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2 years ago

What type of ossification occurs to form an immovable joint

Chemistry

1 answer:

ipn [44]2 years ago

5 0

Answer:

The type of ossification that occurs is the endomembrane or endochondral.

Explanation:

In this type of joints called SYNARTROSIS, there is no intermediate formation of ligaments or cartilage, with which the bone does not grow from a cartilage ossification, but from undifferentiated cells that will calcify and form bone cells and the main bone units that are the OSTEONAS. The bones that present this type of joints are the skull, in the area of the temporal bone (Temporary Suture) that grow abruptly throughout life and by mechanisms different from those bones that present a joint with movement.

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Problem Page A chemist measures the amount of bromine liquid produced during an experiment. He finds that of bromine liquid is p

masha68 [24]

The question is incomplete, here is the complete question:

A chemist measures the amount of bromine liquid produced during an experiment. She finds that 766.g of bromine liquid is produced. Calculate the number of moles of bromine liquid produced. Round your answer to 3 significant digits.

<u>Answer:</u> The amount of liquid bromine produced is 4.79 moles.

<u>Explanation:</u>

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

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

We are given:

Given mass of liquid bromine = 766. g

Molar mass of liquid bromine, $(Br_2)$ = 159.8 g/mol

Putting values in above equation, we get:

$\text{Moles of liquid bromine}=\frac{766.g}{159.8g/mol}=4.79mol$

Hence, the amount of liquid bromine produced is 4.79 moles.

6 0

1 year ago

A teacher wants to figure out The amount of a solution that is needed for a titration experiment in a class lab! the class has 1

AnnZ [28]

The teacher needs 1500 mL for the whole class.

5 0

1 year ago

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Which statement below correctly identifies the number of subatomic particles inside and outside the nucleus of an argon atom?

Juliette [100K]

Answer:

Explanation:

6 0

2 years ago

Check all the boxes that describe the electron sea model. It is the simplest metal bonding model. It is the most complicated met

Mrac [35]

Answer:

Another view could be from this definition: (Electron Sea Model) : The metallic bond consists of a series of metals atoms that have all donated their valence electrons to an electron cloud that permeates the structure. This electron cloud is frequently referred to as an electron sea. It might help to visualize the electron sea model as if it were a box of marbles that are surrounded by water. The marbles represent the metal atoms and the water represents the electron sea.

Explanation:

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2 years ago

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One of the commercial uses of sulfuric acid is the production of calcium sulfate and phosphoric acid. If 19.9 g of Ca₃(PO₄)₂ rea

Natasha_Volkova [10]

57.5 % is the percent yield if 10.9 g of H₃PO₄ is formed.

Explanation:

Balanced equation for the reaction:

Ca₃(PO₄)₂ (s) + 2H₂SO₄ (aq) → H₃PO₄ (aq) + 2CaSO₄ (aq)

data given:

mass of Ca₃(PO₄)₂ = 19.9 grams

mass of H₂SO₄, = 54.3 grams

mass of H₃PO₄ produced = 10.9 grams (actual yield)

percent yield=?

atomic mass of Ca₃(PO₄)₂ = 310.17 grams/mole

atomic mass of H₂SO₄ = 98.07 grams/mole

number of moles is calculated as:

number of moles = $\frac{mass}{atomic mass of one mole}$

putting the values in the above equation:

for Ca₃(PO₄)₂ = $\frac{19.9}{310.17}$

= 0.064 moles

moles of H₂SO₄ = $\frac{54.3}{98.07}$

= 0.55 moles

from the reaction, it can be found that the limiting reagent is Ca₃(PO₄)₂

1 mole of Ca₃(PO₄)₂ reacts to form 1 mole of phosphoric acid

0.064 moles of Ca₃(PO₄)₂ will produce x moles of phosphoric acid

0.064 moles of phosphoric acid produced

mass = number of moles x atomic mass of H3PO4

=0.064 x 97.994

= 6.27 grams (theoretical yield)

FORMULA FOR PERCENT YIELD:

percent yield = $\frac{actual yield}{theoretical yield}$ x 100

= $\frac{6.27}{10.9}$ x 100

= 57.5 %

7 0

2 years ago