All categories

3 years ago

Which model of the atom first introduced the idea of a nucleus?

Chemistry

2 answers:

ankoles [38]3 years ago

5 0

The answer is c.) Rutherford model

liq [111]3 years ago

5 0

the answer to your question is c. rutherfords model

You might be interested in

What is the hydridization of TeBr4

Debora [2.8K]

It's sp3d according to the lewis dot structure (four bonds, one lone pair).

5 0

3 years ago

When light rays appear to be coming from behind a mirror, what type of image has been formed?

Setler79 [48]

Plane is the right answer I thinkkkkk

4 0

3 years ago

Read 2 more answers

The electron configuration of an element is 1s2 2s2. How many valence electrons does the element have?

faust18 [17]

Answer:

Two valence electrons

Explanation:

7 0

3 years ago

What is produced by a neutralization reaction between an arrhenius acid and an arrhenius base?

stepladder [879]

Answer: Salt and Water

Explanation:

An Arrhenius acid (HCl) can best be defined as any substance that when added to water increases the concentration of H+ ions.

While an Arrhenius base (KOH) is any substance that when added to water increases the concentration of OH- ions.

When an Arrhenius acid such as HCl reacts with an Arrhenius base such as KOH, the end products will be salt and water, in a process called Neutralization Reaction.

HCl (aq) + KOH (aq) -------> KCl (aq) + H2O (l)

4 0

3 years ago

A 8.00 L tank at 26.9 C is filled with 5.53 g of dinitrogen difluoride gas and 17.3 g of sulfur hexafluoride gas. You can assume

pshichka [43]

Answer:

mole fraction of N_2 O = 0.330

mole of fraction SF_4 = 0.669

PRESSURE OF N_2 O = 39127.053 Pa

pressure of SF_4 = 792126.36

Total pressure = 118253.413 Pa

Explanation:

Given data:

volume of tank 8 L

Weight of dinitrogen difluoride gas 5.53 g

weight of sulphur hexafluoride gas 17.3 g

Amount of $N_2 O = \frac{5.53}{14*2 + 16} = 0.1256 mol$

amount of $SF_4 = \frac{17.3}{32.1 + 19*4} = 0.254 mol$

mole fraction of $N_2 O = \frac{0.1256}{0.1256 + 0.254} = 0.330$

mole of fraction $SF_4 = \frac{0.254}{0.1256 + 0.254} = 0.669$

PV = nRT

P of N_2 O $= \frac{0.1256 *8.31 (273 + 26.9}{0.008} = 39127.053 Pa$

mole of SF_4 $=\frac{0.254 *8.31*(273+26.9)}{.008} = 79126.36 Pa$

Total pressure = 39127.053 + 79126.36 = 118253.413 Pa

6 0

3 years ago