CHEMISTRY COLLEGE

Suppose that you are a scientist who studies climate changes. While examining the rings of tree trunks, you notice several very large tree rings. What can you conclude about the climate during those years?

Answers

Answer 1

Answer:

Answer:

The climate was wet and cold

Explanation:

Answer 2

Answer:

Answer:

The large tree rings allow you to conclude that the climate was either very warm or wet during those growing seasons, because greater than normal growth occurred.

Explanation: It is the edge sample response

Related Questions

After 56.0 min, 40.0% of a compound has decomposed. What is the half‑life of this reaction assuming first‑order kinetics?
Consider the following reaction at 298K.I2 (s) + Pb (s) = 2 I- (aq) + Pb2+ (aq)Which of the following statements are correct?Choose all that apply.ΔGo > 0The reaction is product-favored.K < 1Eocell > 0n = 2 mol electronsB-
What is the mole fraction of calcium chloride in 3.35 m CaCl2(aq)? The molar mass of CaCl2 is 111.0 g/mol and the molar mass of water is 18.02 g/mol.
How many mL of 0.100 M NaOH are needed to neutralize 50.00 mL of a 0.150 M solution of CH3CO2H, a monoprotic acid? How many mL of 0.100 M NaOH are needed to neutralize 50.00 mL of a 0.150 M solution of CH3CO2H, a monoprotic acid? a. 37.50 mL b. 50.00 mL c. 75.00 mL d. 100.00 mL e. 25.00 mL
The Michael reaction is a conjugate addition process wherein a nucleophilic enolate anion (the donor) reacts with an α,β-unsaturated carbonyl compound (the acceptor). The best Michael reactions are those that take place when a particularly stable enolate anion is formed via treatment of the donor with a strong base. Alternatively, milder conditions can be used if an enamine is chosen as the donor, this variant is termed the Stork reaction. In the second step, the donor adds to the β-carbon of the acceptor in a conjugate addition, generating a new enolate. The enolate abstracts a proton from solvent or from a new donor molecule to give the conjugate addition product. Draw curved arrows to show the movement of electrons in this step of the mechanism.

Scaled Synthesis of Alum. Show your calculations for:a.the experimental scaling factor giving rise to a 15.0 g theoretical yield;b.the corrected volumes of KOH and H2SO4; andc.the theoretical yield of alum based on the actual amount of Al used.Make sure you carefully show each step for these calculations.

Answers

Answer:

Explanation:

You don't give your experimental data, so I shall assume:

Mass of Al = 1.07 g

20 mL of 3 mol·L⁻¹ KOH

20 mL of 9 mol·L⁻¹ H₂SO₄

The overall equation for the reaction is

Mᵣ: 26.98 474.39

2Al + 2KOH +4H₂SO₄ + 22H₂O ⟶ 2K[Al(SO₄)₂]·12H₂O + 3H₂

m/g: 1.07

(c) Theoretical yield of alum

(i) Moles of Al

$\text{Moles of Al} = \text{1.07 g Al} * \frac{\text{1 mol Al}}{\text{26.98 g Al}} = \text{0.039 66 mol Al}$

(ii) Moles of alum

$\text{Moles of alum} = \text{0.039 66 mol Al} * \frac{\text{2 mol alum }}{\text{2 mol Al}} = \text{0.039 66 mol alum \n}$

(iii) Theoretical yield of alum

$\text{Mass of alum} = \text{0.039 66 mol alum} * \frac{\text{474.39 g alum}}{\text{1 mol alum}} = \textbf{18.8 g alum}$

(a) Scaling factor for 15.0 g alum

You want a theoretical yield of 15.0 g, so you must scale down the reaction.

$\text{Scale factor} = (15.0)/(18.8) = \mathbf{0.798}$

(b) Corrected volumes of NaOH and H₂SO₄

V = 0.798 × 20 mL = 16 mL

At 1 atm, an unknown sample melts at 49.9 °C and boils at 209.5 °C. If the temperature is 0°C, what is the state of matter for the sample?

Answers

Answer:

The correct answer is solid.

Explanation:

Based on the given information, it is evident that at 1 atm pressure and 49.9 degrees C the melting of the sample takes place, that is, the unknown sample transforms into the liquid at the mentioned temperature. It can also be said that below 49.9 degrees C, the sample stays at solid-state. From all this, we can also state that at temperature 49.9 degrees C, both the liquid and the solid-state of the sample stays at equilibrium.

As one goes higher, that is, above 49.9 degrees C and up to 209.5 degrees C, the sample remains at liquid state. However, the boiling point of the sample is 209.5 degrees C, which shows that the sample becomes gas above 209.5 degrees C. Thus, the sample remains at solid-state below 49.9 degrees C, at liquid state between 49.9 degrees C to 209.5 degrees C, and a gaseous state above 209.5 degrees C. Hence, if the temperature is 0 degrees C, then solid will be the state of matter for the sample.

Rank the following elements from smallest to
largest atomic radius: Fr, F, Ge, Ru?

Answers

Answer:

1. Flourine, 2 Ruthenium, 3 Germanium, 4 Francium

Explanation:

4, 1, 3, 2 is the order it’ll go in by what you’ve given us

A piston confines 0.200 mol Ne(g) in 1.20 at 25 degree C. Two experiments are performed. (a) The gas is allowed to expand through an additional 1.20 L against a constant of 1.00atm. (b) The gas is allowed to expand reversibly and isothermally to the same final volume. Please calculate the work done by the gas system in these two processes, respectively. Which process does more work? (revised from 6/e exercise 8.11) Please show calculation details.

Answers

Answer:

The second experiment (reversible path) does more work

Explanation:

Step 1:

A piston confines 0.200 mol Ne(g) in 1.20L at 25 degree °C

(a) The gas is allowed to expand through an additional 1.20 L against a constant of 1.00atm

Irreversible path: w =-Pex*ΔV

⇒ with Pex = 1.00 atm

⇒ with ΔV = 1.20 L

W = -(1.00 atm) * 1.20 L

W = -1.20L*atm *101.325 J /1 L*atm = -121.59 J

(b) The gas is allowed to expand reversibly and isothermally to the same final volume.

W = -nRTln(Vfinal/Vinitial)

⇒ with n = the number of moles = 0.200

⇒ with R = gas constant = 8.3145 J/K*mol

⇒ with T = 298 Kelvin

⇒ with Vfinal/Vinitial = 2.40/1.20 = 2

W = -(0.200mol) * 8.3145 J/K*mol *298K *ln(2.4/1.2)

W = -343.5 J

The second experiment (reversible path) does more work

Which of the following is an oxidation-reduction reaction? Which of the following is an oxidation-reduction reaction? Mg(s) + 2 HCl(aq) → MgCl2(aq) + H2(g) HCl(aq) + LiOH(aq) → LiCl(aq) + H2O(l) NaCl(aq) + AgNO3(aq) → AgCl(s) + NaNO3(aq) Pb(C2H3O2)2(aq) + 2 NaCl(aq) → PbCl2(s) + 2 NaC2H3O2(aq) All of the above are oxidation-reduction reactions.

Answers

Answer:

NONE OF THE ABOVE

Explanation:

None of the above are examples of an oxidation - reduction or a redox reaction . This is because there is no change in the oxidation state of any of the elements in the reaction when the reaction happens .

For a redox reaction , transfer of electrons must take place.
The oxidation states must increase or decrease in atleast 2 of the elements of a compound.
For example :

$Zn^(2+)+Cu$ ⇒ $Cu^(2+)+Zn$

Which of the following characteristics is common to both acids and bases? They produce ions when dissolved in water They absorb oxygen ions when dissolved in water They increase hydrogen ions when dissolved in a solution They increase the hydroxide ions when dissolved in a solution

Answers

Answer:

They produce ions when dissolved in water.

Explanation:

Acids and bases have the characteristic in common to each other. Both of them have the property of reacting and dissolving in the water. Both acids and bases lead to the production of the ions when they are placed in a water solution. Acids produce Hydrogen ions when they are dissolved in water. Bases produce hydroxide ion when they are dissolved in water.

Other Questions

Assuming binding is fast relative to subsequent catalytic steps, what relative effect does substitute the various active site residues have on catalysis versus substrate-binding? The various substitutions affecta. substrate-binding more than catalytic turnover. b. catalytic turnover more than substrate binding. c. substrate binding, but not catalytic turnover. d. catalytic turnover, but not substrate binding.

Why does the mixture of Fe and S have different reactions than the compound FeS?

Write the equation for the alpha decay of C-11. Write the equation for the beta decay of Be-10HELP HELP HELP

A scientist performs an experiment involving the collision of two masses on a flat surface. She believes she has made a startling discovery: the total kinetic energy is not conserved. What might she have overlooked?A.Some gravitational potential energy may have been gained.B.Some gravitational potential energy may have been lost.C.Some energy may have been gained due to friction and/or sound.D.Some energy may have been lost due to friction and/or sound.