The picture is above^ someone help me please :)

When might neutralization reactions be used in a laboratory setting?

Ipatiy [6.2K]

Neutralization reactions can be used in a laboratory setting in order to dispose of chemicals. When spills happens, for instance an acid is on the floor, you can use a base to neutralize the spill. Hope this answers the question. Have a nice day.

6 0

2 years ago

Tritium is an isotope of hydrogen (H) that has:

Anika [276]

D. 1 proton and 1 neutron

7 0

3 years ago

Read 2 more answers

The bonding found in calcium chloride is

photoshop1234 [79]

The bonding found in calcium chloride is ionic bonds.

I hope this helps!

4 0

3 years ago

2NOCl(g) ↔ 2NO(g) + Cl2(g) with K = 1.6 x10–5. In an experiment, 1.00 mole of pure NOCl and 1.00 mole of pure Cl2 are placed in

poizon [28]

Answer: $3.8\times 10^{-3}M$

Explanation:

Moles of $NOCl$ = 1 mole

Moles of $Cl_2$ = 1 mole

Volume of solution = 1 L

Initial concentration of $NOCl$ = 1 M

Initial concentration of $Cl_2$ = 1 M

The given balanced equilibrium reaction is,

$2NOCl(g)\rightleftharpoons 2NO(g)+Cl_2(g)$

Initial conc. 1 M 0M 1 M

At eqm. conc. (1-2x) M (2x) M (1+x) M

The expression for equilibrium constant for this reaction will be,

$K_c=\frac{[NO]^2[Cl_2]}{[NOCl]^2}$

The $K_c$ = $1.6\times 10^{-5}}$

Now put all the given values in this expression, we get :

${1.6\times 10^{-5}}=\frac{(2x)^2\times (1+x)}{(1-2x)^2}$

By solving the term 'x', we get :

$x=0.0019$

Concentration of $NO$ at equilibrium= (2x) M = $2\times 0.0019=3.8\times 10^{-3}M$

6 0

2 years ago

Five calcite, CaCO3 (MM 100.085 g/mol), samples of equal mass have a total mass of 12.2±0.1 g . What is the average mass and abs

Westkost [7]

Answer:

The average mass of calcium in each sample is: 0.978 g

The absolute uncertainty is: 0.008 g

Explanation:

The absolute uncertainty of the total samples indicated in the statement is ± 0.1 g.

When you multiply or divide quantities with uncertainties, you calculate the final uncertanty by adding the relative uncertainties together.

The relative uncertainty is the absolute uncertainty divided by the quantity:

Relative uncertainty = 0.1g / 12.2 g = 0.008

The average mass of calcium is calculated using proportions, along with the molar masses:

Molar mass of calcium: 40.078 g/ mol (from a periodic table)

Molar mass of calcite: 100.085 g/mol (given)

Proportion:

40.078 g of calcium / 100.085 g of calcite = x / 12.2 g of calcite

x = 12.2 × 40.078 / 100.085 g = 4.89 g calcium

So the total mass of calcium in the five samples is 4.89 g, and the average mass in each sample is:

Average mass = total mass of five samples / number of samples

Average mass = 4.89 g / 5 = 0.978 g of calcium

So, the first answer is that the average mass of calcium in each sample is 0.978 g ( keep 3 signficant figures, such as the quntitiy 12.2 shows, as you have only used multiplication and division).

The absolute uncertainty of each sample is the relative uncertainty multiplied by the average mass of calcium of the five samples, rounded to one decimal:

Absolute uncertainty = 0.978 g × 0.008 ≈ 0.008 g

The answer to the secon question is that the absolute uncertaingy of calcium in each sample is 0.008 g.

7 0

3 years ago