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SVETLANKA909090 [29]
2 years ago
3

You have a sample of zinc (Zn) and a sample of aluminum (Al). You have an equal number of atoms in each sample. Which of the fol

lowing statements concerning the masses of the samples is true?
A. The mass of the zinc sample is more than the mass of the aluminum sample, but it is not twice as great.

B. The masses of each sample are equal.

C. The mass of the aluminum sample is more than the mass of the zinc sample, but it is not twice as great.

D. The mass of the zinc sample is more than twice as great as the mass of the aluminum sample.

E. The mass of the aluminum sample is more than twice as great as the mass of the zinc sample.
Chemistry
1 answer:
Readme [11.4K]2 years ago
5 0

Answer:

ans-D

Explanation:

As given in question

number of atoms in each sample are equal,

Suppose number of atoms in each sample=X;

number of mole = number of molecule(N)/Avogadro number(NA);

Number of mole of Zn=X/NA;

Number of mole of Al=X/NA;

Where NA=avogadro number;

Mass of Zn sample=mole * molecular weight of Zn

Mass of Zn sample=(X/NA)*65

Mass of Al sample=mole * molecular weight of Al

Mass of Al sample=(X/NA)*27;

Thus masses are:

Mass of Zn sample=(X/NA)*65

Mass of Al sample=(X/NA)*27

from above value it is clearly that mass of Zn is more than twice of mass of Al

that is Massof Zn>2*Mass of Al

hence option D is correct

You might be interested in
The number of p atoms in 1.0 g of ba3(po4)2 is:
marta [7]
First, we determine the number of moles of barium phosphate. This is done using:

Moles = mass / Mr
Moles = 1 / 602
Moles = 0.002

Now, we see from the formula of the compound that each mole of barium phosphate has 2 moles of phosphorus (P). Therefore, the moles of phosphorus are:
0.002 * 2 = 0.004

The number of particles in one mole of substance is 6.02 x 10²³. The number of phosphorus atoms will be:
0.004 * 6.02 x 10²³

2.41 x 10²¹ atoms of phosphorus are present
3 0
2 years ago
A student assistant is cleaning up after a chemistry laboratory exercise and finds three one-liter bottles containing alcohol so
Svetllana [295]

The approximate alcohol content is 210 ml.

Explanation:

It can be deduced from the question that each bottle is of 1000ml or 1 litre.

The first bottle is one half full means it has 500 ml of solution and it has 20% alcohol in it. So volume of alcohol in the solution is

20/100*500

=100 ml

The first bottle is one fifth full, so the volume of mixture is 1/5th of 1000ml

so it is 200ml having 30% alcohol

30/100*200

= 60 ml

The third bottle is one tenth full so its volume is 1/10*1000

100 ml.  having 50% of alcohol

50/100*100

50 ml.

The alcohol content obtained from all these 3 litres is:

100+60+50

= 210 ml of alchohol is obtained from 800 ml of mixture.

3 0
2 years ago
How many grams of NH3 can be prepared from the synthesis of 77.3 grams of nitrogen and 14.2 grams of hydrogen gas?
lbvjy [14]

Answer:

80.41 g

Explanation:

Data Given:

Mass of Nitrogen (N₂) = 77.3 g

Mass of Hydrogen (H₂) = 14.2 g

many grams of NH₃ = ?

Solution:

First we look at the balanced synthesis reaction

              N₂   +    3 H₂  ------—> 2 NH₃

             1 mol      3 mol

As 1 mole of Nitrogen react with 3 mole of hydrogen

Convert moles to mass

molar mass of N₂ = 2(14) = 28 g/mol

molar mass of H₂ = 2(1) + 2 g/mol

Now

                     N₂             +           3 H₂        ------—>      2 NH₃

             1 mol (28 g/mol)     3 mol(2g/mol)

                    28 g                        6 g

28 grams of N₂ react with 6 g of H₂  

So

if 28 grams of N₂ produces 6 g of H₂  so how many grams of N₂ will react with 14.2 g of H₂.

Apply Unity Formula

                 28 g of N₂ ≅ 6 g of H₂

                 X g of N₂ ≅ 14.2 g of H₂

Do cross multiply

                X g of N₂ = 28 g x 14.2 g / 6 g

                X g of N₂ = 66.3 g

As we have given with 77. 3 g of N₂ but from this calculation we come to know that 66.3 g will react with 14.2 g of hydrogen and the remaining 10 g N₂ will be in excess

So, Hydrogen is limiting reactant in this reaction and the amount of NH₃ depends on the amount of hydrogen.

Now

To find mass of NH₃ we will do following calculation

Look at the reaction

As we Know

                     N₂             +           3 H₂        ------—>      2 NH₃

                                                   6 g                            2 mol

So, 6 g of hydrogen gives 2 moles of NH₃, then how many moles of NH₃ will be produce by 14.2 g

Apply Unity Formula

                 6 g of H₂ ≅ 2 mol of NH₃

                14.2 g of H₂ ≅ X mol of NH₃

Do cross multiply

               X mol of NH₃= 14.2 g x 2 mol / 6 g

                X mol of NH₃ = 4.73 mol

So, 14.2 g of hydrogen gives 4.73 moles of NH₃

Now

Convert moles of NH₃ to mass

Formula will be used

        mass in grams = no. of moles x molar mass . . . . . . (2)

Molar mass of  NH₃

Molar mass of  NH₃ = 14 + 3(1)

Molar mass of  NH₃ = 14 + 3 = 17 g/mol

Put values in equation 2

        mass in grams = 4.73 mole x 17 g/mol

        mass in grams =  80.41 g

mass of NH₃=  80.41 g

3 0
2 years ago
4. Convert the following: a. 4g mol of MgCl2 to g b. 2 lb mol of C3H8 to g c. 16 g of N2 to lb mol d. 3 lb of C2H6O to g mol
Nuetrik [128]

Answer:

a) 381.2 g

b) 39916 g

c) 0.0013 lb mol

d) 29.6 g mol

Explanation:

The molecular weight (mw) of a compound is the mass of it per mole, so it's the ratio of the mass (m) per mole (n).

a) The molecular weight of one mol is found at the periodic table. So, for Mg, mw = 24.3 g/mol, for Cl = 35.5 g/mol, so for MgCl2, mw = 24.3 + 2*35.5 = 95.3 g/mol. The g mol is the mass divided by the molecular weight:

g mol = m/mw

4 = m/95.3

m = 381.2 g

b) The pound (lb) is a unity of mass, and the lb mol is a unity of the mass divided by the molecular weight. So, by the periodic table, the molecular weight of C3H8 is 3*12 (of C) + 8*1 (of H) = 44 lb/mol.

lb mol = m/mw

2 = m/44

m = 88 lb

1 lb = 453.592 g

So, m = 88*453.592 = 39916 g

c) The molecular weight of N2 is 2*14 (of N) = 28 lb/mol.

m = 16/453.592 = 0.0353 lb

lb mol = m/mw

lb mol = 0.0353/28

lb mol = 0.0013 lb mol

d) The molecular weight is 2*12 (of C) + 6*1(of H) + 1*16(of O) = 46 g/mol

3 lb = 1360.78 g

g mol = m/mw

g mol = 1360.78/46

g mol = 29.6 g mol

6 0
2 years ago
The plant food contains nh4)3po4 what tests would you run to verify the presence of the nh4 ion and the po4 ion
algol13

For the presence of ammonium ion, there is a need to add sodium hydroxide solution to the water and warm the mixture. Test any vapor that gets produced with damp red litmus paper. It should turn blue as ammonia gas is discharged, which is alkaline. The ionic equation for the reaction is:  

NH₄⁺ + OH⁻ ⇒ NH₃ + H₂O

For the presence of phosphate ions, the addition of barium ions is done. The ionic equation is:  

3Ba₂⁺ + 2PO4³⁻ ⇒ Ba₃ (PO₄)₂ (precipitate)


8 0
2 years ago
Read 2 more answers
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