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2 years ago

Write down five laboratory safety rules.

Chemistry

1 answer:

OLEGan [10]2 years ago

5 0

Answer:

Five Laboratory Safety Rules:

1). Do not eat in the laboratory.

2). Do not touch any chemical or reagent unless you are told to do so.

3). Neither play in lab, nor sit on the table.

4). Don't remove labels on any reagent.

5). Don't taste anything in the laboratory, no matter how familiar it appears.

Hope it helps.

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Explain the volume of ideal gasses as related to the molar concept.

horrorfan [7]

Answer:

n = Initial volume/22.4L

Explanation:

The molar concept is simply one that is used to find the Number of moles and explain the relationship it has with avogadro's number, molecular mass, molar mass e.t.c.

Now, in terms of molar mass, number of moles is given by the formula;

n = mass of the sample/molar mass

In terms of avogadro's number, number of moles is;

1 mole = avogadro's number = 6.02 × 10^(23)

Now, when dealing with ideal gases, the molar volume of an ideal gas is 22.4 L.

Now the relationship between this volume and the mole concept is that the number of moles is gotten by dividing the initial volume by this molar volume.

Thus;

n = Initial volume/22.4L

6 0

2 years ago

If the mass of a box is 140 g, and the volume is 8 cm3, then the density of the box = ? socratic.org

Ksivusya [100]

Answer:

17.5 g/cm³

Explanation:

We can solve this particular problem by keeping in mind the definition of density:

Density = mass / volume

As the problem gives us both the mass and the volume of the box, we can now proceed to calculate the density:

Density = 140 g / 8 cm³

Density = 17.5 g/cm³

The density of the box is 17.5 g/cm³.

4 0

2 years ago

41 g of MgCl^2 is how many moles of magnesium chloride?

kari74 [83]

Answer:

0.43 mol.

Explanation:

Knowing that the no. of moles can be calculated using the relation:

no. of moles (n) = mass/molar mass

mass of MgCl₂ = 41.0 g & molar mass of MgCl₂ = 95.211 g/mol.

∴ n = mass/molar mass = (41.0 g)/(95.211 g/mol) = 0.43 mol.

6 0

2 years ago

Nitric acid, a major industrial and laboratory acid, is produced commercially by the multi-step Ostwald process, which begins wi

Marizza181 [45]

Answer: 3.178 Kg of ammonia must be used to produce 7.839 kg of nitric acid and The equations are not balanced. Explanation: Ostwald process is a multi-step for manufacturing Nitric Acid. First, We must check if our equations are balanced or not, by checking the amount of each element before and after the arrow. Step 1: NH3(g) + O2(g) ⟶ NO(g) + H2O(l) Step 2: NO(g) + O2(g) ⟶ NO2(g) Step 3: NO2(g) + H2O(l) ⟶ HNO3(l) + NO(g) For example in Step 1, the amount of Hydrogen atoms are different on both sides. On left side we can count 3 hydrogen atoms while on the right side we can count only 2 hydrogen atoms. In the Step 2, the amount of Oxygen atoms are different on both sides too. On the Left side we can count 3 oxygen atoms while on the right side we can count only 2 oxygen atoms and in step 3 the amount of Nitrogen atoms are different on both sides too. On the left side we can count 1 nitrogen atom while on the right side we can count 2 nitrogen atoms. So, Let´s balance equations by inspection, just adding coefficients to each compound, until the amount of atoms are equal on both sides. Step 1: 4NH3(g) + 5O2(g) ⟶4 NO(g) +6 H2O(l) Step 2: 2NO(g) + O2(g) ⟶ 2NO2(g) Step 3: 3NO2(g) + H2O(l) ⟶ 2HNO3(l) + NO(g) Second we gather the information what we are going to use in our calculations. Final Mass of HNO3 = 7,839Kg and Molecular Weigth = 63,01g/mol NH3 Molecular Weight= 17,031 g/mol Third we start discoverying the amount of NH3 that reacted completed to generate 7,839Kg of HNO3, using the giving equation and its respective molecular weights. 7,839Kg of HNO3 x 1000g / 1Kg x 1mol HNO3 / 63,01g HNO3 x 3mol NO2 / 2 mol HNO3 x 2 mol NO/ 2 mol NO2 x 4 mol NH3/4 mol NO x 17,031g NH3/1 mol NH3 x 1 Kg / 1000g= 3,178 Kg NH3 The amount of NH3 that is required to produce 7,839 Kg of HNO3 is 3,178 Kg NH3

7 0

2 years ago

How many Grams would he present if 1 mole sample of sodium were collected?

Zigmanuir [339]

Mass = moles x Mr
So in this case the moles =1
And the Mr equals =11
So you would do - 11x1= 11grams

8 0

2 years ago