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2 years ago

The ions Ca2+ and PO43- form salt with the formula

Chemistry

2 answers:

marta [7]2 years ago

7 0

Ca3(PO4)2 is the answer.

Minchanka [31]2 years ago

3 0

Answer : The formula for the compound is $Ca_3(PO_4)_2$

Explanation :

For formation of a neutral ionic compound, the charges on cation and anion must be balanced.

The cation is formed by loss of electrons by metals and anions are formed by gain of electrons by non-metals.

In the given question, the calcium metal forms an ionic compound with polyatomic anion (phosphate) $PO_4^{3-}$ .

Here, calcium is having an oxidation state of +2 called as $Ca^{2+}$ cation and phosphate $PO_4^{3-}$ is an anion with oxidation state of -3. The charges are not balanced. So, the charges are balanced by the cris-cross method. Thus, the compound formed will be, $Ca_3(PO_4)_2$

Hence, the formula for the compound is $Ca_3(PO_4)_2$

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A mineral sample is analyzed for its cobalt and calcium content. A sample is dissolved, and then the cobalt and calcium are prec

bekas [8.4K]

Answer:

9.88

Explanation:

As higher is the Ksp, more soluble is the compound. So, Co(OH)₂ is the less soluble hydroxide.

The maximum concentration of it must be 1x10⁻⁶ M, and the reaction is:

Co(OH)₂(s) ⇄ Co⁺²(aq) + 2OH⁻(aq)

So, [Co⁺²] = 1x10⁻⁶M

Ksp = [Co⁺²] *[OH⁻]²

[OH⁻]² = 5.9x10⁻¹⁵/1x10⁻⁶

[OH⁻] = √(5.9x10⁻⁹)

[OH⁻] = 7.6811x10⁻⁵

pOH = -log[OH⁻]

pOH = -log(7.6811x10⁻⁵)

pOH = 4.11

Knowing that pH + pOH = 14

pH = 14 - 4.11

pH = 9.88

7 0

1 year ago

Read 2 more answers

How many atoms of Mg are present in 97.22 grams of Mg?

Lostsunrise [7]

Answer: Option (b) is the correct answer.

Explanation:

It is given that mass of Mg is 97.22 g and it is known that molar mass of Mg is 24.305 g/mol.

So, calculate the number of moles as follows.

No. of moles = $\frac{mass given in grams}{Molar mass}$

= $\frac{97.22 g}{24.305 g/mol}$

= 4 mol

Also, it is known that 1 mole has $6.023 \times 10^{23}$ atoms/mol. Therefore, calculate the number of atoms in 4 mol as follows.

$4 mol \times 6.023 \times 10^{23} atoms/mol$

= $24.08 \times 10^{23}$ atoms

or, = $2.408 \times 10^{23}$ atoms

Thus, we can conclude that there are $2.408 \times 10^{23}$ atoms in 97.22 grams of Mg.

7 0

2 years ago

A student mixed 20.00 grams of calcium nitrate, 10.00 grams of sodium nitrate, and 50.00 grams of aluminum nitrate in a 5.00 Lit

My name is Ann [436]

Answer:

$M=0.213M$

Explanation:

Hello,

In this case, for each nitrate-based salt, we compute the nitrate moles as shown below:

$n_{NO_3^-}=20.00gCa(NO_3)_2*\frac{1molCa(NO_3)_2}{164.088 gCa(NO_3)_2} *\frac{2molNO_3^-}{1molCa(NO_3)_2} =0.244molNO_3^-$

$n_{NO_3^-}=10.00gNaNO_3*\frac{1molNaNO_3}{84.9947 gNaNO_3} *\frac{1molNO_3^-}{1molNaNO_3} =0.118molNO_3^-$

$n_{NO_3^-}=50.00gAl(NO_3)_3*\frac{1molAl(NO_3)_3}{212.996gAl(NO_3)_3} *\frac{3molNO_3^-}{1molAl(NO_3)_3} =0.704molNO_3^-$

We notice calcium nitrate has two moles of nitrate ion, sodium nitrate has one and aluminium nitrate has three. Hence we add the moles to obtain the total moles nitrate ion:

$n_{NO_3^-}^{Tot}=0.244+0.118+0.704=1.066molNO_3^-$

Finally, we compute the molarity:

$M=\frac{1.066molNO_3^-}{5.00L} \\\\M=0.213M$

Regards.

6 0

2 years ago

Even though we are well into the course Dumbledore notices some of the students are still having trouble with significant figure

Sphinxa [80]

Answer:

B) 2

Explanation:

Significant figures : The figures in a number which express the value -the magnitude of a quantity to a specific degree of accuracy is known as significant digits.

The rule apply for the multiplication and division is :

The least number of significant figures in any number of the problem determines the number of significant figures in the answer.

The rule apply for the addition and subtraction is :

The least precise number present after the decimal point determines the number of significant figures in the answer.

(3.478-2.31) = 1.168 ≅ 1.17 (Rounded to least decimal digit)

(4.428-3.56) = 0.868 ≅ 0.87 (Rounded to least decimal digit)

So,

1.17 * 0.87 = 1.0 (Rounded to least significant)

Answer - two significant digits

5 0

1 year ago

A gas sample at stp contains 1.15 g oxygen gas and 1.55 g nitrogen gas.what is the volume of the gas sample?

ss7ja [257]

O2=32 g/ mol
1.15/32=0.035
N2=28 g/mol
1.55/28=0.055
in STP every 22.4 litters is 1 mol

4 0

2 years ago

Read 2 more answers