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2 years ago

The ratio of oxygen-16 and oxygen-18 isotopes in plankton fossils in deep-sea sediments can be used to determine ________.

2 answers:

3 0

The ratio of oxygen-16 and oxygen-18 isotopes in plankton fossils in deep-sea sediments can be used to determine past temperatures.

kolbaska11 [484]2 years ago

3 0

Answer;

-past temperatures

The ratio of oxygen-16 and oxygen-18 isotopes in plankton fossils in deep-sea sediments can be used to determine past temperatures.

Explanation;

-O-16 will evaporate more readily than O-18 since it is lighter, therefore; during a warm period, the relative amount of O-18 will increase in the ocean waters since more of the O-16 is evaporating.

-Hence, looking at the ratio of O16 to O18 in the past can give clues about global temperatures.

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A boy kicks a ball with a force of 40 N. At exactly the same moment, a gust of wind blows in the opposite direction of the kick

aliya0001 [1]

B, the opposing forces are the same, thus, the ball doesn't move back or forward.

8 0

3 years ago

Read 2 more answers

1. An oxide of chromium is found to have the following % composition: 68.4% Cr

abruzzese [7]

Answer:

Empirical formula is Cr₂O₃.

Explanation:

Given data:

Percentage of Cr = 68.4%

Percentage of O = 31.6%

Empirical formula = ?

Solution:

Number of gram atoms of Cr = 68.4 / 52 = 1.3 2

Number of gram atoms of O = 31.6 / 16 = 1.98

Atomic ratio:

Cr : O

1.32/1.32 : 1.98/1.32

1 : 1.5

Cr : O = 1 : 1.5

Cr : O = 2(1 : 1.5)

Empirical formula is Cr₂O₃.

6 0

3 years ago

One gram of liquid benzene is burned in a bomb calorimeter. The temperature before ignition was 20.826 C, and the temperature af

Licemer1 [7]

Answer:

fH = - 3,255.7 kJ/mol

Explanation:

Because the bomb calorimeter is adiabatic (q =0), there'is no heat inside or outside it, so the heat flow from the combustion plus the heat flow of the system (bomb, water, and the contents) must be 0.

Qsystem + Qcombustion = 0

Qsystem = heat capacity*ΔT

10000*(25.000 - 20.826) + Qc = 0

Qcombustion = - 41,740 J = - 41.74 kJ

So, the enthaply of formation of benzene (fH) at 298.15 K (25.000 ºC) is the heat of the combustion, divided by the number of moles of it. The molar mass od benzene is: 6x12 g/mol of C + 6x1 g/mol of H = 78 g/mol, and:

n = mass/molar mass = 1/ 78

n = 0.01282 mol

fH = -41.74/0.01282

fH = - 3,255.7 kJ/mol

4 0

3 years ago

A tank at is filled with of sulfur tetrafluoride gas and of sulfur hexafluoride gas. You can assume both gases behave as ideal g

dangina [55]

The question is incomplete, the complete question is:

A 7.00 L tank at $21.4^oC$ is filled with 5.43 g of sulfur hexafluoride gas and 14.2 g of sulfur tetrafluoride gas. You can assume both gases behave as ideal gases under these conditions. Calculate the mole fraction and partial pressure of each gas. Round each of your answers to significant digits.

Answer: The mole fraction of sulfur hexafluoride is 0.221 and that of sulfur tetrafluoride is 0.779

Explanation:

The number of moles is defined as the ratio of the mass of a substance to its molar mass. The equation used is:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ ......(1)

For sulfur hexafluoride:

Given mass of sulfur hexafluoride = 5.43 g

Molar mass of sulfur hexafluoride = 146.06 g/mol

Putting values in equation 1, we get:

$\text{Moles of sulfur hexafluoride}=\frac{5.43g}{146.06g/mol}=0.0372mol$

For sulfur tetrafluoride:

Given mass of sulfur tetrafluoride = 14.2 g

Molar mass of sulfur tetrafluoride = 108.07 g/mol

Putting values in equation 1, we get:

$\text{Moles of sulfur tetrafluoride }=\frac{14.2g}{108.07g/mol}=0.1314mol$

Total moles of gas in the tank = [0.0372+ 0.1314] mol = 0.1686 mol

Mole fraction is defined as the moles of a component present in the total moles of a solution. It is given by the equation:

$\chi_A=\frac{n_A}{n_A+n_B}$ .....(2)

where n is the number of moles

Putting values in equation 2, we get:

$\chi_{SF_6}=\frac{0.0372}{0.1686}=0.221$

$\chi_{SF_4}=\frac{0.1314}{0.1686}=0.779$

Hence, the mole fraction of sulfur hexafluoride is 0.221 and that of sulfur tetrafluoride is 0.779

7 0

3 years ago

effect. RbClO(s) → Rb+(aq) + ClO−(aq) HClO(aq) + H2O(l) equilibrium reaction arrow H3O+(aq) + ClO−(aq) The degree of dissociatio

Lemur [1.5K]

Explanation:

Common ion effect is defined as the effect which occurs on equilibrium when a common ion (an ion which is already present in the solution) is added to a solution. This effect generally decreases the solubility of a solute.

Equilibrium reaction of strontium sulfate and sodium sulfate follows the equation:

$RbClO(s)\rightarrow Rb^{+}(aq.)+ClO^{-}(aq.)$

$HClO(aq)+H_2O\rightleftharpoons H_3O^+(aq.)+ClO^{-}(aq.)$

According to Le-Chateliers principle: If there is any change in the variables of the reaction, the equilibrium will shift in the direction in order to minimize the effect.

In the equilibrium reactions, hypochlorite ion is getting increased on the product side, so the equilibrium will shift in the direction to minimize this effect, which is in the direction of hydrogen hypochlorite.

Thus, the addition hypochlorite ions will shift the equilibrium in the left direction.

The dissociation of hydrogen hypochlorite is suppressed due to the common ion effect.

8 0

3 years ago