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uranmaximum [27]
3 years ago
9

Can you please help me?

Chemistry
1 answer:
lorasvet [3.4K]3 years ago
4 0

Answer:

60 moles of NaF

Explanation:

The balanced equation for the reaction is given below:

Al(NO3)3 + 3NaF —> 3NaNO3 + AlF3

From the balanced equation above,

3 moles of NaF reacted to produce 1 mole of AlF3.

Therefore, Xmol of NaF will react to produce 20 moles of AlF3 i.e

Xmol of NaF = 3 x 20

Xmol of NaF = 60 moles

Therefore, 60 moles of NaF are required to produce 20 moles of AlF3.

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marshall27 [118]

Answer: I think it’s the first one

6 0
3 years ago
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What quantity of heat is required to raise the temperature of 460g of aluminum from 15C to 85C?
Hoochie [10]

Answer:

Q = 28.9 kJ

Explanation:

Given that,

Mass of Aluminium, m = 460 g

Initial temperature, T_i=15^{\circ} C

Final temperature, T_f=85^{\circ}

We know that the specific heat of Aluminium is 0.9 J/g°C. The heat required to raise the temperature is given by :

Q=mc\Delta T\\\\Q=mc(T_f-T_i)\\\\Q=460\ g\times 0.9\ J/g^{\circ} C\times (85-15)^{\circ} C\\\\Q=28980\ J\\\\\text{or}\\\\Q=28.9\ kJ

So, 28.9 kJ of heat is required to raise the temperature.

6 0
3 years ago
In which compound do the atoms have the greatest difference in electrongativity?
weqwewe [10]

Explanation:

cesium fluoride is one of the compound

5 0
3 years ago
A chemical reaction produces 653 550 kj of energy as heat in 142.3min. calculate the rate of energy transferred in kj per minute
miskamm [114]

We have the value of  

Total energy produced in the chemical reaction=653 550 KJ  

Time needed=142.3min  

To calculate the rate of energy transfer, that is the amount of energy produced per minute.  

Rate of energy transfer=\frac{Total energy produced}{Time needed}

=\frac{653 550}{142.3}

=4592.76 KJ min⁻¹

So, the rate of energy transfer is 4592.76 KJ min⁻¹.

6 0
3 years ago
A student ran the following reaction in the laboratory at 671 K: 2NH3(g) N2(g) + 3H2(g) When she introduced 7.33×10-2 moles of N
vaieri [72.5K]

Answer:

Kc = 8.05x10⁻³

Explanation:

This is the equilibrium:

                 2NH₃(g)   ⇄     N₂(g)     +     3H₂(g)

Initially       0.0733

React         0.0733α          α/2                3/2α

Eq     0.0733 - 0.0733α    α/2                0.103

We introduced 0.0733 moles of ammonia, initially. So in the reaction "α" amount react, as the ratio is 2:1, and 2:3, we can know the moles that formed products.

Now we were told that in equilibrum we have a [H₂] of 0.103, so this data can help us to calculate α.

3/2α = 0.103

α = 0.103 . 2/3 ⇒ 0.0686

So, concentration in equilibrium are

NH₃ = 0.0733 - 0.0733 . 0.0686 = 0.0682

N₂ = 0.0686/2 = 0.0343

So this moles, are in a volume of 1L, so they are molar concentrations.

Let's make Kc expression:

Kc= [N₂] . [H₂]³ / [NH₃]²

Kc = 0.0343 . 0.103³ / 0.0682² = 8.05x10⁻³

3 0
3 years ago
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