# The change in velocity over a specific amount of time Includes speeding up. slowing down or changing direction. This is known as O speed acceleration changing speed​

Acceleration

Explanation:

Acceleration =

## Related Questions

The metal thallium becomes superconducting at temperatures below 2.39K. Calculate the temperature at which thallium becomes superconducting in degrees Celsius. Round your answer to decimal places.

-270.76°C

Explanation:

Given that metal Thallium becomes superconducting below the temperature of 2.39 kelvin i.e. this temperature is critical temperature for Thallium and below critical temperature a metal offers no resistance to the flow of electric current. Also the metal below its critical temperature expels the magnetic field in such a way that they do not penetrate the metal and pass through its surface only.

We have the relation between kelvin scale and degree Celsius scale of temperature measurement as:

An order is given to administer methylprednisolone, an anti-inflammatory drug, by IV at a rate of 36 mg every 30. min . The IV bag contains 125 mg of methylprednisolone in every 2.0 mL . What should the flow rate be in milliliters per minute (mL/min)?

0.0192 mL per min.

Explanation:

IV rate = 36 mg per 30 min.

IV concentration = 125 mg per 2.0 mL

36 mg per 30 min. IV rate = 36/30 = 1.2 mg per min

If 125 mg methylprednisolone is present in 2.0 mL of the IV nag, how many mL would contain 1.2 mg?

= 2x1.2/125

= 0.0192 mL

Therefore, the flow rate of the IV must be 0.0192 mL per min. in order to be able to deliver 36 mg per 30 min.

Consider the dissolution of AB(s):AB(s)⇌A+(aq)+B−(aq)Le Châtelier's principle tells us that an increase in either [A+] or [B−] will shift this equilibrium to the left, reducing the solubility of AB. In other words, AB is more soluble in pure water than in a solution that already contains A+ or B− ions. This is an example of the common-ion effect.The generic metal hydroxide M(OH)2 has Ksp = 1.05×10−18. (NOTE: In this particular problem, because of the magnitude of the Ksp and the stoichiometry of the compound, the contribution of OH− from water can be ignored. However, this may not always be the case.)What is the solubility of M(OH)2 in pure water?

S = 6.40 × 10⁻⁷ M

Explanation:

In order to calculate the solubility (S) of M(OH)₂ in pure water we will use an ICE Chart. We recognize 3 stages: Initial, Change and Equilibrium, and we complete each row with the concentration or change in concentration.

M(OH)₂(s) ⇄ M²⁺(aq) + 2 OH⁻(aq)

I                                   0                  0

C                                 +S               +2S

E                                   S                 2S

The solubility product (Kps) is:

Kps = 1.05 × 10⁻¹⁸ = [M²⁺].[OH⁻]²=S.(2S)²

1.05 × 10⁻¹⁸ = 4S³

S = 6.40 × 10⁻⁷ M

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

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

-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.

If you want to melt ice into liquid water, does the ice absorbs heat or release heat? Is this endothermic or exothermic?

Endothermic

Explanation:

This would be an endothermic reaction, due to the fact that the water would be signifigantly warmer then the ice cube, which causes it to melt.

absorbs heat, which makes it endothermic.

Explanation:

Basically melting ice is it in windows or Mac reaction makes the ice absorbs heat energy which causes a change to occur

. Explain why, in the sample calculations, 0.1 g of the unknown produced a GREATER freezing point depression than~e same mass of naphthalene.

Naphthalene is a non electrolyte

If the unknown compound is an electrolyte it gives 2 or more ions in solution

( NaCl >> Na+ + Cl- => 2 ions

Ca(NO3)2 >> Ca2+ + 2 NO3- => 3 ions)

the f.p. lowering is directly proportional to the molal concentration of dissolved ions in the solution )

For naphthalene

delta T = 1.86 x m

for a salt that gives 2 ions

delta T = 1.86 x m x 2

hence the lowering in freezion point of unkown is greater then napthalene