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

Answers

Answer 1
Answer: The ratio of oxygen-16 and oxygen-18 isotopes in plankton fossils in deep-sea sediments can be used to determine past temperatures.
Answer 2
Answer:

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.


Related Questions

Help me question 5 ASAP
Which one of these could be in the unknown anion "X" in this acid: H3XcarbonatefluoratenitrogennitriteCould you explain how to find this? The process?
Why is it important to keep your apparatus dry what reaction will occur between the grignard reagent and water?
For the following reaction at equilibrium, which gives a change that will shift the position of equilibrium to favor formation of more products? 2NOBr(g) 2NO(g) + Br 2(g), ΔHº rxn = 30 kJ/mol
Draw 1,2,3,4,5,6-hexachlorocyclohexane with :a. all the chloro groups in axial positions. b. all the chloro groups in equatorial positions.

A chemist prepares a solution of potassium permanganate (KMnO4) by measuring out 3.8 umol of potassium permanganate into a 100 mL volumetric flask and filling the flask to the mark with water. Calculate the concentration in mol/L of the chemist's potassium permanganate solution. Round your answer to 2 significant digits. x 5 ? Explanation Check

Answers

Answer:

3,8×10⁻⁵ mol/L of potassium permanganate solution

Explanation:

To calculate concentration in mol/L you must convert the 3,8 umol to moles and 100 mL to liters, knowing 1 umol are 1×10⁻⁶mol and 1L are 1000 mL.

3,8 umol × (1×10⁻⁶mol / 1 umol ) = 3,8×10⁻⁶mol of potassium permanganate.

100 mL × ( 1L / 1000 mL) = 0,100 L

Thus, concentration in mol/L is:

3,8×10⁻⁶mol / 0,100 L = 3,8×10⁻⁵ mol/L of potassium permanganate solution

I hope it helps!

How many formula units are in a 7.3 x 10-3 gram sample of lithium chloride, LiCI?

Answers

Answer: 1.0 * 10^(20) formula units

Explanation:

so lets start off by looking at what we have. we have 7.3 * 10^(-3) g of LiCl

which is called Lithium Chloride. in order to convert g to moles, we divide the g by the molar mass of Lithium Chloride.

whip out that HANDY DANDY PERIODIC TABLE man the PERIODIC TABLE WILL SAVE YOUR LIFE SOME DAY! someone will walk up to you all mean, and youll be like, "what, you tryna MUG me?" and then you whack 'em with the periodic table like BAM! GOTTEM!

okay so lets look at the periodic table and we notice that the atomic mass of Lithium is 6.941 and the atomic mass of Chlorine is 35.453. notice that in LiCl there is only one of each. so lets add 6.941 + 35.453 = 42.394 g/mol.

now look at what we were given: converting the given quantity to standard format instead of scientific format, we have 0.0073 grams of lithium chloride. we can convert this to moles by dividing it by its molar mass which is 42.394. 0.0073 / 42.394 = 1.72 * 10^(-4)

now lets use AVOCADOS NUMBER i mean AVOGADROS NUMBER!! which is 6.02 * 10^(23)

multiply (1.72 * 10^(-4)) * (6.02 * 10^(23)) and we get 10.35 * 10^(19) formula units.

if you want to be specific about the significant figures, notice that the given quantity in the question only has two significant figures. so we can alter our final answer to only have two sig figs. lets change it: 1.0 * 10^(20) formula units

Convert 50g of calcium carbonate, CaCO3, into moles

Answers

Answer:
            Moles  =  0.5 mol

Solution:

Moles is related to mass as follow,

                                       Moles  =  Mass / M.mass   ----- (1)

Where;
           Mass  =  50 g

           M.mass  =  Ca (40) + C (12) + O₃ (16)₃  =  100 g/mol

Putting values in equation 1,

                                        Moles  =  50 g ÷ 100 g.mol⁻¹

                                        Moles  =  0.5 mol

Given that e = 9.0 v , r = 98 ω and c = 23 μf , how much charge is on the capacitor at time t = 4.0 ms

Answers

Let charge across the capacitor be Q, current through the circuit be I.
Voltage difference across the resistor = rI
Voltage difference across the capacitor = Q/c
Loop rule: net voltage change through a loop must be zero, so
9 = rI + Q/c. Since I = dQ/dt,
r dQ/dt + Q/c = 9
Solving, Q = 9c (1 - e^(t/rc)). Plug in the numbers from the problem for the numerical answer.

An observation that deals with numbers is a(n)A. Prediction

B. Quantitive observation

C. Evaluation

D. Qualitative observation

Answers

Answer:

B (Quantitative)

Explanation:

a clever way to remember the difference between quantitative and qualitative is that there is a nin quantitative which is the first letter to the word number.So quantitative is an oberservation with numbers

the answer is b like maggie said

suspension are mixtures composed of materials that are visible to the naked eyes true or false? in science subject​

Answers

Answer:

True

Explanation:

Suspensions are referred to as mixtures composed of materials which are visible to the naked eyes . They are usually insoluble in the solvent .

The mixtures which are composed of materials and at the same time non visible to the naked eyes due to their very tiny size are referred to as colloids or colloidal particles in the field of science.