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3 years ago

Enter the formula for the compound cobalt(II) phosphate. Express your answer as a chemical formula.

Chemistry

1 answer:

alexira [117]3 years ago

5 0

Answer:

$Co_3(PO_4)_2$

Explanation:

Naming of the ionic compounds:-

The name of the cation is written first and the the name of the anion is written after the name of the cation separated by single space.
The negative ion is written next and a suffix is added at the end of the negative ion. The suffix written is '-ide'.
In case of transition metals, the oxidation state are written in roman numerals in bracket in front of positive ions.

Hence, given ionic compound:-

Cobalt(II) phosphate

So, Cobalt will have a positive charge of +2

Phosphate is $PO_4^{3-}$

So, the formula is :-

Co $PO_4^{3-}$

2 3

$Co_3(PO_4)_2$

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andrezito [222]

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4 0

3 years ago

if you take 2.0mL of a 7.0 ppm NaOH solution and dilute it to 250.0mL, calculate the final concentration in ppm.

Oduvanchick [21]

Answer:

C₂ = 0.056 ppm

Explanation:

Given data:

Initial volume = 2.0 mL

Initial concentration = 7.0 ppm

Final volume = 250.0 mL

Final concentration = ?

Solution:

Formula:

C₁V₁ = C₂V₂

C₁ = Initial concentration

V₁ = Initial volume

C₂ = Final concentration

V₂ = Final volume

Now we will put the values in formula.

C₁V₁ = C₂V₂

7.0 ppm × 2.0 mL = C₂ × 250.0 mL

C₂ = 14.0 ppm.mL /250.0 mL

C₂ = 0.056 ppm

4 0

3 years ago

Arsenic(III) sulfide sublimes readily, even below its melting point of 320 °C. The molecules of the vapor phase are found to eff

dezoksy [38]

The molecular formula : As₄S₆

<h3>Further explanation</h3>

Given

Rate of effusion of arsenic(III) sulfide = 0.28 times the rate of effusion of Ar atoms

Required

The molecular formula

Solution

Graham's law: the rate of effusion of a gas is inversely proportional to the square root of its molar masses or

the effusion rates of two gases = the square root of the inverse of their molar masses:

$\rm \dfrac{r_1}{r_2}=\sqrt{\dfrac{M_2}{M_1} }$

$\rm M_1\times r_1^2=M_2\times r_2^2$

Input the value :

1 = Arsenic(III) sulfide

2 = Ar

MM Ar = 40 g/mol

0.28 = √(40/M₁)

M₁=40 : 0.28²

M₁=510 g/mol

The empirical formula of arsenic(III) sulfide = As₂S₃

(Empirical formula)n = molecular formula

( As₂S₃)n = 510 g/mol

(246.02 g/mol)n = 510 g/mol

n = 2

So the molecular formula : As₄S₆

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3 years ago

Match the mountain range with its formation.

zimovet [89]

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4 years ago

A sample of a compound contains only the elements sodium, sulfur, and oxygen. it is found by analysis to contain 0.979 g na, 1.3

Aloiza [94]

To determine the empirical formula for the compound that contains <span>0.979 g Na, 1.365 g S, and 1.021 g O, we convert these to mole units. The molar masses to be used are:

Molar mass of Na = 23 g/mol
</span>Molar mass of S = 32 g/mol
Molar mass of O = 16 g/ mol

The number of moles is obtained using the molar mass for each element.

moles Na = 0.979 g Na/ 23 g/mol Na = 0.04256
moles S = 1.365 g Na/ 32 g/mol Na = 0.04265
moles O = 1.021 g O/ 16 g/mol Na = 0.06326

We then divide each with the smallest number of moles obtained.

Na: 0.04256/ 0.04256 = 1
S: 0.04265/ 0.04256 = 1.002 ≈ 1
O: 0.06326/ 0.04256 = 1.49 ≈ 1.5

We then have an empirical formula of NaSO₁.₅. However, chemical formulas must have only integers as subscripts, thus, we multiply each to 2. The empirical formula is then Na₂S₂O₃ also known as sodium thiosulfate.

4 0

3 years ago

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