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3 years ago

6

A chemist makes a solution of NaOH for use in an experiment. She dissolves 0.00500 mol NaOH in 1.00 L H2O. The NaOH completely d

issociates to form a solution with [OH− ] = 0.00500 M. What is the pOH of this solution?

2 answers:

Mekhanik [1.2K]3 years ago

6 0

Answer:

C. 2.30

Explanation:

babymother [125]3 years ago

6 0

Answer:

C(2.30)

Explanation:

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3. If all the waves were measured for a period of 5 seconds, which wave had the highest frequency?

Vanyuwa [196]

Answer:

Correct answer is A.

Explanation:

Frequency is the number of oscillations that a wave have per unit time. Since time is measured in seconds, the wave with the highest frequency must register the highest number of oscillation per second. Hence, correct answer is A.

8 0

2 years ago

A student found that the titration had taken 10.00 ml of 0.1002 m naoh to titration 0.132 g of aspirin, a monoprotic acid. calcu

Harman [31]

The balanced equation for the reaction between NaOH and aspirin is as follows;
NaOH + C₉H₈O₄ --> C₉H₇O₄Na + H₂O
stoichiometry of NaOH to C₉H₈O₄ is 1:1
The number of NaOH moles reacted - 0.1002 M / 1000 mL/L x 10.00 mL
Number of NaOH moles - 0.001002 mol
Therefore number of moles of aspirin - 0.001002 mol
Mass of aspirin reacted - 0.001002 mol x 180.2 g/mol = 0.18 g
However the mass of the aspirin sample is 0.132 g but 0.18 g of aspirin has reacted, therefore this question is not correct.

4 0

3 years ago

Read 2 more answers

Calculate the standard enthalpy change for the reaction at 25 ∘ 25 ∘ C. Standard enthalpy of formation values can be found in th

WINSTONCH [101]

<u>Answer:</u> The standard enthalpy change of the reaction is coming out to be -16.3 kJ

<u>Explanation:</u>

Enthalpy change is defined as the difference in enthalpies of all the product and the reactants each multiplied with their respective number of moles. It is represented as $\Delta H$

The equation used to calculate enthalpy change is of a reaction is:

$\Delta H_{rxn}=\sum [n\times \Delta H_f(product)]-\sum [n\times \Delta H_f(reactant)]$

For the given chemical reaction:

$Mg(OH)_2(s)+2HCl(g)\rightarrow MgCl_2(s)+2H_2O(g)$

The equation for the enthalpy change of the above reaction is:

$\Delta H_{rxn}=[(1\times \Delta H_f_{(MgCl_2(s))})+(2\times \Delta H_f_{(H_2O(g))})]-[(1\times \Delta H_f_{(Mg(OH)_2(s))})+(2\times \Delta H_f_{(HCl(g))})]$

We are given:

$\Delta H_f_{(Mg(OH)_2(s))}=-924.5kJ/mol\\\Delta H_f_{(HCl(g))}=-92.30kJ/mol\\\Delta H_f_{(MgCl_2(s))}=-641.8kJ/mol\\\Delta H_f_{(H_2O(g))}=-241.8kJ/mol$

Putting values in above equation, we get:

$\Delta H_{rxn}=[(1\times (-641.8))+(2\times (-241.8))]-[(1\times (-924.5))+(2\times (-92.30))]\\\\\Delta H_{rxn}=-16.3kJ$

Hence, the standard enthalpy change of the reaction is coming out to be -16.3 kJ

6 0

3 years ago

son4ous [18]

Answer: The molar enthalpy change is 73.04 kJ/mol

Explanation:

$HCl+NaOH\rightarrow NaCl+H_2O$

moles of HCl= $molarity\times {\text {vol in L}}=0.415mol/L\times 0.1=0.0415mol$

As NaOH is in excess 0.0415 moles of HCl reacts with 0.0415 moles of NaOH.

volume of water = 100.0 ml + 50.0 ml = 150.0 ml

density of water = 1.0 g/ml

mass of water = $volume \times density=150.0ml\times 1.0g/ml=150.0g$

$q=m\times c\times \Delta T$

q = heat released

m = mass = 150.0 g

c = specific heat = $4.184J/g^0C$

$\Delta T$ = change in temperature = $4.83^0C$

$q=150.0\times 4.184\times 4.83$

$q=3031.3J$

Thus 0.0415 mol of HCl produces heat = 3031.3 J

1 mol of HCL produces heat = $\frac{3031.3}{0.0415}\times 1=73043.3J=73.04kJ$

Thus molar enthalpy change is 73.04 kJ/mol

8 0

3 years ago

Which mixture can be classified as a homogeneous mixture?

Reika [66]

Anything can be homogenous as long as you can only see the same type of liquid
think about it like this
orange juice with pulp is Hetero
orange juice with no pulp is homo

3 0

3 years ago

Read 2 more answers