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3 years ago

If two atoms have the same atomic numbers but different mass numbers, then they can be considered atoms of the same element.

Chemistry

2 answers:

stiks02 [169]3 years ago

8 0

Answer:

True.

Explanation:

If two atoms have the same atomic number, they have the same number of protons in the nucleus.

The number of protons derermines the identity of an atom.

So, two atoms with the same atomic number are the same element.

castortr0y [4]3 years ago

7 0

Same elements can have different nuclei. So, true.

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Paleontologists have discovered several ancestors of the modern horse.

ExtremeBDS [4]

Answer:

hyracotherium

Explanation:

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3 years ago

Read 2 more answers

How many grams of hydrochloric acid are produced when 15.0 grams NaCl react with excess H2SO4 in the reaction

e-lub [12.9K]

Answer:

11.65 g

Explanation:

Amount of NaCl = 15.0 grams

amount of H₂SO₄ = Excess

mass of hydrochloric acid (HCl) = ?

Solution:

To solve this problem first we will look for the reaction that NaCl react with H₂SO₄

Reaction:

2NaCl + H₂SO₄ --------> 2 HCl + Na₂SO₄

As the H₂SO₄ is in excess so the amount of hydrochloric acid (HCl) depends on the amount of NaCl as it its act as limiting reactant.

Now if we look at the reaction

2NaCl + H₂SO₄ --------> 2HCl + Na₂SO₄

1 mol 2 mol

Now convert moles to mass

Then

Molar mass of NaCl = 23 + 35.5 = 58.5 g/mol

Molar mass of HCl = 1 + 35.5 = 36.5 g/mol

So,

2NaCl + H₂SO₄ --------> 2HCl + Na₂SO₄

2 mol (58.5 g/mol) 2 mol (36.5 g/mol)

94 g 73 g

So if we look at the reaction; 94 g of NaCl gives with 73 g of hydrochloric acid (HCl) in a this reaction, then how many grams of hydrochloric acid (HCl) will produce from 15.0 g of NaCl

For this apply unity formula

94 g of NaCl ≅ 73 g of HCl

15.0 g of NaCl ≅ X g of HCl

By Doing cross multiplication

X g of HCl = 73 g x 15.0 g / 94 g

X g of HCl = 11.65 g

11.65 g of hydrochloric acid (HCl) will produce by 15.0 g of NaCl

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3 years ago

Why is drinking urine unhealthy? whats in it?

Kryger [21]

This is by far probably the best question I've answered all day.
It's not unhealthy to drink it.
Urine consists of 95% water and 5% nutrients

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3 years ago

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If you have 16 g of manganese (II) nitrate tetrahydrate, how much water is required to prepare 0.16 M solution from this amount

nirvana33 [79]

Answer: The volume of water required is 398 mL

Explanation:

To calculate the molarity of solution, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}$

We are given:

Mass of solute (manganese (II) nitrate tetrahydrate) = 16 g

Molar mass of manganese (II) nitrate tetrahydrate = 251 g/mol

Molarity of solution = 0.16 M

Putting values in above equation, we get:

$0.16M=\frac{16g\times 1000}{251g/mol\times \text{Volume of solution}}\\\\\text{Volume of solution}=398mL$

Hence, the volume of water required is 398 mL

7 0

3 years ago

What is the molar solubility of marble (i.e., [ca2 ] in a saturated solution in normal rainwater, for which ph=5.60? express you

Brut [27]

Missing in your question :

Ksp of(CaCO3)= 4.5 x 10 -9

Ka1 for (H2CO3) = 4.7 x 10^-7

Ka2 for (H2CO3) = 5.6 x 10 ^-11

1) equation 1 for Ksp = 4.5 x 10^-9

CaCO3(s)→ Ca +2(aq) + CO3-2(aq)

2) equation 2 for Ka1 = 4.7 x 10^-7

H2CO3 + H2O → HCO3- + H3O+

3) equation 3 for Ka2 = 5.6 x 10^-11

HCO3-(aq) + H2O(l) → CO3-2 (aq) + H3O+(aq)

so, form equation 1& 2&3 we can get the overall equation:
CaCO3(s) + H+(aq) → Ca2+(aq) + HCO3-(aq)

note: you could get the overall equation by adding equation 1 to the inverse of equation 3 as the following:
when the inverse of equation 3 is :

CO3-2 (aq) + H3O+ (aq) ↔ HCO3- (aq) + H2O(l) Ka2^-1 = 1.79 x 10^10
when we add it to equation 1
CaCO3(s) ↔ Ca2+(aq) + CO3-2(aq) Ksp = 4.5 x 10^-9

∴ the overall equation will be as we have mentioned before:
when H3O+ = H+

CaCO3(s) + H+(aq) ↔ Ca2+ (aq) + HCO3-(aq) K= 80.55

from the overall equation:

∴K = [Ca2+][HCO3-] / [H+]

when we have [Ca2+] = [HCO3-] so we can assume both = X

∴K = X^2 / [H+]

when we have the PH = 5.6 so we can get [H+]

PH = - ㏒[H+]
5.6 = -㏒[H]
∴[H] = 2.5 x 10^-6

so, by substitution on K expression:

∴ 80.55 = X^2 / (2.5 x10^-6)

∴X = 0.0142

∴[Ca2+] = X = 0.0142

6 0

3 years ago