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Lana71 [14]
3 years ago
14

How many protons are in a hydrogen Atom

Chemistry
2 answers:
andre [41]3 years ago
7 0

Answer:

1

Explanation:

Its periodic number is 1.

Nataly [62]3 years ago
3 0

Answer:

There are 1 plus that is the periodic number

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Please help me this is my fourth attempt.
Vaselesa [24]

Explanation:

CH4 + 4S ---> CS2 + 2H2S

4) 0.75 mol S × (1 mol CS2/4 mol S) = 0.19 mol CS2

5) 3 mol H2S × (1 mol CH4/2 mol H2S) = 1.5 mol CH4

Fe2O3 + 2Al ---> 2Fe + Al2O3

6) 25 g FeO3 × (1 mol Fe2O3/159.69 g Fe2O3) = 0.16 mol Fe2O3

0.16 mol Fe2O3 × (2 mol Al/1 mol Fe2O3) = 0.32 mol Al

0.32 mol Al × (26.98 g Al/1 mol Al) = 8.6 g Al

7) Given:

45 g Al × (1 mol Al/26.98 g Al) = 1.6 mol Al

85 g Fe2O3 ×(1 mol Fe2O3/159.69 g Fe2O3)

= 0.53 mol Fe2O3

Let's look at how much Fe each reactant will produce:

1.6 mol Al × (2 mol Fe/2 mol Al) = 1.6 mol Fe

0.53 mol Fe2O3 × (2 mol Fe/1 mol Fe2O3) = 1.1 mol Fe

Note that the given amount of Fe2O3 will give us fewer Fe. Therefore, Fe2O3 is the limiting reactant.

8) Al will produce 1.6 mol Fe × (55.845 g Fe/1 mol Fe)

= 89 g Fe

Fe2O3 will produce 1.1 mol Fe × (55.845 Fr/1 mol Fe)

= 61 g Fe

9) Since Fe2O3 is the limiting reactant, the ideal yield of Fe for the reaction is 61 g. If the actual reaction only gave us 25 g Fe. then the percent yield of Fe is

%yield = (25 g Fe/61 g Fe) × 100% = 41%

10) If we only got 25 g Fe, then the amount of Al actually used in the reaction is

25 g Fe × (1 mol Fe/55.845 g Fe) = 0.45 mol Fe

0.45 mol Fe × (2 mol Al/2 mol Fe) = 0.45 mol Al

0.45 mol Al × (26.98 g Al/1 mol Al) = 12 g

Therefore, the leftover amount of Al is

25 g Al - 12 g Al = 13 g Al

8 0
2 years ago
1. Calculate the concentration of hydronium ion of both buffer solutions at their starting pHs. Calculate the moles of hydronium
lilavasa [31]

Answer:

This question is incomplete, here's the complete question:

1. Calculate the concentration of hydronium ion of both buffer solutions at their starting pHs. Calculate the moles of hydronium ion present in 20.0 mL of each buffer.

Buffer A

Mass of sodium acetate used: 0.3730 g

Actual ph of the buffer 5.27

volume of the buffer used in buffer capacity titration 20.0 mL

Concentration of standardized NaOH 0.100M

moles of Naoh needed to change the ph by 1 unit for the buffer 0.00095mol

the buffer capacity 0.0475 M

Buffer B

Mass of sodium acetate used 1.12 g

Actual pH of the buffer 5.34

Volume of the buffer used in buffer capacity titration 20.0 mL

Concentration if standardized NaOH 0.100 M

moles of Naoh needed to change the ph by 1 unit 0.0019 mol

the buffer capacity 0.095 M

2.) A change of pH by 1 unit means a change in hydronium ion concentration by a factor of 10. Calculate the number of moles of NaOH that would theoretically be needed to decrease the moles of hydronium you calculated in #1 by a factor of 10 for each buffer. Are there any differences between your experimental results and the theoretical calculation?

3.) which buffer had a higher buffer capacity? Why?

Explanation:

Formula,

moles = grams/molar mass

molarity = moles/L of solution

1. Buffer A

molarity of NaC2H3O2 = 0.3731 g/82.03 g/mol x 0.02 L = 0.23 M

molarity of HC2H3O2 = 0. 1 M

Initial pH

pH = pKa + log(base/acid)

= 4.74 + log(0.23/0.1)

= 5.10

pH = -log[H3O+]

[H3O+] = 7.91 x 10^-6 M

In 20 ml buffer,

moles of H3O+ = 7.91 x 10^-6 M x 0.02 L

= 1.58 x 10^-7 mol

Buffer B

molarity of NaC2H3O2 = 1.12 g/82.03 g/mol x 0.02 L = 0.68 M

molarity of HC2H3O2 = 0.3 M

Initial pH

pH = pKa + log(base/acid)

= 4.74 + log(0.68/0.3)

= 5.10

pH = -log[H3O+]

[H3O+] = 7.91 x 10^-6 M

In 20 ml buffer,

moles of H3O+ = 7.91 x 10^-6 M x 0.02 L

= 1.58 x 10^-7 mol

2. let x moles of NaOH is added,

Buffer A,

pH = 5.10

[H3O+] = 7.91 x 10^-6 M

new pH = 4.10

new [H3O+] = 7.91 x 10^-5 M

moles of NaOH to be added = (7.91 x 10^-5 - 7.91 x 10^-6) x 0.02 L

= 1.42 x 10^-6 mol

3. Buffer B with greater concentration of NaC2H3O2 and HC2H3O2 has higher buffer capacity as it resists pH change to a wider range due to addition of acid or base to the system as compared to low concentration of Buffer A

5 0
3 years ago
What are some good movies to watch on HBOMAX
Marina CMI [18]

Answer:

My Neighbor Totoro

Princess Mononoke

Pom Poko

Only Yesterday

Kiki's Delivery Service

Nausicaä of the Valley of the Wind

7 0
3 years ago
Read 2 more answers
In 25 words or fewer, what do you think a scientist would want to explain about this soda?
Gnesinka [82]

Answer:

This soda is brown, sweet, has high carbonation, and is about 200 calories when consumed.

Explanation:

:)

8 0
2 years ago
Which physical method can be used for obtaining a sample of salt from a sm
lana [24]

The physical method that can be used for obtaining a sample of salt from a small beaker of salt and water would be evaporation.

<h3>Separation of salt and water</h3>

A mixture of salt and water can be separated by a method known as evaporation. This is based on the assumption that the salt in question is a water-soluble salt.

In order to separate the salt/water mixture:

  • Place the mixture in a suitable evaporating dish
  • Boil the mixture until all the water evaporates.
  • The remaining residue would be the salt

Care should be taken not to overheat the residue in order to avoid melting. Evaporation is generally used to separate a mixture of water and soluble salt. If the salt is insoluble, filtration using a suitable filter paper will filter off the salt while the water is collected as the filtrate.

More on evaporation can be found here: brainly.com/question/1097783

#SPJ1

Which physical method can be used for obtaining a sample of salt from a small beaker of salt water?

6 0
1 year ago
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