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nikitadnepr [17]
3 years ago
5

How many cm is equal to 1450 nm?

Chemistry
1 answer:
grigory [225]3 years ago
3 0

Answer:

The answer is D; 1.45 x 10^-4

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if a gas sample in a balloon occupies 1.5 L at atmospheric pressure, what would be the pressure (in mmHg) if the volume was redu
Nata [24]

Answer:

1425 mmHg.

Explanation:

The following data were obtained from the question:

Initial volume (V1) = 1.5 L

Initial pressure (P1) = 1 atm

Final volume (V2) = 0.8 L

Final pressure (P2) =?

Next, we shall determine the final pressure of the gas by using the Boyle's law equation as follow:

P1V1 = P2V2

1 × 1.5 = P2 × 0.8

1.5 = P2 × 0.8

Divide both side by 0.8

P2 = 1.5/0.8

P2 = 1.875 atm

Finally, we shall convert 1.875 atm to mmHg.

This can be obtained as follow:

1 atm = 760 mmHg

Therefore,

1.875 atm = 1.875 × 760 = 1425 mmHg.

Therefore, the new pressure of the gas is 1425 mmHg.

3 0
3 years ago
i am begging anyone to help me with this! (all tutors i've asked said they can't solve it but i need someone to help me out) - i
9966 [12]

First, we need to calculate how much energy we will get from this combustion.

Assuming the combustion is complete, we have the octane reacting with O₂ to form only water and CO₂, so:

C_8H_{18}+O_2\to CO_2+H_2O

We need to balance the reaction. Carbon only appear on two parts, so, we can start by it:

C_8H_{18}+O_2\to8CO_2+H_2O

Now, we balance the hydrogen:

C_8H_{18}+O_2\to8CO_2+9H_2O

And in the end, the oxygen:

C_8H_{18}+\frac{25}{2}O_2\to8CO_2+9H_2O

We can multiply all coefficients by 2 to get integer ones:

2C_8H_{18}+25O_2\to16CO_2+18H_2O

Now, we need to use the enthalpies of formation to get the enthalpy of reaction of this reaction.

The enthalpy of reaction can be calculated by adding the enthalpies of formation of the products multiplied by their stoichiometric coefficients and substracting the sum of enthalpies of formation of the reactants multiplied by their stoichiometric coefficients.

For the reactants, we have (the enthalpy of formation of pure compounds is zero, which is the case for O₂):

\begin{gathered} \Delta H\mleft\lbrace reactants\mright\rbrace=2\cdot\Delta H\mleft\lbrace C_8H_{18}\mright\rbrace+25\cdot\Delta H\mleft\lbrace O_2\mright\rbrace \\ \Delta H\lbrace reactants\rbrace=2\cdot(-250.1kJ)+25\cdot0kJ \\ \Delta H\lbrace reactants\rbrace=-500.2kJ+0kJ \\ \Delta H\lbrace reactants\rbrace=-500.2kJ \end{gathered}

For the products, we have:

\begin{gathered} \Delta H_{}\mleft\lbrace product\mright\rbrace=16\cdot\Delta H\lbrace CO_2\rbrace+18\cdot\Delta H\lbrace H_2O\rbrace \\ \Delta H_{}\lbrace product\rbrace=16\cdot(-393.5kJ)+18\cdot(-285.5kJ) \\ \Delta H_{}\lbrace product\rbrace=-6296kJ-5139kJ \\ \Delta H_{}\lbrace product\rbrace=-11435kJ \end{gathered}

Now, we substract the rectants from the produtcs:

\begin{gathered} \Delta H_r=\Delta H_{}\lbrace product\rbrace-\Delta H\lbrace reactants\rbrace \\ \Delta H_r=-11435kJ-(-500.2kJ) \\ \Delta H_r=-10934.8kJ \end{gathered}

Now, this enthalpy of reaction is for 2 moles of C₈H₁₈, so for 1 mol of C₈H₁₈ we have half this value:

\Delta H_c=\frac{1}{2}\Delta H_r=\frac{1}{2}\cdot(-10934.8kJ)=-5467.4kJ

Now, we have 100 g of C₈H₁₈, and its molar weight is approximately 114.22852 g/mol, so the number of moles in 100 g of C₈H₁₈ is:

\begin{gathered} M_{C_8H_{18}}=\frac{m_{C_8H_{18}}}{n_{C_8H_{18}}} \\ n_{C_8H_{18}}=\frac{m_{C_8H_{18}}}{M_{C_8H_{18}}}=\frac{100g}{114.22852g/mol}\approx0.875438mol \end{gathered}

Since we have approximately 0.875438 mol, and 1 mol releases -5467.4kJ when combusted, we have:

Q=-5467.4kJ/mol\cdot0.875438mol\approx-4786.37kJ

Now, for the other part, we need to calculate how much heat it is necessary to melt a mass, <em>m</em>.

First, we have to heat the ice to 0 °C, so:

\begin{gathered} Q_1=m\cdot2.010J/g.\degree C\cdot(0-(-10))\degree C \\ Q_1=m\cdot2.010J/g\cdot10 \\ Q_1=m\cdot20.10J/g \end{gathered}

Then, we need to melt all this mass, so we use the latent heat now:

Q_2=n\cdot6.03kJ/mol

Converting mass to number of moles of water we have:

\begin{gathered} M=\frac{m}{n} \\ n=\frac{m}{M}=\frac{m}{18.01528g/mol} \end{gathered}

So:

Q_2=\frac{m}{18.01528g/mol}_{}\cdot6.03kJ/mol\approx m\cdot0.334716kJ/g

Adding them, we have a total heat of:

\begin{gathered} Q_T=m\cdot20.10J/g+m\cdot0.334716kJ/g \\ Q_T=m\cdot0.02010kJ/g+m\cdot0.334716kJ/g \\ Q_T=m\cdot0.354816kJ/g \end{gathered}

Since we have a heat of 4786.37 kJ form the combustion, we input that to get the mass (the negative sign is removed because it only means that the heat is released from the reaction, but now it is absorbed by the ice):

\begin{gathered} 4786.37kJ=m\cdot0.354816kJ/g \\ m=\frac{4786.37kJ}{0.354816kJ/g}\approx13489g\approx13.5\operatorname{kg} \end{gathered}

Since we have a total of 20kg of ice, we can clculate the percent using it:

P=\frac{13.5\operatorname{kg}}{20\operatorname{kg}}=0.675=67.5\%

5 0
1 year ago
What is 0.738 × 104? Enter your answer in the box.<br> _____
lisov135 [29]
The answer would be 76.752 .
4 0
3 years ago
Select the missing words to complete the definition of buffer capacity. Buffer capacity is the _____________ of acid or base a b
horsena [70]

Answer:

amount, pH value.

Explanation:

The buffer range is the pH range in which the buffer performs optimally, i.e., neutralizes even when a strong acid or base is introduced to it and resists any major change in its pH value.

The buffer capacity is the amount of acid or base that can be added before the pH of the buffer solution changes significantly.

Thus, the final statement becomes,

Buffer capacity is the amount of acid or base a buffer can handle before pushing the pH value outside of the buffer range.

8 0
3 years ago
How do you use the changes in the phase of water to keep you cool?
sineoko [7]

Answer:

by freezing

Explanation:

8 0
3 years ago
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