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3 years ago

WHEN IN FRICTION not useful

Chemistry

1 answer:

Paladinen [302]3 years ago

6 0

Friction is not useful in Machinery sometimes, Since it leads to the wear and tear of machine parts.

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Why does it matter if atoms are divisible ???

Andreyy89

Atoms are divisible contrary to the early beliefs that the smallest "indivisible" matter is an atom. When an atom loses its identity it means that they are divisible. Atoms chemically react with other kinds of atoms thus changing their activity.

They certainly are not that important to our lives, but it’s good to know :)

5 0

2 years ago

Identify the following changes as physical or chemical.

andrezito [222]

Answer:

physical

Explanation:

5 0

2 years ago

When 0.620 gMngMn is combined with enough hydrochloric acid to make 100.0 mLmL of solution in a coffee-cup calorimeter, all of t

OleMash [197]

Answer:

The enthalpy change during the reaction is -199. kJ/mol.

Explanation:

$Mn(s)+2HCl(aq)\rightarrow MnCl_2(aq)+H_2(g)$

Mass of solution = m

Volume of solution = 100.0 mL

Density of solution = d = 1.00 g/mL

$m=1.00 g/mL\times 100.0 mL = 100 g$

First we have to calculate the heat gained by the solution in coffee-cup calorimeter.

$q=m\times c\times (T_{final}-T_{initial})$

where,

m = mass of solution = 100 g

q = heat gained = ?

c = specific heat = $4.18 J/^oC$

$T_{final}$ = final temperature = $23.1^oC$

$T_{initial}$ = initial temperature = $28.9^oC$

Now put all the given values in the above formula, we get:

$q=100 g \times 4.18 J/^oC\times (28.9-23.1)^oC$

$q=2,242.4 J=2.242 kJ$

Now we have to calculate the enthalpy change during the reaction.

$\Delta H=-\frac{q}{n}$

where,

$\Delta H$ = enthalpy change = ?

q = heat gained = 2.242 kJ

n = number of moles fructose = $\frac{\text{Mass of manganese}}{\text{Molar mass of manganese}}=\frac{0.620 g}{54.94 g/mol}=0.0113 mol$

$\Delta H=-\frac{2.242 kJ}{0.0113 mol }=-199. kJ/mol$

Therefore, the enthalpy change during the reaction is -199. kJ/mol.

8 0

2 years ago

What is the mass of a block with a volume of 18 cm^3, and a density of 9.2 g/cm^3?

Semmy [17]

Answer: 165.6grams

Explanation:

Mass of a block = ?

Volume of block = 18 cm^3

Density of block = 9.2 g/cm^3

The density of any object depends on its mass and volume.

i.e Density of block = Mass / volume

9.2 g/cm^3 = Mass / 18 cm^3

Mass = 9.2 g/cm^3 x 18 cm^3

= 165.6 g

Thus, the mass of the block is 165.6grams

6 0

3 years ago

What volume of water must be added

stellarik [79]

Answer:

litre.50000665432158900643212lo

3 0

3 years ago