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3 years ago

How do i solve this? and please show out the working

Chemistry

2 answers:

irina1246 [14]3 years ago

4 0

What does it say maybe I can help u

Mrrafil [7]3 years ago

3 0

I don't know what does it means after “calculate the”

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What is the chemical fluoride for carbon tetrafluoride?

amid [387]

The chemical formula for carbon tetrafluoride is CF4

6 0

3 years ago

Read 2 more answers

Causes and effects about climate change?

icang [17]

Explanation:

Causes of climate change are:

1.)Smoke from Vehicles and industrial activities

2.) Burning of polythenes.

3.) Because of green house effect

Effects are:

1.) Extinction of birds,plants and animals

2.) Melting of glaciers

3.) Rise in global temperature

6 0

3 years ago

El superóxido de potasio, KO2, se emplea en máscaras de respiración para generar

Orlov [11]

Answer:

Reactivo límite: Superóxido de potasio.

Moles de oxígeno producidas: $n_{O_2}=0.11molO_2$

Explanation:

Hola,

En este caso, considerando la reacción química llevada a cabo:

$4KO_2(s) + 2H_2O(l) \rightarrow 4KOH(s) + 3O_2(g)$

Es posible identificar el reactivo límite calculando las moles de superóxido de potasio que serían consumidas por 0.10 mol de agua por medio de la relación molar 4 a 2 que hay entre ellos:

$n_{KO_2}^{consumido\ por\ agua}=0.10molH_2O*\frac{4molKO_2}{2molH_2O} =0.2molKO_2$

Así, dado que solo hay 0.15 mol the superóxido de potasio, podemos decir que este es el reactivo límite. Luego, calculamos las moles de oxígeno producidas, considerando la relación molar 4 a 3 que hay entre el superóxido y el oxígeno:

$n_{O_2}=0.15molKO_2*\frac{3molO_2}{4molKO_2} \\\\n_{O_2}=0.11molO_2$

Best regards.

4 0

3 years ago

How many moles of sodium hydroxide are needed to make 250 mL of a 3.4 M NaOH solution?

ELEN [110]

.85 moles is the answer

7 0

3 years ago

Calculate e°cell for a silver-aluminum cell in which the cell reaction is al(s) + 3ag+(aq) → al3+(aq) + 3ag(s) –2.46 v 0.86 v –0

tekilochka [14]

When E° cell is an electrochemical cell which comprises of two half cells.

So,

when we have the balanced equation of this half cell :

Al3+(aq) + 3e- → Al(s) and E°1 = -1.66 V

and we have also this balanced equation of this half cell :

Ag+(aq) + e- → Ag(s) and E°2 = 0.8 V

so, we can get E° in Al(s) + 3Ag (aq) → Al3+(aq) + 3Ag(s)

when E° = E°2 - E°1

∴E° =0.8 - (-1.66)

= 2.46 V

∴ the correct answer is 2.46 V

6 0

4 years ago