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Neko [114]
2 years ago
7

6) A 0.20 ml CO2 bubble in a cake batter is at 27°C. In the oven it gets

Chemistry
1 answer:
Nata [24]2 years ago
5 0

Answer: The new volume of cake is 1.31 mL.

Explanation:

Given: V_{1} = 0.20 mL,         T_{1} = 27^{o}C

V_{2} = ?,                    T_{2} = 177^{o}C

Formula used to calculate new volume is as follows.

\frac{V_{1}}{T_{1}} = \frac{V_{2}}{T_{2}}

Substitute the values into above formula as follows.

\frac{V_{1}}{T_{1}} = \frac{V_{2}}{T_{2}}\\\frac{0.20 mL}{27^{o}C} = \frac{V_{2}}{177^{o}C}\\V_{2} = 1.31 mL

Thus, we can conclude that the new volume of cake is 1.31 mL.

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matrenka [14]
The  ratio  of  the diffusion  rate  of Cl2  and   O2   is 1.5

    calculation
rate  of diffusion Cl2/ rate   of  diffusion  O2 =√  molar mass of Cl2/ molar mass  of O2

molar  mass of Cl2  = 35.5 x2 = 71  g/mol
molar  mass O2  = 16 x2 =32g/mol

that  is    rate  of diffusion Cl2/ rate  of  diffusion of O2  =√ 71/ 32=  1.5
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A 600.0 mL sample of 0.20 MHF is titrated with 0.10 MNaOH. Determine the pH of the solution after the addition of 600.0 mL of Na
Leona [35]

Answer: pH=12.69

Explanation:

{\text{Moles of HF}=Molarity\times {\text{Volume of solution in liters}}

{\text{Moles of HF}=0.20M\times 0.6L=0.12 moles

HF\rightarrow H^++F^-

Initial 0.12               0       0

Eqm   0.12-x           x        x

K_a=\frac{[H^+][F^-]}{[HF]}

3.5\times 10^{-4}=\frac{x^2}{0.12-x}  

(neglecting small value of x in comparison to 0.12)

x=4.2\times 10^{-5}

Moles of H^+=4.2\times 10^{-5}

NaOH\rightarrow Na^++OH^-

{\text{Molesof NaOH}}=Molarity\times {\text{Volume of solution in liters}}

{\text{Moles of NaOH}}=0.10M\times 0.6L=0.06 moles

0.06 moles of NaOH will give 0.06 moles of [OH^-]

Now 4.2\times 10^{-5} moles of OH^- will be neutralized by 4.2\times 10^{-5} moles of H^+ and (0.06-4.2\times 10^{-5})=0.059 moles of OH^- will be left.

Molarity of OH^-=\frac{0.059moles}{1.2L}=0.049M

pOH=-\log[OH^-]=-\log[0.049]=1.31

pH = 14 - pOH= 14 - 1.31 = 12.69

5 0
2 years ago
A 103.8g sample of nitric acid solution that is 70.0% HNO3 contains by mass
soldi70 [24.7K]
w/w percentage <span>
               = mass of the pure compound / total mass of the sample x 100%

70% HNO₃ contains by mass means every 100 g of sample has 70 g of HNO₃.</span><span>

The mass of solution = 103.8 g
Hence the mass of HNO₃ = 103.8 g x 70%</span><span>
                                         = 103.8 g x (70 / 100)
<span>                                         = 72.66 g = 72.7 g.</span></span>
3 0
2 years ago
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