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3 years ago

5

BONUS QUESTION - It takes 194 mL of a 0.5 M NaOH solution to neutralize 300

1 answer:

Strike441 [17]3 years ago

6 0

Answer:

1.7 mol/dm3

Explanation:

working out in the image

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What is a correct expression of the law of conservation mass

nekit [7.7K]

Law of conservation of mass- The total mass after a chemical reaction is exactly the same as the mass before.

5 0

3 years ago

Molecules can be either positively or negatively charged depending upon their elemental arrangement

Nuetrik [128]

They are positively or negatively charged based on their electrical configuration of electrons*
For example an electronic configuration of 2,8,3 would have a negative charge if +3 since it needs to lose 3 electrons to gain the electrical configuration of a noble gas
2,8,1 would have a charge of +1 for the same reason
2,8,6 would be -2 since it is easier to gain 2 electrons that lose 6 electrons

Hope this helped :))

8 0

3 years ago

When a solution containing 1.4000 g of Ba(NO3)2 and 2.4000 g of HSO3NH2 is boiled, a precipitate forms. One possible identity fo

Georgia [21]

Answer:

See explanation for detailed solution

Explanation:

The balanced reaction equation is Ba(NO3)2 + 2HSO3NH2 → Ba(SO3NH2)2 + 2HNO3

Number of moles of Ba(NO3)2 = 1.4 g/ 261.337 g/mol = 5.36 × 10^-3 moles

From the reaction equation;

1 mole of Ba(NO3)2 yields 1 mole of Ba(SO3NH2)2

5.36 × 10^-3 moles of Ba(NO3)2 yields 5.36 × 10^-3 moles of Ba(SO3NH2)2

For HSO3NH2

Number of moles = 2.4g/97.10 g/mol =0.0247 moles

2 moles of HSO3NH2 yields 1 mole of Ba(SO3NH2)2

0.0247 moles of HSO3NH2 yields 0.0247 ×1/2 = 0.0137 moles

Hence, Ba(NO3)2 is the limiting reactant

The theoretical yield of Ba(SO3NH2)2 is 5.36 × 10^-3 moles × 329.4986 g/mol = 1.766 g

b)

Number of moles = mass/ molar mass

Molar mass = mass/ number of moles

Molar mass = 1.6925 g/5.36 × 10^-3 moles = 315.76 g

3 0

3 years ago

alexandr1967 [171]

Answer:

The products of self-ionization of water are OH⁻ and H⁺.

Explanation:

The water is self ionized according to the equation:

<em>H₂O → OH⁻ + H⁺. </em>

<em></em>

The ionic product for water (Kw) = [OH⁻][H⁺] = 10⁻¹⁴.

Kw is also called "self-ionization constant" or "auto-ionization constant".

6 0

4 years ago

Suppose that 25.0 mL of 0.10 M CH3COOH (aq) is titrated with 0.10 M NaOH (aq). What is the pH after the addition of 10.0 mL of 0

olya-2409 [2.1K]

Answer:

$pH=-1.37$

Explanation:

We are given that 25 mL of 0.10 M $CH_3COOH$ is titrated with 0.10 M NaOH(aq).

We have to find the pH of solution

Volume of $CH_3COOH=25mL=0.025 L$

Volume of NaoH=0.01 L

Volume of solution =25 +10=35 mL= $\frac{35}{1000}=0.035 L$

Because 1 L=1000 mL

Molarity of NaOH=Concentration OH-=0.10M

Concentration of H+= Molarity of $CH_3COOH$ =0.10 M

Number of moles of H+=Molarity multiply by volume of given acid

Number of moles of H+= $0.10\times 0.025$ =0.0025 moles

Number of moles of $OH^-=0.10\times 0.01$ =0.001mole

Number of moles of H+ remaining after adding 10 mL base = 0.0025-0.001=0.0015 moles

Concentration of H+= $\frac{0.0015}{0.035}=4.28\times 10^{-2} m/L$

pH=-log [H+]=-log [4.28 $\times 10^{-2}$ ]=-log4.28+2 log 10=-0.631+2

$pH=-1.37$

6 0

3 years ago