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xeze [42]
3 years ago
15

The temperature of a system rises by 45°C during a heating process. Express this rise in temperature in Kelvin. (Round the final

answer to the nearest whole number.)
Chemistry
1 answer:
postnew [5]3 years ago
4 0

Answer:

45K

Explanation:

Rise in temperature = Final - initial temperature.

temperature in K = Temperature in Celsius + 273

for Celsius; T2 -T1 =45°C

for kelvin; T2+273 -(T1+273) = ?

                T2+273 -T1-273 =?

                T2-T1 = ?

               T2-T1 =45k

hence ΔT(K) = ΔT(°C) (temperature difference in Celsius is equal to temperature difference in kelvin)

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How many milliliters of a 17% benzalkonium chloride stock solution would be needed to prepare a liter of a 1:200 solution of ben
Nuetrik [128]

Here is the complete question.

Benzalkonium Chloride Solution ------------> 250ml

Make solution such that when 10ml is diluted to a total volume of 1 liter a 1:200 is produced.

Sig: Dilute 10ml to a liter and apply to affected area twice daily

How many milliliters of a 17% benzalkonium chloride stock solution would be needed to prepare a liter of a 1:200 solution of benzalkonium chloride?

(A) 1700 mL

(B) 29.4 mL

(C) 17 mL

(D) 294 mL

Answer:

(B) 29.4 mL

Explanation:

1 L  =   1000 mL

1:200 solution implies the \frac{weight}{volume} in 200 mL solution.

200 mL of solution = 1g of Benzalkonium chloride

1000 mL will be \frac{1000mL}{200mL}=\frac{1g}{xg}

200mL × 1g = 1000 mL × x(g)

x(g) = \frac{200mL*1g}{1000mL}

x(g) = 0.2 g

That is to say, 0.2 g of benzalkonium chloride in 1000mL of diluted solution of 1;200 is also the amount in 10mL of the stock solution to be prepared.

∴ \frac{10mL}{250mL}=\frac{0.2g}{y(g)}

y(g) = \frac{250mL*0.2g}{10mL}

y(g) = 5g of benzalkonium chloride.

Now, at 17% \frac{weight}{volume} concentrate contains 17g/100ml:

∴  the number of milliliters of a 17% benzalkonium chloride stock solution that is needed to prepare a liter of a 1:200 solution of benzalkonium chloride will be;

= \frac{17g}{5g} = \frac{100mL}{z(mL)}

z(mL) = \frac{100mL*5g}{17g}

z(mL) = 29.41176 mL

≅ 29.4 mL

Therefore, there are 29.4 mL of a 17% benzalkonium chloride stock solution that is required to prepare a liter of a 1:200 solution of benzalkonium chloride

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3 years ago
The value of Ka for nitrous acid (HNO2) at 25 ∘C is 4.5×10−4 .a. Write the chemical equation for the equilibrium that correspond
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Answers and Explanation:

a)- The chemical equation for the corresponden equilibrium of Ka1 is:

2. HNO2(aq)⇌H+(aq)+NO−2

Because Ka1 correspond to a dissociation equilibrium. Nitrous acid (HNO₂) losses a proton (H⁺) and gives the monovalent anion NO₂⁻.

b)- The relation between Ka and the free energy change (ΔG) is given by the following equation:

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Where T is the temperature (T= 25ºc= 298 K) and R is the gases constant (8.314 J/K.mol)

At the equilibrium: ΔG=0 and Q= Ka. So, we can calculate ΔGº by introducing the value of Ka:

⇒ 0 = ΔGº + RT ln Ka

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c)- According to the previous demonstation, at equilibrium ΔG= 0.

d)- In a non-equilibrium condition, we have Q which is calculated with the concentrations of products and reactions in a non equilibrium state:

ΔG= ΔGº + RT ln Q

Q= ((H⁺) (NO₂⁻))/(HNO₂)

Q= ( (5.9 10⁻² M) x (6.7 10⁻⁴ M) ) / (0.21 M)

Q= 1.88 10⁻⁴

We know that   ΔGº= 19092.8 J/mol, so:

ΔG= ΔGº + RT ln Q

ΔG= 19092.8 J/mol + (8.314 J/K.mol x 298 K x ln (1.88 10⁻⁴)

ΔG= -2162.4 J/mol

Notice that ΔG<0, so the process is spontaneous in that direction.

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