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2 years ago

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Sea arches and sea stacks provide evidence of weathering and erosion. In three to five sentences, explain the roles weathering a

nd erosion play in creating these landforms.(4 points)

1 answer:

Montano1993 [528]2 years ago

4 0

Answer:

?

Explanation:

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Combustion analysis of a 13.42-g sample of the unknown organic compound (which contains only carbon, hydrogen, and oxygen) produ

kirza4 [7]

<u>Answer:</u> The molecular formula for the given organic compound is $C_{18}H_{20}O_2$

<u>Explanation:</u>

The chemical equation for the combustion of hydrocarbon having carbon, hydrogen and oxygen follows:

$C_xH_yO_z+O_2\rightarrow CO_2+H_2O$

where, 'x', 'y' and 'z' are the subscripts of Carbon, hydrogen and oxygen respectively.

We are given:

Mass of $CO_2=39.61g$

Mass of $H_2O=9.01g$

We know that:

Molar mass of carbon dioxide = 44 g/mol

Molar mass of water = 18 g/mol

<u>For calculating the mass of carbon:</u>

In 44 g of carbon dioxide, 12 g of carbon is contained.

So, in 39.61 g of carbon dioxide, $\frac{12}{44}\times 39.61=10.80g$ of carbon will be contained.

<u>For calculating the mass of hydrogen:</u>

In 18 g of water, 2 g of hydrogen is contained.

So, in 9.01 g of water, $\frac{2}{18}\times 9.01=1.00g$ of hydrogen will be contained.

Mass of oxygen in the compound = (13.42) - (10.80 + 1.00) = 1.62 g

To formulate the empirical formula, we need to follow some steps:

<u>Step 1:</u> Converting the given masses into moles.

Moles of Carbon = $\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{10.80g}{12g/mole}=0.9moles$

Moles of Hydrogen = $\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{1g}{1g/mole}=1moles$

Moles of Oxygen = $\frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{1.62g}{16g/mole}=0.10moles$

<u>Step 2:</u> Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 0.10 moles.

For Carbon = $\frac{0.9}{0.10}=9$

For Hydrogen = $\frac{1}{0.10}=10$

For Oxygen = $\frac{0.10}{0.10}=1$

<u>Step 3:</u> Taking the mole ratio as their subscripts.

The ratio of C : H : O = 9 : 10 : 1

Hence, the empirical formula for the given compound is $C_9H_{10}O$

For determining the molecular formula, we need to determine the valency which is multiplied by each element to get the molecular formula.

The equation used to calculate the valency is :

$n=\frac{\text{Molecular mass}}{\text{Empirical mass}}$

We are given:

Mass of molecular formula = 268.34 g/mol

Mass of empirical formula = 134 g/mol

Putting values in above equation, we get:

$n=\frac{268.34g/mol}{134g/mol}=2$

Multiplying this valency by the subscript of every element of empirical formula, we get:

$C_{(9\times 2)}H_{(10\times 2)}O_{(1\times 2)}=C_{18}H_{20}O_2$

Thus, the molecular formula for the given organic compound is $C_{18}H_{20}O_2$ .

3 0

2 years ago

Consider a general reaction A ( aq ) enzyme ⇌ B ( aq ) A(aq)⇌enzymeB(aq) The Δ G ° ′ ΔG°′ of the reaction is − 5.980 kJ ⋅ mol −

Grace [21]

Answer : The value of $K_{eq}$ is, 11.2

The value of $\Delta G_{rxn}$ is -9.04 kJ/mol

Explanation :

The relation between the equilibrium constant and standard Gibbs free energy is:

$\Delta G^o=-RT\times \ln K_{eq}$

where,

$\Delta G^o$ = standard Gibbs free energy = -5.980 kJ/mol = -5980 J/mol

R = gas constant = 8.314 J/K.mol

T = temperature = $25^oC=273+25=298K$

$K_{eq}$ = equilibrium constant = ?

Now put all the given values in the above formula, we get:

$\Delta G^o=-RT\times \ln K_{eq}$

$-5980J/mol=-(8.314J/K.mol)\times (298K)\times \ln K_{eq}$

$K_{eq}=11.2$

Thus, the value of $K_{eq}$ is, 11.2

Now we have to calculate the $\Delta G_{rxn}$ .

The formula used for $\Delta G_{rxn}$ is:

The given reaction is:

$A(aq)\rightleftharpoons B(aq)$

$\Delta G_{rxn}=\Delta G^o+RT\ln Q$

$\Delta G_{rxn}=\Delta G^o+RT\ln \frac{[B]}{[A]}$ ............(1)

where,

$\Delta G_{rxn}$ = Gibbs free energy for the reaction = ?

$\Delta G_^o$ = standard Gibbs free energy = -30.5 kJ/mol

R = gas constant = $8.314\times 10^{-3}kJ/mole.K$

T = temperature = $37.0^oC=273+37.0=310K$

Q = reaction quotient

[A] = concentration of A = 1.8 M

[B] = concentration of B = 0.55 M

Now put all the given values in the above formula 1, we get:

$\Delta G_{rxn}=(-5980J/mol)+[(8.314J/mole.K)\times (310K)\times \ln (\frac{0.55}{1.8})$

$\Delta G_{rxn}=-9035.75J/mol=-9.04kJ/mol$

Therefore, the value of $\Delta G_{rxn}$ is -9.04 kJ/mol

3 0

2 years ago

Describe the difference in structure between beH2 and CaH2

katen-ka-za [31]

Hello!

BeH₂ is a linear molecule, while CaH₂ is an angular molecule.

The difference between these two molecules is given by the number of electrons they have. Be is in the 2nd period of the Periodic table, and the ion Be⁺² doesn't have any free electron pairs when bonding to H. Ca is in the 4 period of the periodic table, meaning that it has more electrons, and the ion Ca⁺² has two free electron pairs when bonding to H that makes the molecule angular by pushing the bonds at an angle by sterical hindrance.

Have a nice day!

7 0

2 years ago

The researcher performed a follow-up experiment to measure the rate of oxygen consumption by muscle and brain cells. Predict the

Naddik [55]

Answer:

Mitochondria are abundantly present in mammalian cells. Their fraction varies from tissue to tissue, ranging from <1% (volume) in white blood cells to 35% in heart muscle cells. However, mitochondria should not be thought of as single entities, but rather a dynamic network that continuously undergoes fission and fusion processes. In skeletal muscle, mitochondria exist as a reticular membrane network. The subsarcolemmal (SS) and intermyofibrillar (IMF) mitochondria are located in distinct subcellular regions, and they possess subtle differences in biochemical and functional properties that are characterized by their anatomical locations. SS mitochondria lie directly beneath the sarcolemmal membrane and the IMF mitochondria are located in close contact with the myofibril. Their different properties are likely to influence their capacity for adaptation. SS mitochondria account for 10-15% of the mitochondrial volume and this population has been shown to be more susceptible to adaptation than the IMF mitochondria. However, the IMF mitochondria were found to have higher rates of protein synthesises, enzyme activities and respiration (1).

Explanation:

0 0

2 years ago

Read 2 more answers

Compared to micellular Compound 1, Compound 2 is structurally more rigid as a result of what type of interaction?

zimovet [89]

Answer:

D. Intramolecular covalent bond

Explanation:

Compound D is structurally more rigid as a result of intramolecular covalent bonding. The forces that hold together atoms within a compound are greater as compared to forces holding two molecules together (intermolecular bonding). On the other hand Hydrogen bonds are weaker as compared to covalent bonds. Covalent bonds involve the sharing of electrons between two atoms and Hydrogen bonds are formed between a highly electronegative atom like oxygen, Flourine,Chlorine to hydrogen.

8 0

2 years ago