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4 years ago

Write the charge balance for a solution that contains h , oh–, mg2 , hso4–, so42–, mg(hso4) , mg(oh) , na , and no3–.

1 answer:

5 0

A solution is when a solute is added to a water solvent. Since water is neutral, the charge of the said solution is dictated by the charge of the ion. This is the subscript of the element. If it is not stated, then we use the common ion form.

1. H: The charge for hydrogen ion is +1.
2. OH⁻: The charge is ⁻1.
3. Mg2: This ion has a charge of +2.
4. HSO₄⁻: -1
5. SO₄²⁻: -2
6. Mg(HSO₄): This is not an ion, but an undissociated compound. So, it has a neutral charge of 7.
7. Mg(OH): Neutral at pH 7
8. Na: Sodium ion has a charge of +1.
9. NO₃⁻: -1.

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Identify the polynomial. x3- y3 + z monomial binomial trinomial

Firdavs [7]

Answer: Option (c) is the correct answer.

Explanation:

A monomial is defined as an algebraic expression in which there is only one term present.

For example, 3x is a monomial.

A binomial is defined as an expression where there will be two terms present. For example, $(x^{2} + y^{2})$ is a binomial.

A polynomial is defined as an algebraic expression which contains more than two terms.

For example, $x^{3} - y^{3} + z$ is a polynomial.

Thus, we can conclude that the given expression $x^{3} - y^{3} + z$ is a polynomial.

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3 years ago

What 6 elements in the periodic table are named after real people?

Romashka [77]

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obelium (No, 102) – Alfred Nobelroentgenium (Rg, 111) – Wilhelm Roentgen (formerly Ununumium)rutherfordium (Rf, 104) – Ernest Rutherford seaborgium (Sg, 106) – Glenn T. Seaborg

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3 years ago

Vanadium (V) and oxygen (O) form a series of compounds with the following compositions: Mass % V 76.10 67.98 61.42 56.02 Mass %

11Alexandr11 [23.1K]

Answer:

For every given mass of Vanadium, the relative number of oxygen atoms present or the mole ratio of Oxygen to Vanadium is:

A. 1:1

B. 3:2

C. 2:1

D. 5:2

Note: The question is stated more clearly below:

Vanadium (V) and oxygen (O) form a series of compounds with the following compositions: Mass % V 76.10 67.98 61.42 56.02 Mass % O 23.90 32.02 38.58 43.98 Compound Mass % N 1 33.28 2 39.94.

What are the relative numbers of atoms of oxygen in the compounds for a given mass of vanadium?

Explanation:

Number of moles in 100 g mass = % mass / molar mass

Molar mass of Vanadium, V = 51 g/mol

Molar mass of oxygen atom, O = 16 g/mol

1. Percentage mass of V and O is 76.10% and 23.90% respectively.

Number of moles of each atom;

V = 76.10/51.0 = 1.5 moles

O = 23.9/16 = 1.5 moles

Mole ratio of oxygen to vanadium = 1.5/1.5 = 1 : 1

2. Percentage mass of V and O is 67.98% and 32.02% respectively

Number of moles of each atom:

V = 67.98/51 = 1.33

O = 32.02/16 = 2

Mole ratio of oxygen to vanadium = 2/1.33 = 1.5 : 1 = 3 : 2

3. Percentage mass of V and O is 61.42% and 38.58% respectively

Number of moles of each atom:

V = 61.42/51 = 1.2

O = 38.58/16 = 2.4

Mole ratio of oxygen to vanadium = 2.4/1.2 = 2 : 1

4. Percentage mass of V and O is 56.02% and 43.98% respectively

Number of moles of each atom:

V = 56.02/51 = 1.10

O = 43.98/16 = 2.75

Mole ratio of oxygen to vanadium = 2.75/1.10 = 2.5 : 1 = 5 : 2

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3 years ago

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Answer:

a blocker?

Explanation:

just a guess

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4 years ago

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Answer:

I think the answer is

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3 years ago