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3 years ago

9

How many electrons are in Hydrogen?

2 answers:

kifflom [539]3 years ago

8 0

1 electron per shell

KonstantinChe [14]3 years ago

5 0

1. Hydrogen has 1 electron.

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F<br>Calculate the pH of 1.00 X 10^-6 M HCI *

amid [387]

Explanation:

A.ph 6 B. correct ph goes in a scale up to 14 below 7 is acidic above 7 is basic and in the middle is newtral when they say pH they are asking how acidic is it when they ask for pOH they are asking how basic it is pH is calculated using logs as is pOH so ans

12.65

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3 years ago

Chemistry student needs of 55g acetone for an experiment. by consulting the crc handbook of chemistry and physics, the student d

sergeinik [125]

Answer:

70.15 cm³

Solution:

Data Given;

Mass = 55 g

Density = 0.784 g.cm⁻³

Required:

Volume = ?

Formula Used:

Density = Mass ÷ Volume

Solving for Volume,

Volume = Mass ÷ Density

Putting values,

Volume = 55 g ÷ 0.784 g.cm⁻³

Volume = 70.15 cm³

5 0

3 years ago

Which of these factors will cause a solid solute to dissolve faster?

Nostrana [21]

I believe the answer you are looking for is the 4th one.

8 0

3 years ago

Read 2 more answers

The net ionic equation for the reaction between aqueous nitric acid and aqueous sodium hydroxide is ________. h+ (aq) + na+ (aq)

Nana76 [90]

HNO3+NaOH ----> H2O
H⁺ +NO3⁻+Na⁺+OH⁻ ---> Na⁺ +NO3⁻ +H2O
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7 0

3 years ago

Linda performed the following trials in an experiment. Trial 1: Heat 30.0 grams of water at 0 °C to a final temperature of 40.0

nexus9112 [7]

<u>Answer:</u> The correct answer is Option b.

<u>Explanation:</u>

To calculate the amount of heat absorbed or released, we use the following equation:

$q=mc\Delta T$ .....(1)

where, q = amount of heat absorbed or released.

m = mass of the substance

c = heat capacity of water = 4.186 J/g ° C

$\Delta T$ = Change in temperature

<u>For Trial 1:</u>

We are given:

$m=30g\\\Delta T=[40-0]^oC=40^oC\\q=?J$

Putting values in equation 1, we get:

$q=30g\times 4.186J/g^oC\times 40^oC$

q = 5023.2 J

<u>For trial 2:</u>

We are given:

$m=40g\\\Delta T=[40-30]^oC=10^oC\\q=?J$

Putting values in equation 1, we get:

$q=40g\times 4.186J/g^oC\times 10^oC$

q = 1674.4 J

Heat gained by Trial 1 than trial 2 = $(5023.2-1674.4)J=3347J$

Hence, the amount of heat gained in Trial 1 about 3347 J more than the heat released in Trial 2.

Thus, the correct answer is Option b.

4 0

3 years ago