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2 years ago

Carbon dioxide, water vapor, and oxygen pass through ---------------

Chemistry

2 answers:

Lelechka [254]2 years ago

5 0

The answer is B) stomata.

rjkz [21]2 years ago

3 0

Answer is B

Stomata allows gases such as carbon dioxide, water vapour and oxygen to diffuse into and out of the internal tissues of the plant.

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5) Calculate the molality of 0.210 mol of KBr dissolved in 0.075kg pure water?

Margaret [11]

Answer:

$\boxed {\boxed {\sf 2.8 \ m }}$

Explanation:

The formula for molality is:

$m=\frac{moles \ of \ solute}{kg \ of \ solvent}$

There are 0.210 moles of KBr and 0.075 kilograms of pure water.

$moles= 0.210 \ mol \\kilograms = 0.075 \ kg$

Substitute the values into the formula.

$m= \frac{ 0.210 \ mol }{0.075 \ kg}$

Divide.

$m= 2.8 \ mol/kg= 2.8 \ m$

The molality is 2.8 moles per kilogram

5 0

2 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:

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

5 0

10 months ago

What is the purpose of doing a cell culture lab?

Anuta_ua [19.1K]

Answer:

Cell culture is one of the most important techniques in cellular and molecular biology since it provides a platform to investigate the biology, biochemistry, physiology (e.g., aging) and metabolism of wild-type cells and diseased cells.

8 0

2 years ago

How many atoms in 5.01 mole of lead?

leva [86]

Answer:

30.17 × 10²³ atoms

Explanation:

Given data:

Number of moles of lead = 5.01 mol

Number of atoms = ?

Solution:

Avogadro number:

The given problem will solve by using Avogadro number.

It is the number of atoms , ions and molecules in one gram atom of element, one gram molecules of compound and one gram ions of a substance.

The number 6.022 × 10²³ is called Avogadro number.

For example,

18 g of water = 1 mole = 6.022 × 10²³ molecules of water

1.008 g of hydrogen = 1 mole = 6.022 × 10²³ atoms of hydrogen

In given question:

1 mole = 6.022 × 10²³ atoms

5.01 mol × 6.022 × 10²³ atoms / 1 mol

30.17 × 10²³ atoms

3 0

2 years ago

Relating ideas explain why ionic crystals melt at much higher temperatures than typical covalent molecular crystals.

gayaneshka [121]

Because of the way the temp is set up

6 0

3 years ago