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1 year ago

Explain why scientists find the particulate matter of theory useful

1 answer:

6 0

This idea has historical significance. The ancient Greek philosopher Democritus (born 460 BCE), who held that everything is composed of small particles moving in empty space, is credited with developing the first hypothesis we have about the microscopic universe. He had some concrete proof for this, such the fact that items like a new loaf of bread or a rose may give off a scent even when they are far from the source. Being a materialist, he thought that these odors originated from actual material particles released by the bread or the rose, rather than being purely a type of magic. He reasoned that these particles must float through the air, with some of them maybe landing in your nose where you can smell them immediately. This still makes sense in modern times. But many of us now have quite different perspectives on these "particles."

Thank you,

Eddie

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Fireworks exploding in the sky and giving off light are an example of a(n) _____.

Serggg [28]

Exothermic change. Because the firework when it exploded, released energy in the form of light. In exothermic changes energy is released, and in endothermic changes energy is absorbed.

- This wouldn't be a physical change, but instead a chemical change. A clue that it is a chemical change is that energy was given off.

7 0

3 years ago

Read 2 more answers

10 points

daser333 [38]

In 1869 he published a table of the elements organized by increasing atomic mass.

Mendeleev is called the "father of the modern periodic table

stated that if the atomic weight of an element caused it to be placed in the wrong group, then the weight must be wrong. (He corrected the atomic masses of Be, In, and U)

was so confident in his table that he used it to predict the physical properties of three elements that were yet unknown.

After the discovery of these unknown elements between 1874 and 1885, and the fact that Mendeleev's predictions for Sc, Ga, and Ge were amazingly close to the actual values, his table was generally accepted.

However, in spite of Mendeleev's great achievement, problems arose when new elements were discovered and more accurate atomic weights determined.

STOP DELETING MY COMMENTS PEOPLE PLEASE!!!!!!!!!

5 0

2 years ago

Read 2 more answers

A thermometer reads an outside air temperature of 35°c. What is the temperature in degrees Fahrenheit

Hoochie [10]

35°c is equal to 95°f

To do this multiply 35 and 1.8

35 x 1.8=63

Now add 32

Resulting in the answer 95

(The equation for to solve for c and f is c1.8+32=f

3 0

2 years ago

The equilibrium constant for the reaction is 1.1 x 106 M. HONO(aq) + CN-(aq) ⇋ HCN(aq) + ONO-(aq) This value indicates that

kakasveta [241]

The given question is incomplete. The complete question is given here :

The equilibrium constant for the reaction is $1.1\times 10^6$ M.

$HONO(aq)+CN^- (aq)\rightleftharpoons HCN(aq)+ONO^-(aq)$

This value indicates that

A. $CN^-$ is a stronger base than $ONO^-$

B. HCN is a stronger acid than HONO

C. The conjugate base of HONO is $ONO^-$

D. The conjugate acid of CN- is HCN

Answer: A. $CN^-$ is a stronger base than $ONO^-$

Explanation:

Equilibrium constant is the ratio of product of the concentration of products to the product of concentration of reactants.

When $K_{p}>1$ ; the reaction is product favoured.

When $K_{p};$ ; the reaction is reactant favored.

$When K_{p}=1$ ; the reaction is in equilibrium.

As, $K_p>>1$ , the reaction will be product favoured and as it is a acid base reaction where $HONO$ acts as acid by donating $H^+$ ions and $CN^-$ acts as base by accepting $H^+$

Thus $HONO$ is a strong acid thus $ONO^-$ will be a weak conjugate base and $CN^-$ is a strong base which has weak $HCN$ conjugate acid.

Thus the high value of K indicates that $CN^-$ is a stronger base than $ONO^-$

7 0

2 years ago

For each reaction, find the value of ΔSo. Report the value with the appropriate sign. (a) 3 NO2(g) + H2O(l) → 2 HNO3(l) + NO(g)

aev [14]

Answer:

ΔS° = -268.13 J/K

Explanation:

Let's consider the following balanced equation.

3 NO₂(g) + H₂O(l) → 2 HNO₃(l) + NO(g)

We can calculate the standard entropy change of a reaction (ΔS°) using the following expression:

ΔS° = ∑np.Sp° - ∑nr.Sr°

where,

ni are the moles of reactants and products

Si are the standard molar entropies of reactants and products

ΔS° = [2 mol × S°(HNO₃(l)) + 1 mol × S°(NO(g))] - [3 mol × S°(NO₂(g)) + 1 mol × S°(H₂O(l))]

ΔS° = [2 mol × 155.6 J/K.mol + 1 mol × 210.76 J/K.mol] - [3 mol × 240.06 J/K.mol + 1 mol × 69.91 J/k.mol]

ΔS° = -268.13 J/K

7 0

3 years ago