Choose the statement that best describes the titration from which this data was collected

Answers

Answer 1

Answer: I think the correct answer is the first option. The analyte was an acid but and the titrant is a base since from the graph the initial pH of the solution is at an acidic pH when there is still no titrant added.

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Select the THREE answers that are correct concerning a saturated solution of PbCl2. PbCl2(s) Pb+2(aq) + 2Cl-(aq)
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In a chemical reaction, the difference between the potential energy of the products and the potential energy of the reactants is equal to the(1) activation energy(2) kinetic energy
(3) heat of reaction
(4) rate of reaction

Answers

3.heat of reaction
hope that helped

A solution containing 1000g of an unknown substance in 12.3g of naphthalene was found to freeze at 1.2∘C. What is the molar mass of the unknown substance?

Answers

Answer: Therefore, the molar mass of the unknown substance is 68.4 g/mol.

Explanation: We can use the freezing point depression equation to solve for the molar mass of the unknown substance:

ΔT = Kf × m

where ΔT is the change in freezing point, Kf is the freezing point depression constant of the solvent (naphthalene), and m is the molality of the solution.

First, we need to calculate the molality of the solution:

molality = moles of solute / mass of solvent (in kg)

We don't know the number of moles of the unknown substance, but we can assume that the naphthalene does not contribute significantly to the total mass of the solution (since its mass is much smaller than the mass of the unknown substance). Therefore, we can use the entire mass of the solution (1000g + 12.3g = 1012.3g) as the mass of solvent.

mass of solute = 1000g

mass of solvent = 12.3g

mass of solution = 1012.3g

molality = (1000g / molar mass) / (12.3g / 1000g) = 81.3 / molar mass

Next, we need to calculate the change in freezing point:

ΔT = 1.2∘C

Finally, we can use the freezing point depression constant of naphthalene to solve for the molar mass of the unknown substance:

Kf for naphthalene = 6.8∘C/m

ΔT = Kf × m

1.2 = 6.8 × (81.3 / molar mass)

molar mass = 68.4 g/mol

Therefore, the molar mass of the unknown substance is 68.4 g/mol.

mixture or pure substance: 1.blood 2.dyes 3.self-raising flour 4.muesli 5.copper wire 6.distilled water 7.table salt 8.milk 9.bronze 10.tea 11.oxygen 12.air

Answers

Answer: Mixture: Blood , Self raising flour,muesli ,dyes, milk, tea, air, bronze

Pure substance: Copper wire, distilled water, table salt, oxygen.

Explanation:

Mixture is a substance which is made up two or more number of compounds which chemically inactive and retain their distinct chemical properties.

Blood , Self raising flour,muesli ,dyes, milk, tea, air, bronze

Pure substance is defined as anything with uniform and unchanging composition is known s pure substance.

Copper wire, distilled water, table salt, oxygen.

Changes to the global climate could include _______.a. changes to temperature
b. changes to pressure
c. changes to precipitation
d. all of the above

Which of the following is not released by trees into the atmosphere?
a. oxygen
b. water vapor
c. ozone
d. none of the above

Answers

1. Answer is d. all of the above.

Climate can be defined as the data collection of weather conditions of a long period of time in a certain area. The weather conditions that affect to the climate are temperature, precipitation, atmospheric pressure, wind and so on. Any changes in weather can change the global climate.

2. Answer is c. ozone.

Trees release oxygen and water vapor into the atmosphere but not ozone. The oxygen gas is released by trees as the by-product of photosynthesis. This is the natural way of production of oxygen gas. Water vapor can be released due to the transpiration of trees.

1. The correct answer is option D. all of the above.

Global climate change will bring changes in temperature such as heat waves, droughts, volcanic eruption, deforestation, extinction of species due to large shift in temperatures, changes in precipitation such as heavy snowfall, ocean acidification, and ozone depletion.

2. The correct answer is option C, ozone.

Ozone is not released by trees into the atmosphere. Plants have chlorophyll and in the presence of sunlight plants uses sunlight and carbon dioxide to form glucose and oxygen.

6CO₂ + 6H₂O -> C₆H₁₂O₆ + 6O₂

During transpiration plants exhale water vapor through the stomata, tiny pores that are found on the surface of the leaves.

Thus trees never release ozone into the atmosphere.

How many grams of calcium bromide are in 50.0 mL of a 0.25 M calcium bromide solution?

Answers

The mass of the compound can be calculated by the molarity. The mass of the calcium bromide in the given solution is 2.5 g.

The mass of the given compound can be calculated by the molarity formula,

$\bold {M = \frac {w}{m* v} * 1000}\n\n\bold {w = \frac {M* m* v}{ 1000}}$

Where,

M- molarity of the solution = 0.25 M

w - given mass =?

m -molar mass of Calcium bromide = 200 g/mol

v-volume in mL= 50 mL

Put the values in the formula,

$\bold {w = \frac {0.25* 200* 50}{ 1000}}\n\n\bold {w = 2.5\ g}$

Therefore, the mass of the calcium bromide in the given solution is 2.5 g.

To know more about Molarity,

brainly.com/question/12127540

$C_(m)=(n)/(V_(r))\n\nm=(m)/(M)\n\nC_(m)=(m)/(MV_(r)) \ \ \ \ \Rightarrow \ \ \ \ m=C_(m)MV_(r)}\n\nC_(m)=0.25(mol)/(L)\nM_{CaBr_(2)}=200(g)/(mol)\nV_(r)=50mL=0.05L\n\nm=0.25(mol)/(L)*200(g)/(mol)*0.05L= \emph{\textbf{2.5g}}$

Which chemicals previously used in products such as refrigerants, aerosol sprays, and cleaning products contributed to the depletion of the ozone layer

Answers

Answer:chlorofluorocarbons (CFCs) could deplete Earth's atmospheric ozone layer, which blocks the sun's damaging ultraviolet rays. When the scientists reported their findings in 1974, CFCs were widely used as refrigerant gases and as propellants in aerosol sprays

Explanation:

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