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2 years ago

How many minutes are in 230 seconds?

Chemistry

2 answers:

DIA [1.3K]2 years ago

8 0

Answer:

230 Seconds = 3.83333333 Minutes

Explanation:

AlexFokin [52]2 years ago

7 0

Answer:

3 minutes and 50 seconds

Explanation:

If there are 60 seconds in 1 minute..

60, 120, 180 (1 minute, 2 minutes, 3 minutes)

230 - 180 = 50

So it's 230 seconds consists of 3 minutes and 50 seconds.

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⦁answer Calculate the density in g/L of 478 mL of krypton at 47° C and 671 mm Hg. ⦁ Determine the molar mass of a gas that has a

STALIN [3.7K]

Answer:

The correct answers are:

- Krypton: density= 2.8 g/L

- Molar Mass= 63.99 g/mol

- Mass of O₂= 15.29 g

Explanation:

The general equation of an ideal gas is the folllowing:

P x V = n x R x T

Where: P= pressure (in atm), V= volume; n= number of moles, R= gas constant (0,082 L.atm/K.mol) and T= temperature (in K).

For krypton:

P= 671 mmHg = 0,882 atm

V= 478 ml x 1000 ml/1 L= 0,478 L

T= 47ºC= 320 K

MM= 83.8 g/mol (from Periodic Table, Kr is an inert gas so it is a monoatomic gas)

P x V = n x R x T

Since the number of moles of a compound can be calculated by dividing the mass of compound (m) into its molar mass (MM):

n= m/MM

We can replace the expression in the first equation to obtain:

$P x V= \frac{m}{MM} x R x T$

m/V= $\frac{P x MM}{R x T}$

Density (d) is equal to the mass per volume (m/V), so we can directly calculate the density:

d= m/V= \frac{P x MM}{R x T}=

= (0.882 atm x 83.8 g/mol)/(0.082 L.atm/K.mol x 320 K)

= 2.81 g/L

For the gas:

d= 2.18 g/L

T= 66ºC= 339 K

P= 720 mmHg= 0.947 atm

d= \frac{P x MM}{R x T}

⇒MM = $\frac{dx R x T}{P}$

= (2.18 g/L x 0.082 L.atm/K.mol x 339 K)/(0.947 atm)

= 63.99 g/mol ≅ 64 g/mol

For the O₂:

V= 5.60 L

P= 1.75 atm

T= 250 K

MM(O₂) = 2 x Atomic Mass O= 2 x 16 g/mol= 32 g/mol

We can use the second equation:

P x V= \frac{m}{MM} x R x T

⇒ m = $\frac{P x V x MM}{R x T}$ = (1.75 atm x 5.6 L x 32 g/mol)/(0.082 L.atm/K.mol x 250 K)

= 15.29 g≅ 16 g

4 0

3 years ago

Read 2 more answers

Consider the statement, “In a chemical reaction, atoms are neither created nor destroyed, only rearranged.” Why is this a statem

Karo-lina-s [1.5K]

Answer: The law of conservation of mass is reffering to the fact that energy cannot be created or destroyed. So when you speak about atoms not being created or destroyed it is the same thing. Unless you're talking about an atomic bomb where the atoms are split.

7 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, m.

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:

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

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