All categories

3 years ago

In chemistry the term molecule was in introduced by

Chemistry

1 answer:

AURORKA [14]3 years ago

3 0

Answer:

Avogadro

Explanation:

Molecule: The term molecule was introduced by Avogadro. Molecule is the smallest particle of matter that exists independently and is formed by the combination of atoms. Molecule is also defined as the smallest particle of matter that can exist and retains all the properties of that substance. Was this answer helpful?

plss mark me bainlest

You might be interested in

When electrons in photopigments become excited, what is the result of the energy input?

forsale [732]

Answer:

The energy can be used in a redox reaction

Explanation:

When electrons in photopigments become excited, what is the result of the energy input?

The energy can be used in a redox reaction. The energy can be transferred to a nearby pigment. They can cause fluorescence or heat emission

Please mark me a brainliest answef

8 0

3 years ago

The change in frequency of a wave acceleration

nikdorinn [45]

Answer:

it decreases the amplitude of the wave as it propagates

Explanation:

5 0

3 years ago

Calculate how many grams of sodium azide (NaN3) are needed to inflate a 25.0 × 25.0 × 20.0 cm bag to a pressure of 1.35 atm at a

Luden [163]

Answer : The mass of $NaN_3$ at temperature $20^oC$ 28.47 g.

The mass of $NaN_3$ at temperature $10^oC$ 29.51 g.

Solution : Given,

Pressure of gas = 1.35 atm

Temperature of gas = $20^oC=273+20=293K$ $(0^oC=273K)$

Volume of gas = $25\times 25\times 20cm=12500cm^3=12.5L$ $(1L=1000cm^3)$

Molar mass of $NaN_3$ = 65 g/mole

Part 1 : First we have to calculate the moles of gas at temperature $20^oC$ . The gas produced in the given reaction is $N_2$ .

Using ideal gas equation,

$PV=nRT$

where,

P = pressure of the gas

V = volume of the gas

T = temperature of the gas

n = number of moles of gas

R = Gas constant = 0.0821 Latm/moleK

Now put all the given values in this formula, we get

$(1.35atm)\times (12.5L)=n\times (0.0821Latm/moleK)\times (293K)$

By rearranging the terms, we get the value of 'n'

$n=0.7015moles$

The moles of $N_2$ = 0.7015 moles

The given balanced reaction is,

$20NaN_3(s)+6SiO_2(s)+4KNO_3(s)\rightarrow 32N_2(g)+5Na_4SiO_4(s)+K_4SiO_4(s)$

As, 32 moles of $N_2$ produced from 20 moles of $NaN_3$

So, 0.7015 moles of $N_2$ produced from $\frac{20}{32}\times 0.7015=0.438$ moles of $NaN_3$

Now we have to calculate the mass of $NaN_3$ .

$\text{ Mass of }NaN_3=\text{ Moles of }NaN_3\times \text{ Molar mass of }NaN_3$

$\text{ Mass of }NaN_3=(0.438moles)\times (65g/mole)=28.47g$

Therefore, the mass of $NaN_3$ needed are 28.47 g.

Part 2 : We have to calculate the moles of gas at temperature $10^oC$ and same volume & pressure.

Using ideal gas equation,

$PV=nRT$

Now put all the given values in this formula, we get

$(1.35atm)\times (12.5L)=n\times (0.0821Latm/moleK)\times (283K)$

By rearranging the terms, we get the value of 'n'

$n=0.726moles$

The moles of $N_2$ = 0.726 moles

The given balanced reaction is,

$20NaN_3(s)+6SiO_2(s)+4KNO_3(s)\rightarrow 32N_2(g)+5Na_4SiO_4(s)+K_4SiO_4(s)$

As, 32 moles of $N_2$ produced from 20 moles of $NaN_3$

So, 0.726 moles of $N_2$ produced from $\frac{20}{32}\times 0.726=0.454$ moles of $NaN_3$

Now we have to calculate the mass of $NaN_3$ .

$\text{ Mass of }NaN_3=\text{ Moles of }NaN_3\times \text{ Molar mass of }NaN_3$

$\text{ Mass of }NaN_3=(0.454moles)\times (65g/mole)=29.51g$

Therefore, the mass of $NaN_3$ needed are 29.51 g.

7 0

4 years ago

What is the pH of a solution with a concentration of 4.2 × 10–5 M H3O+? 2.31 4.38 5.62 6.87

Eva8 [605]

<h3><u>Answer</u>;</h3>

pH = 4.38

<h3><u>Explanation;</u></h3>

Consider that ;

[H3O+] = 10-pH mol/L

Therefore; the pH is the -log of the [hydronium ion].

pH = - log [H3O+]

Thus;

pH = - log (4.2 × 10^–5)

= 4.38

3 0

3 years ago

Read 2 more answers

What is a limitation of solar power?

kodGreya [7K]

A is correct .

please vote my answer brainliest. thanks!

3 0

3 years ago

In chemistry the term molecule was in introduced by​

In chemistry the term molecule was in introduced by