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kvv77 [185]
3 years ago
8

What are the types of connective tissue? Describe each type, and give an example of each.

Chemistry
1 answer:
In-s [12.5K]3 years ago
3 0

Answer:

he major types of connective tissue are connective tissue proper, supportive tissue, and fluid tissue. Loose connective tissue proper includes adipose tissue, areolar tissue, and reticular tissue.

Explanation:

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The table below shows the number of sub-atomic particles in an atom of magnesium.
Fittoniya [83]

Answer:

24

Explanation:

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Why is the electrical charge of an atom zero or neutral?
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Because the atom isn't a isotope. The electrons, (negative charge) , and the protons, (positive charge), balance each other out. 
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The distinctive odor of vinegar is due to acetic acid, HC₂H₃O₂, which reacts with sodium hydroxide. If a 3.45 mL sample of vineg
Vilka [71]

Answer:

Explanation:

CH₃COOH + NaOH = CH₃COONa + H₂O .

42.5 mL of .115 M of NaOH will contain .0425 x .115 moles of NaOH

= 48.875 x 10⁻⁴ moles NaOH

It will react with same number of moles of acetic acid

So number of moles of acetic acid in 3.45 mL = 48.875 x 10⁻⁴

number of moles of acetic acid in 1000 mL = 48.875 x 10⁻⁴ x 10³ / 3.45 moles

= 1.4167 moles

= 1.4167 x 60 gram

= 85 grams .

So 85 grams of acetic acid will be contained in one litre of acetic acid.

8 0
3 years ago
Which of earths spheres contains most of its mass?
zysi [14]
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8 0
3 years ago
Read 2 more answers
If the sample contained 2.0 moles of KClO3 at a temperature of 214.0 °C, determine the mass of the oxygen gas produced in grams
Westkost [7]

Answer : The mass of the oxygen gas produced in grams and the pressure exerted by the gas against the container walls is, 96 grams and 1.78 atm respectively.

Explanation : Given,

Moles of KCl_3 = 2.0 moles

Molar mass of O_2 = 32 g/mole

Now we have to calculate the moles of MgO

The balanced chemical reaction is,

2KClO_3\rightarrow 2KCl+3O_2

From the balanced reaction we conclude that

As, 2 mole of KClO_3 react to give 3 mole of O_2

So, 2.0 moles of KClO_3 react to give \frac{2.0}{2}\times 3=3.0 moles of O_2

Now we have to calculate the mass of O_2

\text{ Mass of }O_2=\text{ Moles of }O_2\times \text{ Molar mass of }O_2

\text{ Mass of }O_2=(3.0moles)\times (32g/mole)=96g

Therefore, the mass of oxygen gas produced is, 96 grams.

Now we have to determine the pressure exerted by the gas against the container walls.

Using ideal gas equation:

PV=nRT\\\\PV=\frac{w}{M}RT\\\\P=\frac{w}{V}\times \frac{RT}{M}\\\\P=\rho\times \frac{RT}{M}

where,

P = pressure of oxygen gas = ?

V = volume of oxygen gas

T = temperature of oxygen gas = 214.0^oC=273+214.0=487K

R = gas constant = 0.0821 L.atm/mole.K

w = mass of oxygen gas

\rho = density of oxygen gas = 1.429 g/L

M = molar mass of oxygen gas = 32 g/mole

Now put all the given values in the ideal gas equation, we get:

P=1.429g/L\times \frac{(0.0821L.atm/mole.K)\times (487K)}{32g/mol}

P=1.78atm

Thus, the pressure exerted by the gas against the container walls is, 1.78 atm.

7 0
3 years ago
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