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3 years ago

A procedure involving removal of the coronal portion of an exposed vital pulp is a

Chemistry

2 answers:

const2013 [10]3 years ago

8 0

Answer:

preserve of the vitality of the remaining portion of the pulp within the root of the tooth. Also referred to as root canal therapy

Explanation:

sesenic [268]3 years ago

5 0

Answer:

ExplanaPulpotomy is a vital pulp therapy in which the coronal portion of the pulp is removed surgically and the remaining radicular pulp is preserved intact. Over the remaining radicular pulp tissue, a suitable material is placed which has the potential to protect the pulp from further insult and initiate healing and repairtion:

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You develop and visualize a silica gel TLC plate and are not satisfied with the results. Which of the following experimental par

borishaifa [10]

Answer:

Explanation:

Toe change the retention factor of a TLC analysis, you can change your solvent for a more or less polar one, depending on your analyte. You can use a mix of solvents too.

You can also change the your method to visualize the spots, you can use fluorescent compounds that can only be seen in black light, you can use Iodine, Bromine and so on.

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4 years ago

1 mole = _____________________ _________________ particles

Artist 52 [7]

1 mole = 6.02 * 10^23 atoms

This is known as Avogadro's Number. This value is approximate.

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3 years ago

Read 2 more answers

What volume of methane gas at 237 K and 101.33 kPa do you have when the volume is decreased to 0.50 L, with a temperature of 300

Alexxandr [17]

Answer: A volume of 0.592 L of methane gas is required at 237 K and 101.33 kPa when the volume is decreased to 0.50 L, with a temperature of 300 K and a pressure of 151.99 kPa.

Explanation:

Given: $T_{1}$ = 237 K, $P_{1}$ = 101.33 kPa, $V_{1}$ = ?

$T_{2}$ = 300 K, $P_{2}$ = 151.99 kPa, $V_{2}$ = 0.50 L

Formula used is as follows.

$\frac{P_{1}V_{1}}{T_{1}} = \frac{P_{2}V_{2}}{T_{2}}$

Substitute the values into above formula as follows.

$\frac{P_{1}V_{1}}{T_{1}} = \frac{P_{2}V_{2}}{T_{2}}\\\frac{101.33 kPa \times V_{1}}{237 K} = \frac{151.99 kPa \times 0.50 L}{300 K}\\V_{1} = 0.592 L$

Thus, we can conclude that a volume of 0.592 L of methane gas is required at 237 K and 101.33 kPa when the volume is decreased to 0.50 L, with a temperature of 300 K and a pressure of 151.99 kPa.

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3 years ago

1) Copper oxide and carbon combine to produce metallic copper. Which of these is the balanced equation for this reaction?

Goshia [24]

The answer to your question is B

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3 years ago

Which is smaller 4 gallons or 33 pints?

fenix001 [56]

4 gallons
1 pint=0.125 gallons
0.125 times 33= 4.125

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3 years ago