= 1

0.4365 mol H/0.4325= 1.009 *rounds to 1

Step 4: use calculations as subscripts

Oxygen = 1 so the subscript will 1 (O)

Hydrogen = 1 so the subscript will 1 (H)

making E.F= OH

Molecular Formula:

Step 1: identify  molecular mass and mass from the E.F

the molecular mass is given 34.00g/mol

the mass of the E.F is

(oxygen mass from periodic table)+(hydrogen mass from periodic table)

16+1.008= 17.008

Step 2:Divide the molecular mass by the mass given by the emipirical formula.

\frac{34.00}{17.008}= 1.999 round to 2

Step 3:Multiply the empirical formula (the subscripts) by this number to get the molecular formula.  ANSWER: M.F=2(OH)- O_{2} H_{2}

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3 years ago
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