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1 year ago

Write an equation for the formation of bf3(g) from its elements in their standard states.

1 answer:

7 0

In nature, boron is monoatomic. Therefore, its formula is B.
On the other hand, fluorine is diatomic. Therefore, its formula is F2

Now, the basic unbalanced equation is:
B + F2 .........> BF3

Now, we need to balance this equation. As you can see, we have two fluorine moles entering the reaction and 3 formed in the products.
Balancing the equation, we will reach the following balanced reaction:
2B + 3F2 .......> 2BF3

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How many moles of lead (ii) chromate are in 51 grams of this substance? answer in units of mol?

maria [59]

Mass of lead (II) chromate is 51 g. The molecular formula is $PbCrO_{4}$ and its molar mass is 323.2 g/mol

Number of moles can be calculated using the following formula:

$n=\frac{m}{M}$

Here, m is mass and M is molar mass.

Putting the values,

$n=\frac{(51 g}{323.1937 g/mol}=0.1578 mol$

Therefore, number of moles of lead (II) chromate will be 0.1578 mol.

5 0

1 year ago

A chemist adds 180.0 mL of a 1.42M sodium carbonate (Na CO,) solution to a reaction flask. Calculate the millimoles of sodium ca

svet-max [94.6K]

Answer: The millimoles of sodium carbonate the chemist has added to the flask are 256

Explanation:

Molarity is defined as the number of moles dissolved per liter of the solution.

To calculate the number of moles for given molarity, we use the equation:

$\text{Molarity of the solution}=\frac{\text{milli moles of solute}}{\text{Volume of solution in ml}}$ .....(1)

Molarity of $BaCl_2$ solution = 1.42 M

Volume of solution = 180.0 mL

Putting values in equation 1, we get:

$1.42M=\frac{\text{milli moles of }BaCl_2}{180.0ml}\\\\\text{milli moles of }BaCl_2}={1.42M\times 180.0ml}=256milli mol$

Thus the millimoles of sodium carbonate the chemist has added to the flask are 256.

6 0

1 year ago

Rank the compounds by the ease with which they ionize under sn1 conditions. rank the compounds from easiest to hardest. to rank

marysya [2.9K]

The rate of Formation of Carbocation mainly depends on two factors'

1) Stability of Carbocation:
The ease of formation of Carbocation mainly depends upon the ionization of substrate. If the forming carbocation id tertiary then it is more stable and hence readily formed as compared to secondary and primary.

2) Ease of detaching of Leaving Group:
The more readily and easily the leaving group leaves the more readily the carbocation is formed and vice versa. In given scenario the carbocation formed is tertiary in all three cases, the difference comes in the leaving group. So, among these three substrates the one containing Iodo group will easily dissociate to form tertiary carbocation because due to its large size Iodine easily leaves the substrate, secondly Chlorine is a good leaving group compared to Fluoride. Hence the order of rate of formation of carbocation is,

R-I > R-Cl > R-F

B > C > A

3 0

1 year ago

Of protons, neutrons and electrons which of these can never change an atom during an ordinary chemical or physical change. Expla

hjlf

Answer:

protons

Explanation:

electron number changes when the atom reacts with another atom to gain a full octet

neutron number changes when it goes through radioactive decay

but proton number never changes

6 0

1 year ago

If a hazardous bottle is labeled 20.2 percent by mass Hydrochloric Acid, HCl, with a density of 1.096 g/mL, calculate the molari

SIZIF [17.4K]

Answer:6M

Explanation:

From Co= 10pd/M

Where Co= molar concentration of raw acid

p= percentage by mass of raw acid=20%

d= density of acid=1.096g/cm3

M= molar mass of acid=36.5

Co= 10×20×1.096/36.5=6M

5 0

1 year ago