Determine molar mass of Mn (ClO3)3

What type of ossification occurs to form an immovable joint

ipn [44]

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.

5 0

2 years ago

Write the empirical formula of at least four binary ionic compounds that could be formed from the following ions: Ca2+, V5+, Br-

Jet001 [13]

Answer:

CaS, CaBr₂, VBr₅, and V₂S₅.

Explanation:

The ionic compound should be neutral; the overall charge of it is equal to zero.

Binary ionic compound is composed of two different ions.

<u>Ca²⁺ can combined with either Br⁻ or S²⁻ to form binary ionic compounds.</u>

CaS can be formed via combining Ca²⁺ with S²⁻ to form the neutral binary ionic compound CaS.

CaBr₂ can be formed via combining 1 mole of Ca²⁺ with 2 moles of Br⁻ to form the neutral binary ionic compound CaBr₂.

<u>V⁵⁺ can combined with either Br⁻ or S²⁻ to form binary ionic compounds.</u>

V₂S₅ can be formed via combining 2 moles of V⁵⁺ with 5 moles of S²⁻ to form the neutral binary ionic compound V₂S₅.

VBr₅ can be formed via combining 1 mole of V⁵⁺ with 5 moles of Br⁻ to form the neutral binary ionic compound VBr₅.

<em>So, the empirical formula of four binary ionic compounds that could be formed is: CaS, CaBr₂, VBr₅, and V₂S₅.</em>

5 0

2 years ago

How long will it take 10.0 mL of Ne gas to effuse through a porous barrier if it has been observed that 125 minutes are required

cupoosta [38]

Answer:

88.8 minutes

Explanation:

Graham's law of diffusion relates rate of difusion by the following formula

Rate1 / rate 2 = √( Mass of argon / Mass of Neon)

Where rate = volume divided by time

Rate 1 = 10 ml / t1

Rate 2 = 10 ml / t2

Rate 1/ rate 2 = 10 ml / t1 ÷ 10 ml/ t2 = t2/ t1

t2/t1 = √(Mass of argon / mass of Neon) = √( 39.984/20.179)

125 / t1 = 1.4026

t1 = 125 / 1.4026 = 88.8 minutes

7 0

2 years ago

A 0.89% (w/v) sodium chloride solution is referred to as physiological saline solution because it has the same concentration of

maks197457 [2]

1) 0.89% m/v = 0.89 grams of NaCl / 100 ml of solution

=> 8.9 grams of NaCl in 1000 ml of solution = 8.9 grams of NaCl in 1 liter of solution

2) Molarity = M = number of moles of solute / liters of solution

=> calculate the number of moles of 8.9 grams of NaCl

3) molar mass of NaCl = 23.0 g /mol + 35.5 g/mol = 58.5 g / mol

4) number of moles of NaCl = mass / molar mass = 8.9 g / 58.5 g / mol = 0.152 mol

5) M = 0.152 mol NaCl / 1 liter solution = 0.152 M

Answer: 0.152 M

4 0

2 years ago

Percentage yield of sodium peroxide if 5 g of sodium oxide produces 5.5 g of sodium peroxide

Rama09 [41]

<h3>Answer:</h3>

87.40 %

<h3>Explanation:</h3>

Concept being tested: Percent yield of a product

We are given;

Mass of Sodium oxide 5 g

Experimental or Actual yield of sodium peroxide IS 5.5 g

We are required to calculate the percent yield of sodium peroxide;

The equation for the reaction that forms sodium peroxide is

2Na₂O + O₂ → 2Na₂O₂

<h3>Step 1; moles of sodium oxide</h3>

Moles = mass ÷ molar mass

Molar mass of sodium oxide is 61.98 g/mol

Therefore;

Moles = 5 g ÷ 61.98 g/mol

= 0.0807 moles

<h3>Step 2: Theoretical moles of sodium peroxide produced </h3>

From the equation, 2 moles of sodium oxide produces 1 mole of sodium peroxide.

Thus, moles of sodium peroxide used is 0.0807 moles

<h3>Step 3: Theoretical mass of sodium peroxide used</h3>

Mass = Number of moles × Molar mass

Molar mass of sodium peroxide = 77.98 g/mol

Therefore;

Theoretical mass = 0.0807 moles × 77.98 g/mol

= 6.293 g

Theoretical mass of Na₂O₂ is 6.293 g

<h3>Step 4: Percent yield of Na₂O₂</h3>

We know that percent yield is given by the ratio of actual yield to theoretical yield expressed as a percentage.

$Percent yield=(\frac{Actual yield}{theoretical yield})100$

$Percent yield(Na_{2}O_{2})=(\frac{5.5g}{6.293g})100$

= 87.40 %

Therefore, the percentage yield of sodium peroxide is 87.4%

8 0

2 years ago