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Archy [21]
4 years ago
11

How much work is required to lift a 0.500 kg block 18.5 m?

Chemistry
1 answer:
kipiarov [429]4 years ago
3 0

Answer:

Work = 90.65 j

Explanation:

Given data:

Mass = 0.500 Kg

Distance = 18.5 m

Work done = ?

Solution:

Work = force . distance

Force = mg

Work = mg.distance

Work = mgh

Work = 0.500 Kg × 9.8 m/s²× 18.5 m

Work = 90.65 Kg .m²/s²

Kg .m²/s² = j

Work = 90.65 j

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The potential energy diagram for a reaction starts at 180 kJ and ends at 300 kJ. What type of reaction does the diagram best rep
quester [9]

Answer: Endothermic reaction

Explanation:

Exothermic reactions are defined as the reactions in which energy of the product is lesser than the energy of the reactants. The total energy is released in the form of heat and \Delta H for the reaction comes out to be negative.

Endothermic reactions are defined as the reactions in which energy of the product is greater than the energy of the reactants. The total energy is absorbed in the form of heat and \Delta H for the reaction comes out to be positive.

As the energy of reactants is 180 kJ and that of products is 300 kJ, the energy of products is greater than that of reactants, which means  the energy has been absorbed and reaction is endothermic.

4 0
4 years ago
While out on a hike, you find an unknown mineral. How would you test the mineral to find out what it is?
Mashcka [7]
Probably look up the descriptions of the element and see if it matches the mineral.
5 0
3 years ago
Introduction: Chemical formulas represent ratios. To make H2O, you need two atoms of H for each atom of O; you would also need t
Nina [5.8K]

Answer:

The answer to your question is given below

Explanation:

To convert from Particles to grams, one must have a clear understanding of Avogadro's hypothesis. From the Avogadro's hypothesis, we can easily convert from mole to particles and to grams.

Avogadro's hypothesis gives us a background understanding that 1 mole of any substance contains 6.02x10^23 particles. With this in mind we can easily convert particles to grams and grams to particles.

Now consider the following example:

1. How many particles are there in 4g of carbon?

Solution:

From Avogadro's hypothesis, 1 mole of any substance contains 6.02x10^23 particles. This implies that 1 mole of carbon contains 6.02x10^23 particles.

1 mole of carbon = 12g

Now we can say that 12g of carbon contains 6.02x10^23 particles.

If 12g of carbon contains 6.02x10^23 particles,

then 4g of carbon will contain = (4x6.02x10^23) /12 = 2.01x10^23 particles.

The next example will teach us how to convert from particles to grams..

2. What mass of helium contains 4.2x10^24 particles.

Solution:

According to Avogadro's hypothesis,

1 mole of He contains 6.02x10^23 particles.

1 mole of He = 4g

We can thus, say that 4g of He contains 6.02x10^23 particles.

1f 4g of He contains 6.02x10^23 particles,

Then Xg of He contains 4.2x10^24 Particles i.e

Xg of He = (4x4.2x10^24)/(6.02x10^23)

Xg of He = 27.91g

Therefore, 27.91g of He contains 4.2x10^24 Particles

With the above illustrations I believe you have understood how to convert from grams to particles and from particles to grams.

7 0
3 years ago
Why is the atom neutrally charged?
V125BC [204]
Atoms are electrically neutral because they have equal numbers of protons (positively charged) and electrons (negatively charged).
3 0
3 years ago
How much heat is absorbed during production of 147 g of NO by the combination of nitrogen and oxygen?
marin [14]

Answer:

\large \boxed{\text{105 kcal}}

Explanation:

MM:                        30.01

            N₂ + O₂ ⟶ 2NO; ΔH = +43 kcal/mol

m/g:                         147

Treat the heat as if it were a reactant in the reaction. Then you can write

N₂ + O₂ + 43 kcal ⟶ 2NO

The conversion factor is then 43 kcal/2 mol NO.

1. Moles of NO

\text{Moles of NO} = \text{147 g NO} \times \dfrac{\text{1 mol NO}}{\text{30.01 g NO}} = \text{4.898 mol NO}

2. Amount of heat

\text{Heat} = \text{4.898 mol NO } \times \dfrac{\text{43 kcal}}{\text{2 mol NO}} = \text{105 kcal}\\\\\text{The reaction absorbs $\large \boxed{\textbf{105 kcal}}$}

7 0
3 years ago
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