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3 years ago

Layers of rock that bend can produce a downward fold known as a(n)

2 answers:

8 0

Layers of rock that bend can produce a downward fold known as a syncline.

This is the type of fold that occurs with the younger layers of rocks remaining closer to the center of the structure. Synclines can be recognized by the fact that the youngest layers of rocks always remain near the folds center.

Hope this helped..

STatiana [176]3 years ago

3 0

Answer:

Syncline

Explanation:

Synclines are the down folds that are constructed due to the compressive forces acting on a layer from both sides. It usually occurs in size ranging from a microscopic scale to a large megascopic scale. The oldest rocks are present at its outermost limb and the rocks become gradually younger toward its hinge area.

In an outcrop, the syncline and the anticline, both form simultaneously. The limbs that dip towards the trough of the fold indicate a syncline whereas the limbs that dips towards the crest are the anticline. The variation in the shape of these foldings (anticline and syncline) depends upon the intensity of the force.

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A chemist uses hot hydrogen gas to convert chromium(III) oxide to

Darya [45]

Answer: 3.024 g grams of hydrogen are needed to convert 76 grams of chromium(III) oxide, $Cr_{2}O_{3}$

Explanation:

The reaction equation for given reaction is as follows.

$Cr_{2}O_{3} + 3H_{2} \rightarrow 2Cr + 3H_{2}O$

Here, 1 mole of $Cr_{2}O_{3}$ reacts with 3 moles of $H_{2}$ .

As mass of chromium (III) oxide is given as 76 g and molar mass of chromium (III) oxide $(Cr_{2}O_{3})$ is 152 g/mol.

Number of moles is the mass of substance divided by its molar mass. So, moles of $Cr_{2}O_{3}$ is calculated as follows.

$No. of moles = \frac{mass}{molar mass}\\= \frac{76 g}{152 g/mol}\\= 0.5 mol$

Now, moles of $H_{2}$ .given by 0.5 mol of $Cr_{2}O_{3}$ is calculated as follows.

$0.5 mol Cr_{2}O_{3} \times \frac{3 mol H_{2}}{1 mol Cr_{2}O_{3}}\\= 1.5 mol H_{2}$

As molar mass of $H_{2}$ is 2.016 g/mol. Therefore, mass of $H_{2}$ is calculated as follows.

$No. of moles = \frac{mass}{molar mass}\\1.5 mol = \frac{mass}{2.016 g/mol}\\mass = 3.024 g$

Thus, we can conclude that 3.024 g grams of hydrogen are needed to convert 76 grams of chromium(III) oxide, $Cr_{2}O_{3}$ .

7 0

3 years ago

Gastric juice is made up of substances secreted from parietal cells, chief cells, and mucous-secreting cells. The cells secrete

neonofarm [45]

Answer:

The amount of energy required to transport hydrogen ions from a cell into the stomach is 37.26KJ/mol.

Explanation:

The free change for the process can be written in terms of its equilibrium constant as:

ΔG° = $-RTInK_(eq)$

where:

R= universal gas constant

T= temperature

$K_eq$ = equilibrum constant for the process

Similarly, free energy change and cell potentia; are related to each other as follows;

ΔG= -nFE°

from above;

F = faraday's constant

n = number of electrons exchanged in the process; and

E = standard cell potential

∴ The amount of energy required for transport of hydrogen ions from a cell into stomach lumen can be calculated as:

ΔG° = $-RTInK_(eq)$

where;

[texK_eq[/tex]= $\frac{[H^+]_(cell)}{[H^+(stomach lumen)]}$

For transport of ions to an internal pH of 7.4, the transport taking place can be given as:

$H^+_{inside}$ ⇒ $H^+_{outside}$

Equilibrum constant for the transport is given as:

$K_{eq}=\frac{[H^+]_{outside}}{[H^+]_{inside}}$

$=\frac{[H^+]_{cell}}{[H^+]_{stomach lumen}}$

$[H^+]_{cell}$ = 10⁻⁷⁴

=3.98 * 10⁻⁸M

$[H^+]_{stomach lumen}$ = 10⁻²¹

=7.94 * 10⁻³M

Hence;

$K_{eq}=\frac{[H^+]_{cell}}{[H^+]_{stomachlumen}}$

= $\frac{3.98*10^{-8}}{7.94*10{-3}}$

= 5.012 × 10⁻⁶

Furthermore, free energy change for this reaction is related to the equilibrium concentration given as:

ΔG° = $-RTInK_(eq)$

If temperature T= 37° C ; in kelvin

=37° C + 273.15K

=310.15K; and

R-= 8.314 j/mol/k

substituting the values into the equation we have;

ΔG₁ = $-(8.314J/mol/K)(310.15)TIn(5.0126*10^{-6})$

= 31467.93Jmol⁻¹

≅ 31.47KJmol⁻¹

If the potential difference across the cell membrane= 60.0mV.

Energy required to cross the cell membrane will be:

ΔG₂ = $-nFE°_{membrane}$

ΔG₂ = $-(1 mol)(96.5KJ/mol/V)(60*10^{-3})$

= 5.79KJ

Therefore, for one mole of electron transfer across the membrane; the energy required is 5.79KJmol⁻¹

Now, we can calculate the total amount of energyy required to transport H⁺ ions across the membrane:

Δ $G_{total} = G_{1}+G_{2}$

= (31.47+5.79) KJmol⁻¹

= 37.26KJmol⁻¹

We can therefore conclude that;

The amount of energy required to transport ions from cell to stomach lumen is 37.26KJmol⁻¹

5 0

3 years ago

2. What is the weight of the hydrochloric acid that fills a 144.5 mL container? The density of

hoa [83]

I don’t understand the question sorry

8 0

3 years ago

What is the molecular formula of a hydrocarbon with m+ = 78?

fiasKO [112]

A molecular formula represents the exact number of atoms present for each element in the compound.

For carbon atom: $\frac{78}{12}$ = $6\frac{6}{12}$ ( as molar mass of carbon is 12 g/mol)

Now, 6 carbon atoms are present and rest are hydrogen atoms i.e. 6

Thus, formula becomes $C_{6}H_{6}$ ( $C_{x}H_{y})$

Now, check for unsaturation:

Degree of unsaturation = $\frac{2x+2-y}{2}$

Substitute the value of x and y,

Degree of unsaturation = $\frac{2(6)+2-6}{2}$

= $\frac{12-4}{2}$

= $4$ implies one ring and three double bonds.

Thus, formula comes out to be $C_{6}H_{6}$ i.e. benzene ring.

4 0

4 years ago

How many O atoms are in 1.25 mol of SO2

Alex73 [517]

The number of O atoms that ate in 1.25 mol of SO2 is 2.5 moles

calculation

moles of O= total number O atoms in SO2 x 1.25 moles

The number of O atoms in SO2 = 1 x2 = 2 atoms

moles is therefore= 2 x 1.25 moles = 2.5 moles

4 0

3 years ago