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3 years ago

How are metals and nonmetals similar

Chemistry

1 answer:

Elza [17]3 years ago

6 0

Answer:

As compared to metals, they have low density and will melt at low temperatures. The shape of nonmetals cannot be changed easily because they are brittle and will break. Elements that have properties of both metals and nonmetals are called metalloids. They can be shiny or dull and their shape is easily changed.

Explanation:

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A person produces two sound waves with a flute, one immediately after the other. Both sound waves have the same pitch, but the s

Nitella [24]

Answer:

A) wavelength

Explanation:

6 0

3 years ago

Does any solid Ag₂CrO₄ form when 2.7x10⁻⁵g of AgNO₃ is dissolved in 15.0 mL of 4.0x10⁻⁴MK₂CrO₄?

e-lub [12.9K]

Molarity of Ag+ is less than the molar solubility thus ppt will not occur.

Balanced reaction-:

Moles of AgNO3=mass(g)molar mass (g/mol) =2.7×10−5g / 169.86 gmol

=1.589⋅10^−7 mol

Molarity of Ag+=moles of solute(L)=1.589⋅10−7 mol0.015 L=1.059⋅10−5M

Ksp of Ag2CrO4

=[Ag+]2[CrO42−]

1.2⋅10−12=[2s]2[s]

4s3=1.2⋅10−12

s=6.69⋅10−5 M

Molarity of Ag+ is less than the molar solubility thus ppt will not occur.

<h3>What is the molarity calculation formula?</h3>

The volume of solvent required to dissolve the provided solute is multiplied by the ratio of the moles of the solute whose molarity has to be computed. (M=frac{n}{V}) The molality of the solution that needs to be computed in this case is M. n is the solute's molecular weight in moles.

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1 year ago

The Haber Process synthesizes ammonia at elevated temperatures and pressures. Suppose you combine 1580 L of nitrogen gas and 351

ikadub [295]

Answer : The volume of reactant measured at STP left over is 409.9 L

Explanation :

First we have to calculate the moles of $N_2$ and $H_2$ by using ideal gas equation.

$PV_{N_2}=n_{N_2}RT$

where,

P = Pressure of gas at STP = 1 atm

V = Volume of $N_2$ gas = 1580 L

n = number of moles $N_2$ = ?

R = Gas constant = $0.0821L.atm/mol.K$

T = Temperature of gas at STP = 273 K

Putting values in above equation, we get:

$1atm\times 1580L=n_{N_2}\times (0.0821L.atm/mol.K)\times 273K$

$n_{N_2}=70.49mole$

$PV_{H_2}=n_{H_2}RT$

where,

P = Pressure of gas at STP = 1 atm

V = Volume of $H_2$ gas = 3510 L

n = number of moles $H_2$ = ?

R = Gas constant = $0.0821L.atm/mol.K$

T = Temperature of gas at STP = 273 K

Putting values in above equation, we get:

$1atm\times 3510L=n_{H_2}\times (0.0821L.atm/mol.K)\times 273K$

$n_{H_2}=156.6mole$

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

$N_2(g)+3H_2(g)\rightarrow 2NH_3(g)$

From the balanced reaction we conclude that

As, 3 mole of $H_2$ react with 1 mole of $N_2$

So, 156.6 moles of $H_2$ react with $\frac{156.6}{3}\times 1=52.2$ moles of $N_2$

From this we conclude that, $N_2$ is an excess reagent because the given moles are greater than the required moles and $H_2$ is a limiting reagent and it limits the formation of product.

Now we have to calculate the excess moles of $N_2$ reactant (unreacted gas).

Excess moles of $N_2$ reactant = 70.49 - 52.2 = 18.29 moles

Now we have to calculate the volume of reactant, measured at STP, is left over.

$PV=nRT$