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Scilla [17]
3 years ago
9

How many ATOMS of OXYGEN are there in the following compound

Chemistry
1 answer:
STatiana [176]3 years ago
5 0
6 Atoms!
Mg = 1 atom.
O = 4 atoms.
A = 1 atom.
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Ice has a density of 0.92g/cm3. It will float in water.
SVEN [57.7K]

The density of pure water is 1 g/cm^3.

Its density is 0.98 g cm 3 at room temperature, in comparison with the handiest zero.92 g cm 3 for ice, a reality that has to be defined through atomic, and molecular concepts. If ice has been no longer much less dense than water, it might sink, having a devastating impact on lake backside ecosystems. believe it or now not, ice is honestly about 9% much less dense than water. for the reason that water is heavier, it displaces the lighter ice, causing the ice to glide to the pinnacle.

The density of ice is about 90 percent that of water, but that could range because ice can contain air, too. meaning that about 10 percent of an ice cube or iceberg will be above the water line. The density of water is maximum at four∘C, and the density of the ice is much less than the water due to its susceptible intermolecular pressure of attraction. as the density of water is more, it's miles heavier than ice. therefore ice floats on the floor of the water. Ice continually floats due to the fact it's far less dense than everyday water. because frozen water molecules shape a crystal.

Learn more about A density here:-brainly.com/question/17780219

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6 0
11 months ago
Sodium metal and water react to form hydrogen and sodium hydroxide. If 1.99 g of sodium react with water to form 0.087 g of hydr
zavuch27 [327]

Answer:

1.56 g of water was involved in the reaction

Explanation:

From the stoichiometric equation

2Na + 2H2O = 2NaOH + H2

NB : Mm Na= 23, Mm H2O = ( 2+16)= 18

2(23) of Na requires 2(18) of water

Hence 1.99g of Na will require 1.99×2×18/2(23) of water = 1.56 g of water

7 0
3 years ago
Read 2 more answers
A chemist prepares a solution of copper(II) fluoride by measuring out of copper(II) fluoride into a volumetric flask and filling
Simora [160]

The question is incomplete, here is the complete question.

A chemist prepares a solution of copper(II) fluoride by measuring out 0.0498 g of copper(II) fluoride into a 100.0mL volumetric flask and filling the flask to the mark with water.

Calculate the concentration in mol/L of the chemist's copper(II) fluoride solution. Round your answer to 3 significant digits.

<u>Answer:</u> The concentration of copper fluoride in the solution is 4.90\times 10^{-3}mol/L

<u>Explanation:</u>

To calculate the molarity of solute, we use the equation:

\text{Molarity of the solution}=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}

We are given:

Given mass of copper (II) fluoride = 0.0498 g

Molar mass of copper (II) fluoride = 101.54 g/mol

Volume of solution = 100.0 mL

Putting values in above equation, we get:

\text{Molarity of copper (II) fluoride)=\frac{0.0498\times 1000}{101.54\times 100.0}\\\\\text{Molarity of copper (II) fluoride}=4.90\times 10^{-3}mol/L

Hence, the concentration of copper fluoride in the solution is 4.90\times 10^{-3}mol/L

4 0
2 years ago
Water droplets forming on the inside of a cold windshield is an example of
8_murik_8 [283]
Condensation I think.
4 0
3 years ago
Read 2 more answers
A first order reaction has a rate constant of 0.543 at 25 c and 6.47 at 47
alukav5142 [94]

The activation energy Ea can be related to rate constant (k) at temperature (T) through the equation:

ln(k2/k1) = Ea/R[1/T1 - 1/T2]

where :

k1 is the rate constant at temperature T1

k2 is the rate constant at temperature T2

R = gas constant = 8.314 J/K-mol

Given data:

k1 = 0.543 s-1; T1 = 25 C = 25+273 = 298 K

k2 = 6.47 s-1; T = 47 C = 47+273 = 320 K

ln(6.47/0.543) = Ea/8.314 [1/298 - 1/320]

2.478 = 2.774 *10^-5 Ea

Ea = 0.8934*10^5 J = 89.3 kJ

5 0
2 years ago
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