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mina [271]
3 years ago
15

Sodium Oxide (Na2O) can react with hydrochloric acid (HCl) to produce sodium chloride (NaCl) and water (H2O) according to the fo

llowing equation: Na2O + 2HCl -> 2NaCl + H2O How many total atoms are present before and after the reaction? A) 2 before and 2 after
B) 3 before and 3 after
C) 5 before and 5 after
D) 7 before and 7 after
Chemistry
2 answers:
nika2105 [10]3 years ago
6 0

The answer is D) 7 before and 7 after

Just had this question

Sonbull [250]3 years ago
5 0

Answer:

D) 7 before and 7 after

Step-by-step explanation:

  Na₂O   +  2HCl    ⟶    2NaCl      +   H₂O

3 atoms + 4 atoms ⟶ 2×2 atoms + 3 atoms

        7 atoms         ⟶    4 atoms   + 3 atoms

        7 atoms         ⟶            7 atoms

There are seven atoms present before and after the reaction.


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Why do igneous rocks not contain fossils in them ?
Mashutka [201]

Answer:

Igneous rocks do not contain any fossils.

Explanation:

This is because any fossils in the original rock will have melted when the rock melted to form magma. Sedimentary rocks are formed from the broken remains of other rocks that become joined together.

6 0
3 years ago
PLEASE HELP!
svetoff [14.1K]

Answer:

period 3 and group 3

Explanation:

I'm saying group 3 because that is how I learnt it at school, but if you count it then it's in group 13.

8 0
3 years ago
How plastic polymer types of compositions, surface hydrophobicity, crystallinity, porosity, size, shapes, aging or degree of wea
Andreas93 [3]

Explanation:

Scientific evidences abound of the occurrence of plastic pollution, from mega- to nano-sized plastics, in virtually all matrixes of the environment. Apart from the direct effects of plastics and microplastics pollution such as entanglement, inflammation of cells and gut blockage due to ingestion, plastics are also able to act as vectors of various chemical contaminants in the aquatic environment. This paper provides a review of the association of plastic additives with environmental microplastics, how the structure and composition of polymers influence sorption capacities and highlights some of the models that have been employed to interpret experimental data from recent sorption studies. The factors that influence the sorption of chemical contaminants such as the degree of crystallinity, surface weathering, and chemical properties of contaminants. and the implications of chemical sorption by plastics for the marine food web and human health are also discussed. It was however observed that most studies relied on pristine or artificially aged plastics rather than field plastic samples for studies on chemical sorption by plastics.

4 0
3 years ago
NaOH (aq) + HCl(aq) H2O + NaCl (aq)
yuradex [85]

Answer:

The answer to your question is 0.005

Explanation:

Data

Volume of NaOH = 25 ml

[NaOH] = 0.2 M

moles of NaOH = ?

To solve this problem is not necessary to have the chemical reaction. Just use the formula of Molarity and solve it for moles.

Formula

Molarity = moles / volume

-Solve for moles

moles = Molarity x volume

-Convert volume to liters

          1000 ml ---------------- 1 l

              25 ml ---------------- x

               x = (25 x 1) / 1000

               x = 0.025 l

-Substitution

moles = 0.2 x 0.025

-Result

moles = 0.005

7 0
3 years ago
Calculate the percent ionization of propionic acid (C2H5COOH) in solutions of each of the following concentrations (Ka is given
tekilochka [14]

Answer:

a) = 0.704%

b) = 1.30%

c) = 2.60%

Explanation:

Given that:

K_a= 1.34*10^{-5

For Part A; where Concentration of A = 0.270 M

Percentage Ionization(∝)  \alpha = \sqrt{\frac{K_a}{C} }

\alpha = \sqrt{\frac{1.34*10^{-5}}{0.270} }

\alpha = \sqrt{4.9629*10^{-5}}

\alpha = 7.044*10^{-3

percentage% (∝) = 7.044*10^{-3}*100

= 0.704%

For Part B; where Concentration of B = 7.84*10^{-2 M

\alpha = \sqrt{\frac{1.34*10^{-5}}{7.84*10^{-2}} }

\alpha = \sqrt{1.709*10^{-4} }

\alpha = 0.0130\\

percentage% (∝) = 0.0130 × 100%

= 1.30%

For Part C; where Concentration of C= 1.92*10^{-2} M

\alpha = \sqrt{\frac{1.34*10^{-5}}{1.97*10^{-2}} }

\alpha = \sqrt{6.802*10^{-4}}

\alpha =0.02608

percentage% (∝) = 0.02608  × 100%

= 2.60%

7 0
3 years ago
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