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3 years ago

What is the chemical name for a compound with the formula ba3n2?

1 answer:

3 0

Barium nitride

copy the name of the metal as it is written in the periodic table and the first part of the name of the nonmetal (nitrogen) and replace the end by "ide"

You might be interested in

Calculate the standard cell potential given the following standard reduction potentials: Al3++3e−→Al;E∘=−1.66 V Cu2++2e−→Cu;E∘=0

Maru [420]

Answer:

Explanation:

To calculate the cell potential we use the relation:

Eº cell = Eº oxidation + Eº reduction

Now in order to determine which of the species is going to be oxidized, we have to remember that the more the value of the reduction potential is negative, the greater its tendency to be oxidized is. In electrochemistry we use the values of the reductions potential in the tables for simplicity because the only thing we need to do is change the sign of the reduction potential for the oxized species .

So the species that is going to be oxidized is the Aluminium, and therefore:

Eº cell = -( -1.66 V ) + 0.340 V = 5.06 V

Equally valid is to write the equation as:

Eº cell = Eº reduction for the reduced species - Eº reduction for the oxidized species

These two expressions are equivalent, choose the one you fell more comfortable but be careful with the signs.

3 0

3 years ago

If you collect 1.75 L of hydrogen gas during a lab experiment when the room temperature is 23oC and the barometric pressure is 1

ziro4ka [17]

Answer:

$n=0.0747mol$

Explanation:

Hello,

In this case, since we can consider hydrogen gas as an ideal gas, we check the volume-pressure-temperature-mole relationship by using the ideal gas equation:

$PV=nRT$

Whereas we are asked to compute the moles given the temperature in Kelvins, thr pressure in atm and volume in L as shown below:

$n=\frac{105kPa*\frac{0.009869atm}{1kPa}*1.75L}{0.082\frac{atm*L}{mol*K}*(23+273.15)K} \\\\n=0.0747mol$

Best regards.

8 0

3 years ago

Help me pleaseee. ty if you dooo :))

Lina20 [59]

$\huge \fcolorbox{black}{red}{♛answer♛}$

Hydrogen Fluoride / HF

$\huge\sf\underline{\underline{\red{❥︎ Thanks}}}$

3 0

2 years ago

A certain gas at 2oC and 1.00 atm pressure fills a 4.0 container. What volume will the gas occupy at 100oC and 780 torr pressure

artcher [175]

The final volume V₂=4.962 L

<h3>Further explanation</h3>

Given

T₁=20 + 273 = 293 K

P₁= 1 atm

V₁ = 4 L

T₂=100+273 = 373 K

P₂=780 torr=1,02632 atm

Required

The final volume

Solution

Combined gas law :

P₁V₁/T₁=P₂V₂/T₂

Input the value :

V₂=(P₁V₁T₂)/(P₂T₁)

V₂=(1 x 4 x 373)/(1.02632 x 293)

V₂=4.962 L

6 0

3 years ago

How many milliliters of an aqueous solution of 0.170 M ammonium carbonate is needed to obtain 16.1 grams of the salt

Citrus2011 [14]

There will be needed 982.35 mL of solution to obtain 16.1 grams of the salt.There will be needed mL of

Why?

In order to calculate how many milliliters are needed to obtain 16.1 grams of the salt given its concentration, we first need to find its chemical formula which is the following:

$(NH_{4})2CO_{3}$

Now that we know the chemical formula of the substance, we need to find its molecular mass. We can do it by the following way:

$N_{2}=14g*2=28g\\\\2H_{4}=2*1g*4=8g\\\\C=12.01g*1=12.01g\\\\O_{3}=15.99g*3=47.97g$

We have that the molecular mass of the substance will be:

$MolecularMass=\frac{28g+8g+12.01g+47.97g}{mol}=95.98\frac{g}{mol}$

Therefore, knowing the molecular mass of the substance, we need to calculate how many mols represents 16.1 grams of the same substance, we can do it by the following way:

$mol_{(NH_{4})2CO_{3}=\frac{mass_{(NH_{4})2CO_{3}}}{molarmass_{(NH_{4})2CO_{3}}}$

$mol_{(NH_{4})2CO_{3}=\frac{16.1g}{95.98\frac{g}{mol}}=0.167mol$

Finally, if we need to calculate how many milliliters are needed, we need to use the following formula:

$M=\frac{moles_{solute}}{volume_{solution}}$

$M=\frac{moles_{solute}}{volume_{solution}}\\\\volume_{solution}=\frac{moles_{solute}}{M}$

Now, substituting and calculating, we have:

$volume_{solution}=\frac{0.167mol}{0.170\frac{mol}{L}}\\\\volume_{solution}=0.982L=0.982L*1000=982.35mL$

Henc, there will be needed 982.35 mL of solution to obtain 16.1 grams of the salt.

Have a nice day!

5 0

3 years ago