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3 years ago

How can you increase the amount of work done when you lift a book explain

Chemistry

1 answer:

Gnoma [55]3 years ago

4 0

Explanation:

In this case, while lifting the book we are working against the force of gravity.

Using the Newton's laws, we can find the force F required for lifting the book having mass (m) and acceleration due to gravity (g) that is ;

F = mg

and, the change in the position of the book that is Δx (Height)

→ Δx = Final position - Initial position

which is only the height, then the amount of work done will be calculated by :

W= mgh

Where W = Work Done

m = Mass of the Body

g = Acceleration due to Gravity

h = Height of Body being displaced

You might be interested in

A solution is made by dissolving 4.87 g of potassium nitrate in water to a final volume of 86.4 mL solution. What is the weight/

lara31 [8.8K]

Answer:

A solution is made by dissolving 4.87 g of potassium nitrate in water to a final volume of 86.4 mL solution. The weight/weight % or percent by mass of the solute is :

2.67%

Explanation:

Note : Look at the density of potassium nitrate in water if given in the question.

You are calculating weight /Volume not weight/weight % or percent by mass of the solute

Here the weight/weight % or percent by mass of the solute is asked : So first convert the VOLUME OF SOLUTION into MASS

Density of potassium nitrate in water KNO3 = 2.11 g/mL

$density=\frac{mass}{volume}$

Density = 2.11 g/mL

Volume of solution = 86.4 mL

$2.11=\frac{mass}{86.4}$

$mass = 2.11\times 86.4$

$mass=182.3grams$

Mass of Solute = 4.87 g

Mass of Solution = 183.2 g

w/w% of the solute =

$= \frac{mass\ of\ solute}{mass\ of\ solution}\times 100$

$=\frac{4.87}{183.2}\times 100$

w/w%=2.67%

8 0

3 years ago

What is the bond energy required to break one mole of carbon-carbon bonds

ArbitrLikvidat [17]

Answer:

<h2>100 kcal of bond energy</h2>

8 0

3 years ago

When a 17.7 mL sample of a 0.368 M aqueous hypochlorous acid solution is titrated with a 0.301 M aqueous barium hydroxide soluti

nordsb [41]

Answer:

pH = 12.98

Explanation:

Step 1: Data given

Volume of aqueous hypochlorous acid solution = 17.7 mL = 0.0177 L

Molarity of aqueous hypochlorous acid solution = 0.368 M

Molarity of aqueous barium hydroxide solution = 0.301 M

Volume of aqueous barium hydroxide solution = 16.2 mL = 0.0162 L

Step 2: The balanced equation

2HCl + Ba(OH)2 → BaCl2 + 2H2O

Step 3: Calculate moles

Moles = molarity * volume

Moles HCl = 0368 M * 0.0177 L

Moles HCl = 0.0065136 moles

Moles Ba(OH)2 = 0.301 M * 0.0162 L

Moles Ba(OH)2 = 0.0048762 moles

Step 4: Calculate the limiting reactant

For 2 moles HCl we need 1 mol Ba(OH)2 to produce 1 mol BaCl2 and 2 moles H2O

HCl is the limiting reactant. It will completely be consumed 0.0065136 moles. Ba(OH)2 is in excess. There will react 0.0065136/2 = 0.0032568‬ moles. There will remain 0.0048762 moles - 0.0032568‬ = 0.0016194 moles

Step 5: Calculate molarity Ba(OH)2

Molarity Ba(OH)2 = moles / volume

Molarity Ba(OH)2 = 0.0016194 moles / 0.0339 L

Molarity Ba(OH)2 = 0.04777 M

Step 6: Calculate [OH-]

Ba(OH)2 → Ba^2+ + 2OH-

For Ba(OH)2 we have 2* [OH-]

[OH-] = 2*0.04777 = 0.09554 M

Step 7: Calculate pOH

pOH = -log[OH-]

pOH = -log(0.09554)

pOH = 1.02

Step 8: Calculate pH

pH = 14 - 1.02

pH = 12.98

8 0

3 years ago

Significant figures

Sergeu [11.5K]

Answer:

go look for that on google

Explanation:

4 0

3 years ago

Using the equations 2 Sr(s) + O₂ (g) → 2 SrO (s) ∆H° = -1184 kJ/mol SrO (s) + CO₂ (g) → SrCO₃ (s) ∆H° = -234 kJ/mol CO₂ (g) → C(

kkurt [141]

Answer: The $\Delta H^o_{rxn}$ for the reaction is 72 kJ.

Explanation:

Hess’s law of constant heat summation states that the amount of heat absorbed or evolved in a given chemical equation remains the same whether the process occurs in one step or several steps.

According to this law, the chemical equation is treated as ordinary algebraic expressions and can be added or subtracted to yield the required equation. This means that the enthalpy change of the overall reaction is equal to the sum of the enthalpy changes of the intermediate reactions.

The given chemical reaction follows:

$2SrCO_3(s)\rightarrow 2Sr(s)+2C(s)+3O_2(g)$ $\Delta H^o_{rxn}=?$