CHEMISTRY COLLEGE

What is the potentual energy of 7 kg object, 12 m off the ground?​

Answers

Answer 1
Answer:

Answer:

823.2 J

Explanation:

PE = mgh

= (7 kg) (9.8 m/s^2) (12 m)

= 823.2 J
Answer 2
Answer:

Answer:

RequiredAnswer:-

Mass=m=7kg

Height=h=12m

Gravitational force=g=10m/s^2

  • As we know that

{\boxed{\sf Potential\:energy{}_((P.E))=mgh}}

  • Substitute the values

{:}\longmapsto\sf P.E=7×12×10

{:}\longmapsto\sf P.E=84×10

{:}\longmapsto\sf Potential\:energy=840Joule

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I need help!! ASAP please..

Answers

Answer:

164 g

Explanation:

Fireworks explode above your
head. Is this exothermic or
endothermic and why?

Answers

I think it is exothermic

6CO2 + 6H20 --> C6H12O6 + 602What is the total number of moles of CO2 needed to make 2 moles of CH1206?

Answers

Answer:

12 mol CO₂

General Formulas and Concepts:

Atomic Structure

  • Compounds
  • Moles
  • Mole Ratio

Stoichiometry

  • Analyzing reactions rxn
  • Using Dimensional Analysis

Explanation:

Step 1: Define

Identify

[rxn] 6CO₂ + 6H₂O → C₆H₁₂O₆ + 6O₂

[Given] 2 mol C₆H₁₂O₆

[Solve] mol CO₂

Step 2: Identify Conversions

[rxn] 6CO₂ → C₆H₁₂O₆

Step 3: Convert

  1. [DA] Set up:                                                                                                   \displaystyle 2 \ mol \ C_6H_(12)O_6((6 \ mol \ CO_2)/(1 \ mol \ C_6H_(12)O_6))
  2. [DA] Multiply [Cancel out units]:                                                                      \displaystyle 12 \ mol \ CO_2

Which of the following statement best defines matter?

Answers

show us the statements

Calculate the number of particles in 5.0 grams of NaCl.

Answers

Answer:

5.2 × 10²² particles NaCl

General Formulas and Concepts:

Chemistry - Atomic Structure

  • Reading a Periodic Table
  • Using Dimensional Analysis
  • Avogadro's Number - 6.022 × 10²³ atoms, molecules, formula units, etc.

Explanation:

Step 1: Define

5.0 g NaCl

Step 2: Identify Conversions

Avogadro's Number

Molar Mass of Na - 22.99 g/mol

Molar Mass of Cl - 35.45 g/mol

Molar Mass of NaCl - 22.99 + 35.45 = 58.44 g/mol

Step 3: Convert

5.0 \ g \ NaCl((1 \ mol \ NaCl)/(58.44 \ g \ NaCl) )((6.022 \cdot 10^(23) \ particles \ NaCl)/(1 \ mol \ NaCl) ) = 5.15229 × 10²² particles NaCl

Step 4: Check

We are given 2 sig figs. Follow sig fig rules and round.

5.15229 × 10²² particles NaCl ≈ 5.2 × 10²² particles NaCl

Zeros laced at the end of the significant number are...

Answers

Answer:

Zeros located at the end of significant figures are significant.

Explanation:

Hope it will help :)
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