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Katyanochek1 [597]
3 years ago
8

Shells constructed from seawater incorporate the 18O/16O ratio of seawater during their lifetime within their CaCO3 shell walls,

providing a paleothermometer that is used to estimate the temperatures of ancient seas.
Chemistry
1 answer:
Ksivusya [100]3 years ago
6 0

Answer:

Hello, the above question is not complete, nonetheless let us check somethings out.

Explanation:

Paleothermometer definition is from two words, that is "Paleo" which means something that is old and ''thermometer" which is an instrument for measuring temperature. So, if we add this up, Paleothermometer is an instrument for measuring "old" temperature, that is temperature. One of the Paleothermometer that is been used is the δ18O which is the one in the question that has isotopic ratio of 18O/16O, and it deals with the measurement of 18O to 16O. The others include Alkenones Paleothermometer, Mg/Ca Paleothermometer, Leaf physiognomy and so on.

If the values of the isotopic ratio that is 18O/16O ratio is low, then the temperature is high. To Calculate the 18O/16O ratio for ancient ocean then we will be using the equation below;

δ18O = (z - 1) × 1000. Where z= [(18O/16O)/( 18O/16O)sm. And sm= standard mean.

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Identify whether each species functions as a Brønsted-Lowry acid or a Brønsted-Lowry base in this net ionic equation. ClO- + H2P
MaRussiya [10]

Answer:

ClO- = bronsted Lowry base

H2PO4- =Bronsted Lowry base

Explanation:

ClO- = is bronsted Lowry base because it accepts a proton in the reaction

H2PO4- =Bronsted Lowry base because it's a proton donor in the reaction

8 0
4 years ago
a sample of a gas has a volume kf 640 cm^{3} at 100°c and 1490 mmhg,what would be its volume at stp?​
scoundrel [369]

Answer:

V₂ = 918.1 cm³

Explanation:

Given data:

Initial volume = 640 cm³

Initial temperature = 100°C (100+273 = 373 K)

Initial pressure = 1490 mmHg (1490 /760 = 1.96 atm)

Final volume = ?

Final temperature = 273 K

Final pressure = 1 atm

Solution:

Formula:  

P₁V₁/T₁ = P₂V₂/T₂  

P₁ = Initial pressure

V₁ = Initial volume

T₁ = Initial temperature

P₂ = Final pressure

V₂ = Final volume

T₂ = Final temperature

now we will put the values in formula.

V₂ = P₁V₁ T₂/ T₁ P₂  

V₂ = 1.96 atm × 640 cm³ × 273 K / 373 K × 1 atm

V₂ = 342451.2 atm .cm³ . K / 373 K. atm

V₂ = 918.1 cm³

3 0
3 years ago
--IF YOU PUSH SOMETHING WITH STRONG FORCE DOES THE MOTION GO SLOWER
zaharov [31]

Answer:

i dont think so

Explanation:

is there a image to explain this ? if so it would help ALOT

8 0
2 years ago
Read 2 more answers
Solid aluminum and gaseous oxygen react in a combination reaction to produce aluminum oxide: 4Al (s) + 3O2 (g) → 2Al2O3 (s) In
jek_recluse [69]

Answer: The percent yield of the reaction is 74 %

Explanation:

To calculate the number of moles, we use the equation:

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

\text{Moles of aluminium}=\frac{2.5g}{27g/mol}=0.092mol

For oxygen gas:

\text{Moles of oxygen gas}=\frac{2.5g}{32g/mol}=0.078mol

The chemical equation for the reaction of titanium and chlorine gas follows:

4Al(s)+3O_2(g)\rightarrow 2Al_2O_3(s)

By Stoichiometry of the reaction:

4 moles of aluminium reacts with 3 moles of oxygen.

So, 0.092 moles of aluminium reacts with = \frac{3}{4}\times 0.092=0.069mol of oxygen

As, given amount of oxygen is more than the required amount. So, it is considered as an excess reagent.

Thus, aluminium is considered as a limiting reagent because it limits the formation of product.

By Stoichiometry of the reaction:

4 moles of aluminium produce = 2 moles of Al_2O_3

So, 0.092 moles of aluminium will produce = \frac{2}{4}\times 0.092=0.046moles of Al_2O_3

Now, calculating the mass of aluminium oxide:

\text{Mass of aluminium oxide}=moles\times {\text {molar mas}}=0.046mol\times 102g/mol=4.7g

To calculate the percentage yield of titanium (IV) chloride, we use the equation:

\%\text{ yield}=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100

Experimental yield  = 3.5 g

Theoretical yield = 4.7 g

Putting values in above equation, we get:

\%\text{ yield of reaction}=\frac{3.5g}{4.7g}\times 100\\\\\% \text{yield of reaction}=74\%

Hence, the percent yield of the reaction is 74 %

6 0
3 years ago
16. Mass = 10g
inessss [21]

Actual volume=Final Volume-initial volume

\\ \sf\longmapsto 50ml-30ml=20ml

Now

\\ \sf\longmapsto Density=\dfrac{Mass}{Volume}

\\ \sf\longmapsto Density=\dfrac{10}{20}

\\ \sf\longmapsto Density=2g/ml

3 0
3 years ago
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