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OlgaM077 [116]
3 years ago
10

Which statement accurately compares the trends in atomic number and atomic mass in the periodic table?

Chemistry
2 answers:
Kruka [31]3 years ago
7 0

Answer : The correct option is, (C) Both the atomic mass and the atomic number increase from left to right.

Explanation :

The general trend of atomic number and atomic mass in the periodic table is,

Both atomic number and atomic mass increase from left to right and decreases from right to left in the periodic table due to the addition of the number of neutrons and the number of protons in the nucleus.

Hence, the correct option is, (C) Both the atomic mass and the atomic number increase from left to right.

slava [35]3 years ago
4 0

C.Both the atomic mass and the atomic number increase from left to right.

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As a heterogeneous mixture
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What is true about all chemical equations?<br> Links are reported
Vanyuwa [196]

Answer:Your answer is A

Explanation:

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6 0
3 years ago
Does a displacement reaction take place in 'magnesium + lead nitrate'? and if so why?
cestrela7 [59]

Answer:

Maybe or maybe not (not sure)

Explanation:

A displacement reaction is a type of reaction where one element is displaced by another from a compound.

In the case of magnesium and lead nitrate, magnesium is more reactive than lead. Therefore, it will displace lead from lead nitrate to form magnesium nitrate and lead.

The reaction can be represented as:

Mg(s) + Pb(NO3)2(aq) → Mg(NO3)2(aq) + Pb(s)


Another answer could be;

A displacement reaction does not take place in 'magnesium + lead nitrate' because magnesium is more reactive than lead.

5 0
2 years ago
I RLLY NEED HELP. Use Image B (picture above) and calculate the density of a ring that has a mass of 32 grams. Read Page 11 "Mea
dimaraw [331]

Answer:

Mass of ring = 32 g

Volume of ring = 4 mL

Density of ring = 8 g/mL

Explanation:

From the question given above, the following data were obtained:

Mass of ring = 32 g

Volume of water = 64 mL

Volume of water + ring = 68 mL

Density of ring =?

Next, we shall determine the volume of the ring. This can be obtained as follow:

Volume of water = 64 mL

Volume of water + ring = 68 mL

Volume of ring =?

Volume of ring= (Volume of water + ring) – (Volume of water)

Volume of ring = 68 – 64

Volume of ring = 4 mL

Finally, we shall determine the density of the ring. This can be obtained as follow:

Mass of ring = 32 g

Volume of ring = 4 mL

Density of ring =?

Density = mass / volume

Density of ring = 32 / 4

Density of ring = 8 g/mL

8 0
3 years ago
How many grams of hydrochloric acid are produced when 15.0 grams NaCl react with excess H2SO4 in the reaction
e-lub [12.9K]

Answer:

11.65 g

Explanation:

Amount of NaCl  = 15.0 grams

amount of H₂SO₄ = Excess

mass of hydrochloric acid (HCl) = ?

Solution:

To solve this problem first we will look for the reaction that NaCl react with  H₂SO₄

Reaction:

         2NaCl + H₂SO₄ --------> 2 HCl + Na₂SO₄

As the  H₂SO₄ is in excess so the amount of hydrochloric acid (HCl) depends on the amount of NaCl as it its act as limiting reactant.

Now if we look at the reaction

         2NaCl + H₂SO₄ --------> 2HCl + Na₂SO₄

           1 mol                              2 mol  

Now convert moles to mass

Then

Molar mass of NaCl = 23 + 35.5 = 58.5 g/mol

Molar mass of HCl = 1 + 35.5 = 36.5 g/mol

So,

            2NaCl           +      H₂SO₄     -------->       2HCl        +     Na₂SO₄

      2 mol (58.5 g/mol)                                    2 mol (36.5 g/mol)

              94 g                                                           73 g

So if we look at the reaction; 94 g of NaCl gives with 73 g of hydrochloric acid (HCl) in a this reaction, then how many grams of hydrochloric acid (HCl) will produce from 15.0 g of NaCl

For this apply unity formula

           94 g of NaCl  ≅ 73 g of HCl

           15.0 g of NaCl  ≅ X g of HCl

By Doing cross multiplication

           X g of HCl = 73 g x 15.0 g / 94 g

           X g of HCl = 11.65 g

11.65 g of hydrochloric acid (HCl) will produce by 15.0 g of NaCl

7 0
3 years ago
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