What results when two or more atoms share electrons equally?

What is the mass of 4 moles of almunium atom

lana [24]

Answer:

108 grams

Explanation:

5 0

3 years ago

The gravitational force exerted by an object is given by F = mg, where F is the force in newtons, m is the mass in kilograms, an

Andre45 [30]

The height of the column is 0.457 m and the mass of the atmosphere is calculated as 1.03 × 10 ⁴ kg.

<h3>What is Pascal? </h3>

Pascal is defined as the force per unit area. It expressed in Newton pr square meter of area.

1 Pa = 1 N / m²

Pressure = force / Area

According to the question, the expression of force is as given below

F = mg

where,

F is the force,

m is the mass,

g is equal to the acceleration due to the gravity

Now, the area of the atmosphere is 1 m².The pressure is 1 atm. Pressure in Pascal.

1 atm = 1.01325 × 10 5pa

Therefore, the expression for pascal become as follows.

1.01325 × 10 5 pa = mg /area

1.01325 × 10 5 pa = m × 9.81 m/s² / 1 m²

M = 1.01325 × 10 5 pa × 1 m² / 9.81 m² × 1 Nm -² /1 pa × 1 kg m-² / 1 N

1.03 × 10 ⁴ kg

Given,

The density is 22.6 g /mL , pressure is 1 atm, and area is 1 m²

The relation between density and pressure can be given as follows.

P = hpg… … …(1 )

were , h is the height of the column

p is the density.

Hpg = 1.01325 × 10 5 pa × 1 N/m² /1 pa

H = 1.01325 × 10 5 N/m² / pg × 1 kg ms-² / 1 N

= 1.01325 × 10 5 kg m-¹ s -² / 22.6 g mL -1 × 1 kg/ 10 ³ g × 1 mL / 10 -6 m ³ × 9.81 m s- ²

= 0.457 m

Therefore, the height of the column is 0.457 m.

Thus, we concluded that the height of the column is 0.457 m and the mass of the atmosphere is calculated as 1.03 × 10 ⁴ kg.

learn more about density:

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7 0

2 years ago

Which net ionic equation best represents the reaction that occurs when an aqueous solution of barium chloride is mixed with an a

vovangra [49]

Answer : The net ionic equation will be:

$Ba^{2+}(aq)+SO_4^{2-}(aq)\rightarrow BaSO_4(s)$

Explanation :

Complete ionic equation : In complete ionic equation, all the substance that are strong electrolyte and present in an aqueous are represented in the form of ions.

Net ionic equation : In the net ionic equations, we are not include the spectator ions in the equations.

Spectator ions : The ions present on reactant and product side which do not participate in a reactions. The same ions present on both the sides.

The balanced molecular equation will be,

$BaCl_2(aq)+Li_2SO_4(aq)\rightarrow 2LiCl(aq)+BaSO_4(s)$

The complete ionic equation in separated aqueous solution will be,

$Ba^{2+}(aq)+2Cl^-(aq)+2Li^+(aq)+SO_4^{2-}(aq)\rightarrow 2Li^+(aq)+2Cl^-(aq)+BaSO_4(s)$

In this equation the species, $Li^+\text{ and }Cl^{-}$ are the spectator ions.

By removing the spectator ions , we get the net ionic equation.

The net ionic equation will be:

$Ba^{2+}(aq)+SO_4^{2-}(aq)\rightarrow BaSO_4(s)$

6 0

3 years ago

HELP PLS !!!! :))Formula unit of Sr and CI<br><br> Formula unit of AI and S

xenn [34]

Answer : The formula unit of Sr & Cl is $SrCl_2$ and the formula unit of Al & S is $Al_2S_3$

Explanation :

Formula unit : it is defined as the lowest number ratio of ions of an elements in an ionic compound or covalent compound.

For the formula unit of Sr & Cl, two chloride ions $(Cl^-)$ are needed to neutralize the one strontium ion $(Sr^{2+})$ .

For the formula unit of Al & S, three sulfide ions $(S^{2-})$ are needed to neutralize the two aluminium ion $(Al^{3+})$ .

The formula unit of $SrCl_2$ and $Al_2S_3$ are shown below.

4 0

3 years ago

Aqueous sulfuric acid reacts with solid sodium hydroxide to produce aqueous sodium sulfate and liquid water . If of sodium sulfa

notka56 [123]

Answer:

27%

Explanation:

Hello,

The following information is missing, but I found it: "1.92 g of sodium sulfate is produced from the reaction of 4.9 g of sulfuric acid and 7.8 g of sodium hydroxide" so the undergoing chemical reaction is:

$2NaOH+H_2SO_4-->Na_2SO_4+2H_2O$

Now, to compute the percent yield, we must first establish the limiting reagent to subsequently determine the theoretical yield of sodium sulfate because the real (1.92g) is already given, thus, we consider the following procedure:

$n_{NaOH}=7.8gNaOH*\frac{1molNaOH}{40gNaOH}=0.2molNaOH\\n_{H_2SO_4}=4.9gH_2SO_4*\frac{1molH_2SO_4}{98gH_2SO_4}=0.050molH_2SO_4\\$

- The moles of sodium hydroxide that completely react with 0.05 moles of sulfuric acid are:

$0.2molNaOH*\frac{1molH_2SO_4}{2molNaOH}=0.098molH_2SO_4$

As this number is higher than the previously computed 0.05 moles of available sulfuric acid, one states that the sulfuric acid is the limiting reagent. Now, the theoretical grams of sodium sulfate are found via:

$0.05molH_2SO_4*\frac{1molNa_2SO_4}{1mol H_2SO_4} *\frac{142.04gNa_2SO_4}{1molNa_2SO_4} =7.1gNa_2SO_4$

Finally, the percent yield turns out into:

$Y=\frac{1.92g}{7.1g} *100$

$Y=27.0$ %

Best regards.

6 0

3 years ago