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Papessa [141]
3 years ago
6

What activity is considered a proper laboratory technique chem?

Chemistry
1 answer:
ale4655 [162]3 years ago
4 0

One of the activity of laboratory is the addition of acid to water.

The addition of acid to water make it easier for the reaction so that the acid gets diluted safely. On the other hand if water is added to acid it will get splash vigrously and may injure the hand of the chemist. The addition of water directly to acid contribute to exothermic reaction and releases more heat which may result in spraying around.

Adding acid to water, result in solution with very dilute solution and releases small heat. This small heat is not sufficient even to vaporizes a single drop of water.

Thus, adding acid to water is safer for dilution.


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NO2+H4➡2 H2O+N What are the elements
Firdavs [7]
They are nitrogen, oxygen, hydrogen
6 0
3 years ago
Read 2 more answers
Find the volume of a gold ring that has a mass of 12.00 grams. The density of gold is 19.30 g/mL. The volume is
kozerog [31]

Answer:

<h3>The answer is 0.622 mL</h3>

Explanation:

The volume of a substance when given the density and mass can be found by using the formula

volume =  \frac{mass}{density}  \\

From the question

mass = 12 g

density = 19.30 g/mL

We have

volume =  \frac{12}{19.30}  \\  = 0.62176165...

We have the final answer as

<h3>0.622 mL</h3>

Hope this helps you

3 0
3 years ago
Which quantity will increase if the temperature is raised?
Irina-Kira [14]
OD.......................
4 0
3 years ago
Suppose 12.0 g of carbon (c) reacts with 70.0 g of sulfur (s) to give 76.0 g of the compound carbon disulfide 1 c s 2 2 . In the
Nitella [24]

Answer is: mass of unused sulfur is 5.87 grams.

Balanced chemical reaction: C + 2S → CS₂.

m(C) = 12.0 g; mass of carbon.

m(S) = 70.0 g; mass of sulfur.

n(C) = m(C) ÷ M(C).

n(C) = 12 g ÷ 12 g/mol.

n(C) = 1 mol; amount of substance.

n(S) = m(S) ÷ M(S).

n(S) = 70 g ÷ 32.065 g/mol.

n(S) = 2.183 mol.

From chemical reaction: n(C) : n₁(S) = 1 : 2.

n₁(S) = 1 mol · 2 = 2 mol.

Δn(S) = n(S) - n₁(S).

Δn(S) = 2.183 mol - 2 mol.

Δn(S) = 0.183 mol; amount of unused sulfur.

Δm(S) = 0.183 mol · 32.065 g/mol.

Δm(S) = 5.87 g.

4 0
3 years ago
If the value of kc at 25oc is 3.7108, and the equilibrium concentrations for n2 and h2 are 0.000105 m and 0.0000542 m, respectiv
frez [133]

Equation of decomposition of ammonia:

N2+3H2->2NH3

Euilibrium constant:

Kc=(NH3)^2/((N2)((H2)^3))

As concentration of N2=0.000105, H2=0.0000542

so equation will become:

3.7=(NH3)^2/(0.000105)*(0.0000542)^3

NH3=√(3.7*0.000105*(0.0000542)^3)

NH3=7.8×10⁻⁹

So concentration of ammonia will be 7.8×10⁻⁹.

5 0
3 years ago
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