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3 years ago

The molar mass of Beryllium (Be)

Chemistry

1 answer:

e-lub [12.9K]3 years ago

8 0

The molar mass of Beryllium is 9.012182 u (symbol I can't put down) 0.000003 U

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What mass of ammonia can be produced if 13.4 grams of nitrogen gas reacted ?

aivan3 [116]

Answer:

If 13.4 grams of nitrogen gas reacts we'll produce 16.3 grams of ammonia

Explanation:

Step 1: Data given

Mass of nitrogen gas (N2) = 13.4 grams

Molar mass of N2 = 28 g/mol

Molar mass of NH3 = 17.03 g/mol

Step 2: The balanced equation

N2 + 3H2 → 2NH3

Step 3: Calculate moles of N2

Moles N2 = Mass N2 / molar mass N2

Moles N2 = 13.4 grams / 28.00 g/mol

Moles N2 = 0.479 moles

Step 4: Calculate moles of NH3

For 1 mol N2 we need 3 moles H2 to produce 2 moles NH3

For 0.479 moles N2 we'll produce 2*0.479 = 0.958 moles

Step 5: Calculate mass of NH3

Mass of NH3 = moles NH3 * molar mass NH3

Mass NH3 = 0.958 moles * 17.03 g/mol

Mass NH3 = 16.3 grams

If 13.4 grams of nitrogen gas reacts we'll produce 16.3 grams of ammonia

3 0

3 years ago

There are some cells that do not have DNA. <br> O True<br> O false

raketka [301]

Answer:

false

Explanation:

8 0

3 years ago

Read 2 more answers

Lithium bettery structure<br> and better charge rate

Zolol [24]

The way its structured is they're layered on top of each other and because of that it has a much faster charge rate

6 0

3 years ago

A sample of water is heated from 10.0 degrees Celcius to 15.0 degrees Celcius by the addition of 125 Joules of heat. What is the

geniusboy [140]

Answer:

125 = 5*4.186*m

m = 125 / 5*4.186 = 5.97g of water.

Explanation:

The water is increased by 5 degrees. This means that for every gram of water, it takes 5*4.186 J to increase the temperature by 5 degrees, aka 5*4.186 J/g

hope this help:)

6 0

3 years ago

3 Ni2+(aq) + 2 Cr(OH)3(s) + 10 OH− (aq) → 3 Ni(s) + 2 CrO42−(aq) + 8 H2O(l) ΔG∘ = +87 kJ/mol Given the standard reduction potent

Tom [10]

Answer : The standard reduction potential, $E^o_{(Cr^{+6}/Cr^{+3})}$ is -0.13 V.

Solution : Given,

$E^o_{(Ni^{2+}/Ni)}=-0.28V$

$\Delta G^o=+87KJ/mole=+87000J/mole$ (1 KJ = 1000 J)

The net reaction is,

$3Ni^{2+}(aq)+2Cr(OH)_3(s)+10OH^-(aq)\rightarrow 3Ni(s)+2CrO^{2-}_4(aq)+8H_2O(l)$

The half cell reactions are :

At cathode : $Ni^{2+}(aq)+2e^-\rightarrow Ni(s)$ $E^o_{(Ni^{2+}/Ni)}=-0.28V$

At anode : $CrO^{2-}_4(aq)+4H_2O(l)+3e^-\rightarrow Cr(OH)_3(s)+5OH^-(aq)$ $E^o_{(Cr^{+6}/Cr^{+3})}=?$

First we have to calculate the $E^o_{cell}$ by using formula,

$\Delta G^o=-nFE^o_{cell}$

where,

$\Delta G^o$ = Gibbs's free energy

n = number of electrons in a net chemical reaction = 6 electrons

F = Faraday constant = 96485 C

$E^o_{cell}$ = standard cell potential

Now put all the given values in this formula, we get

$+87000KJ/mole=-6\times (96485)\times E^o_{cell}\\E^o_{cell}=-0.15V$

Now we have to calculate the $E^o_{(Cr^{+6}/Cr^{+3})}$ by using formula,

$E^o_{cell}=E^o_{cathode}-E^o_{anode}$

$E^o_{cell}=E^o_{(Ni^{2+}/Ni)}-E^o_{(Cr^{+6}/Cr^{+3})}$

Now put all the given values in this formula, we get

$-0.15V=-0.28V-E^o_{(Cr^{+6}/Cr^{+3})}$

$E^o_{(Cr^{+6}/Cr^{+3})}=-0.13V$

Therefore, the standard reduction potential, $E^o_{(Cr^{+6}/Cr^{+3})}$ is -0.13 V.

4 0

4 years ago