Cr(PO4)2 =<br> name the compound

2. Use the Henderson-Hasselbalch equation to estimate the pH of a buffer solution that is composed of 20 mL of 10 M sodium forma

Mkey [24]

Answer:

pH of buffer =4.75

Explanation:

The pH of buffer solution is calculated using Henderson Hassalbalch's equation:

$pH=pKa+log[\frac{[salt]}{[acid]}$

Given:

pKa = 3.75

concentration of acid = concentration of formic acid = 1 M

concentration of salt = concentration of sodium formate = 10 M

$pH=3.75+log[\frac{10}{1}]=3.75+1=4.75$

pH of buffer =4.75

7 0

3 years ago

In the winter , people often buy large bags of rock salt to sprinkle on their walkways . Why do people do this?

slamgirl [31]

The salt causes the water to freeze at a lower temperature. When a solute, aka salt, is introduced to the system, the freezing point is lowered. This makes the water freeze at a lower temperature.

6 0

3 years ago

How many hydrogen atoms are in 5.80 mol of ammonium sulfide

IRINA_888 [86]

Moles of ammonium sulfide = 5.80 mol

The formula of ammonium sulfide is (NH₄)₂S

So each molecule of ammonium sulfide has (4 x 2) or 8 atoms of H

One mole of ammonium sulfide has 8 moles of H

5.80 mol of ammonium sulfide has (8 x 5.8) or 46.4 moles of H

As per the definition of Avogadro's number, 1 mole = 6.022 x 10²³ atoms

46.4 moles of H x (6.022 x 10²³ atoms/ 1 mole of H)

= 2.8 x 10²⁵ H atoms

Therefore, 2.8 x 10²⁵ H atoms are in 5.80 mol of ammonium sulfide.

5 0

2 years ago

Be sure to answer all parts. The standard enthalpy of formation and the standard entropy of gaseous benzene are 82.93 kJ/mol and

skelet666 [1.2K]

Answer : The values of $\Delta H^o,\Delta S^o\text{ and }\Delta G^o$ are $33.89kJ,95.94J/K\text{ and }5.299kJ/mol$ respectively.

Explanation :

The given balanced chemical reaction is,

$C_6H_6(l)\rightarrow C_6H_6(g)$

First we have to calculate the enthalpy of reaction $(\Delta H^o)$ .

$\Delta H^o=H_f_{product}-H_f_{reactant}$

$\Delta H^o=[n_{C_6H_6(g)}\times \Delta H_f^0_{(C_6H_6(g))}]-[n_{C_6H_6(l)}\times \Delta H_f^0_{(C_6H_6(l))}]$

where,

$\Delta H^o$ = enthalpy of reaction = ?

n = number of moles

$\Delta H_f^0_{(C_6H_6(g))}$ = standard enthalpy of formation of gaseous benzene = 82.93 kJ/mol

$\Delta H_f^0_{(C_6H_6(l))}$ = standard enthalpy of formation of liquid benzene = 49.04 kJ/mol

Now put all the given values in this expression, we get:

$\Delta H^o=[1mole\times (82.93kJ/mol)]-[1mole\times (49.04J/mol)]$

$\Delta H^o=33.89kJ/mol=33890J/mol$

Now we have to calculate the entropy of reaction $(\Delta S^o)$ .

$\Delta S^o=S_f_{product}-S_f_{reactant}$

$\Delta S^o=[n_{C_6H_6(g)}\times \Delta S^0_{(C_6H_6(g))}]-[n_{C_6H_6(l)}\times \Delta S^0_{(C_6H_6(l))}]$

where,

$\Delta S^o$ = entropy of reaction = ?

n = number of moles

$\Delta S^0_{(C_6H_6(g))}$ = standard entropy of formation of gaseous benzene = 269.2 J/K.mol

$\Delta S^0_{(C_6H_6(l))}$ = standard entropy of formation of liquid benzene = 173.26 J/K.mol

Now put all the given values in this expression, we get:

$\Delta S^o=[1mole\times (269.2J/K.mol)]-[1mole\times (173.26J/K.mol)]$

$\Delta S^o=95.94J/K.mol$

Now we have to calculate the Gibbs free energy of reaction $(\Delta G^o)$ .

As we know that,

$\Delta G^o=\Delta H^o-T\Delta S^o$

At room temperature, the temperature is $25^oC\text{ or }298K$ .

$\Delta G^o=(33890J)-(298K\times 95.94J/K)$

$\Delta G^o=5299.88J/mol=5.299kJ/mol$

Therefore, the values of $\Delta H^o,\Delta S^o\text{ and }\Delta G^o$ are $33.89kJ,95.94J/K\text{ and }5.299kJ/mol$ respectively.

6 0

2 years ago

Read 2 more answers

a solution contains water (h20), hydrochloric acid and a wire of magnesium (mg). It takes about 60 seconds for the chemical reac

Crazy boy [7]

The reaction should go faster

8 0

3 years ago

Cr(PO4)2 = name the compound