All categories

2 years ago

What is the name of Br6F10 ?

Chemistry

1 answer:

astraxan [27]2 years ago

3 0

Answer:

idfk

Explanation:

You might be interested in

What mass of copper is required to replace silver from 4.00g of silver nitrate dissolved in water?

larisa [96]

The balanced chemical reaction is written as:

Cu +2AgNO3 → Cu(NO3)2 + 2Ag

We are given the amount of silver nitrate to be used for the reaction. This value will be the starting point of our calculations. It is as follows:

4.00 g AgNO3 ( 1 mol / 169.87 g ) ( 1 mol Cu / 2 mol AgNO3 ) ( 63.456 g / 1 mol ) = 0.747 g Cu

3 0

2 years ago

in collecting the precipitate, why would it be inappropriate to heat the reacted mixture and evaporate off the water?

djverab [1.8K]

In collecting the precipitate, it is inappropriate to heat the reacted mixture and evaporate off the water because it is possible that the mixture contains other substances that precipitates as well when the mixture is being heated so you will not be able to collect what you want.

6 0

2 years ago

When substances go through chemical changes, which of the following will always happen? A. Exactly one new substance will form a

tiny-mole [99]

I think b is the answer

4 0

2 years ago

Read 2 more answers

A solution of 0.108 M cysteine is titrated with 0.0540 M HNO3 . The pKa values for cysteine are 1.70 , 8.36 , and 10.74 , corres

kirill [66]

Answer:

2.698

Explanation:

Cysteine + H+ <==> cysteineH+

at the equivalence point

cosidering 1 L of cysteine solution

volume of HNO3 to reach equivalence point = 0.108 M x 1 L/0.0540 M = 2.0 L

[CysteineH+] formed = (0.108 M x 1 L)/(1.0 + 2.0) L = 0.036 M

hydrolysis of salt

cysteineH+ + H_2O <==> cysteine + H_3O+

let x amount hydrolyzed

Ka1 = [cysteine][H3O+]/[cysteineH+]

pKa1 = -log[Ka1] = 1.70

Ka = 0.02

so,

0.02 = x^2/(0.036-x)

x^2 + 0.02x - 0.00072 = 0

x = 0.002

pH at equivalence point = -log(0.002) = 2.698

5 0

2 years ago

Calculate the concentration of all ions present in each of the following solutions of strong electrolytes.a. 0.0200 mol sodium p

JulsSmile [24]

Answer:

For a: The concentration of $Na^+\text{ and }PO_4^{3-}$ ions in the solution are 6 M and 2 M respectively.

For b: The concentration of $Ba^{2+}\text{ and }NO_3^{-}$ ions in the solution are 0.5 M and 1.0 M respectively.

For c: The concentration of $K^{+}\text{ and }Cl^{-}$ ions in the solution are 0.051 M and 0.051 M respectively.

For d: The concentration of $NH_4^{+}\text{ and }SO_4^{2-}$ ions in the solution are 1.34 M and 0.67 M respectively.

Explanation:

To calculate the molarity of solution, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}\times 1000}{\text{Volume of solution (in mL)}}$ ......(1)

Or,

$\text{Molarity of the solution}=\frac{\text{Mass of solute}}{\text{Molar mass of solute}\times \text{Volume of solution (in L)}}$ ...(2)

For the given options:

For a:

The chemical formula of sodium phosphate is $Na_3PO_4$

Moles of sodium phosphate = 0.0200 moles

Volume of solution = 10.0 mL

Putting values in equation 1, we get:

$\text{Molarity of the sodium phosphate}=\frac{0.0200\times 1000}{10.0}=2M$

1 mole of sodium phosphate produces 3 moles of $Na^+$ ions and 1 mole of $PO_4^{3-}$ ions

So, concentration of $Na^+\text{ ions}=(3\times 2)=6M$

Concentration of $PO_4^{3-}\text{ ions}=(1\times 2)=2M$

Hence, the concentration of $Na^+\text{ and }PO_4^{3-}$ ions in the solution are 6 M and 2 M respectively.

For b:

The chemical formula of barium nitrate is $Ba(NO_3)_2$

Moles of barium nitrate = 0.300 moles

Volume of solution = 600.0 mL

Putting values in equation 1, we get:

$\text{Molarity of the barium nitrate}=\frac{0.300\times 1000}{600.0}=0.5M$

1 mole of barium nitrate produces 1 mole of $Ba^{2+}$ ions and 2 mole of $NO_3^{-}$ ions

So, concentration of $Ba^{2+}\text{ ions}=(1\times 0.5)=0.5M$

Concentration of $NO_3^{-}\text{ ions}=(2\times 0.5)=1M$

Hence, the concentration of $Ba^{2+}\text{ and }NO_3^{-}$ ions in the solution are 0.5 M and 1.0 M respectively.

For c:

The chemical formula of potassium chloride is KCl

Given mass of potassium chloride = 1.00 g

Molar mass of potassium chloride = 39 g/mol

Volume of solution = 0.500 L

Putting values in equation 1, we get:

$\text{Molarity of the potassium chloride}=\frac{1.00}{39\times 0.500}=0.051M$

1 mole of potassium chloride produces 1 mole of $K^{+}$ ions and 1 mole of $Cl^{-}$ ions

So, concentration of $K^{+}\text{ ions}=(1\times 0.051)=0.051M$

Concentration of $Cl^{-}\text{ ions}=(1\times 0.051)=0.051M$

Hence, the concentration of $K^{+}\text{ and }Cl^{-}$ ions in the solution are 0.051 M and 0.051 M respectively.

For d:

The chemical formula of ammonium sulfate is $(NH_4)_2SO_4$

Given mass of ammonium sulfate = 132 g

Molar mass of ammonium sulfate = 132 g/mol

Volume of solution = 1.50 L

Putting values in equation 1, we get:

$\text{Molarity of the ammonium sulfate}=\frac{132}{132\times 1.50}=0.67M$

1 mole of ammonium sulfate produces 2 moles of $NH_4^{+}$ ions and 1 mole of $SO_4^{2-}$ ions

So, concentration of $NH_4^{+}\text{ ions}=(2\times 0.67)=1.34M$

Concentration of $SO_4^{2-}\text{ ions}=(1\times 0.67)=0.67M$

Hence, the concentration of $NH_4^{+}\text{ and }SO_4^{2-}$ ions in the solution are 1.34 M and 0.67 M respectively.

4 0

2 years ago