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2 years ago

Approximately how many formula units of Co2(HPO4)3 are found in 175. g of hydrogenphosphate ion?

Chemistry

1 answer:

marshall27 [118]2 years ago

4 0

Answer:

This question lacks options, the options are:

A. 2.89 x 1022

B. 8.66 x 1022

C. 3.66 x 1023

D. 1.10 x 1024

E. 3.29 x 1024

The answer is C. 3.66 × 10^23

Explanation:

In the chemical compound given as follows: Co2(HPO4)3, hydrogenphosphate ion is (HPO4)3.

Number of moles of hydrogenphosphate ion can be calculated using the formula

mole (n) = mass (m) /molar mass (MM)

Molar mass of (HPO4)3;

{1 + 31 + 16(4)}3

= {32 + 64)3

= (96)3

= 288g/mol

mole = 175/288

mole = 0.6076

mole = 0.607mol

The formula of units of hydrogenphosphate ion can be calculated by multiplying number of moles by Avogadro's number (6.022 × 10²³)

6.022 × 10²³ × 0.607

= 3.655 × 10²³

= 3.66 × 10²³

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Why are infants and children often more susceptible to the effects of toxic substances than adult? (Explain)

puteri [66]

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2 years ago

Determine if each statement below is True or False regarding Arrhenius and Br∅nsted-Lowry definitions for acids and bases.

Vlad1618 [11]

Answer:

All statements are true

Explanation:

Arrhenius theory:-

The Arrhenius theory was introduced introduced by Swedish scientist named Svante Arrhenius in 1887.

According to the theory, acids are the substances which dissociate in the aqueous medium to produce electrically charged atoms ( may be molecule). Out of these species furnished, one must be a proton or the hydrogen ion, $H^+$ .

Hence, statement 1 is correct.

Bronsted-Lowry base is the specie which accepts the proton form the medium. For example, Benzoate ion is the conjugate base of the benzoic acid. It is a Bronsted-Lowry base and the dissociation of benzoate ion can be shown as:-

$C_6H_5COO^-_{(aq)} + H_2O_{(l)}\rightleftharpoons C_6H_5COOH_{(aq)} + OH^-_{(aq)}$

Hence, statement 2 is correct.

Acetic acid will dissociate as:-

$CH_3COOH\rightleftharpoons CH_3COO^-+H^+$

Hence, statement 3 is also correct.

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2 years ago

A 50.0 mL solution of 0.141 M KOH is titrated with 0.282 M HCl . Calculate the pH of the solution after the addition of each of

Kobotan [32]

Answer:

pH =1 2.84

Explanation:

First we have to start with the reaction between HCl and KOH:

$HCl~+~KOH->~H_2O~+~KCl$

Now for example, we can use a volume of 10 mL of HCl. So, we can calculate the moles using the molarity equation:

$M=\frac{mol}{L}$

We know that $10mL=0.01L$ and we have the concentration of the HCl $0.282M$ , when we plug the values into the equation we got:

$0.282M=\frac{mol}{0.01L}$

$mol=0.282*0.01$

$mol=0.00282$

We can do the same for the KOH values ( $50mL=0.05L$ and $0.141M$ ).

$0.141M=\frac{mol}{0.05L}$

$mol=0.141*0.05$

$mol=0.00705$

So, we have so far 0.00282 mol of HCl and 0.00705 mol of KOH. If we check the reaction we have a molar ratio 1:1, therefore if we have 0.00282 mol of HCl we will need 0.00282 mol of KOH, so we will have an excess of KOH. This excess can be calculated if we substract the amount of moles:

$0.00705-0.00282=0.00423mol~of~KOH$

Now, if we want to calculate the pH value we will need a concentration, in this case KOH is in excess, so we have to calculate the concentration of KOH. For this, we already have the moles of KOH that remains left, now we need the total volume:

$Total~volume=50mL+10mL=60mL$