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ki77a [65]
3 years ago
12

How many atoms are in 5.00 grams of aluminum foil.​

Chemistry
1 answer:
sasho [114]3 years ago
3 0

= 30802.53 im pretty sure if im wrong let me know its the best i can do :/

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_al(s)+_HCL(aq) AlCl3(aq)+_H2(g)
Novay_Z [31]
2Al + 6HCl -> 2AlCl3 + 3H2
5 0
3 years ago
What is the empirical formula of an oxide of nitrogen containing 63.61% by mass of nitrogen and 36.69% by mass of oxygen?
katen-ka-za [31]

Answer:

D. N₂O

Explanation:

Let's assume we have 100 g of the compound.  That means it consists of 63.61 grams of nitrogen and 36.69 grams of oxygen.

Converting masses to moles:

63.61 g N × (1 mol N / 14.01 g N) = 4.540 mol N

36.69 g O × (1 mol O / 16.00 g O) = 2.293 mol O

Normalize by dividing by the smallest:

4.540 / 2.293 = 1.980 mol N

2.293 / 2.293 = 1.000 mol O

So there is approximately twice as many N atoms as O atoms.  The empirical formula is therefore N₂O.

8 0
3 years ago
Two gas-filled tanks have the same volume, temperature, and pressure. They are identical in every way except that one is filled
Tems11 [23]

Answer:

The tank with O₂ weighs more.

Explanation:

We can find the mass of gas using the ideal gas equation.

P.V=n.R.T=\frac{m}{M} .R.T\\m=\frac{P.V.M}{R.T}

Considering the pressure (P), volume (V), temperature (T) and ideal gas constant (R) are the same, we can establish that:

m ∝ M

The mass is directly proportional to the molar mass. The molar mass of O₂ (32 g/mol) is higher than the molar mass of N₂ (28 g/mol). Therefore, the tank with O₂ weighs more.

8 0
3 years ago
PH & POH
Mashcka [7]

Answer:

<h2>4.0 </h2>

Explanation:

The pH of a solution can be found by using the formula

pH = - log [ { H}^{+}]

From the question we have

ph =  -  log(9.25 \times  {10}^{ - 5} )  \\  = 4.03385

We have the final answer as

<h3>4.0</h3>

Hope this helps you

5 0
3 years ago
Calculate the standard cell potential at 25 ∘c for the reaction x(s)+2y+(aq)→x2+(aq)+2y(s) where δh∘ = -793 kj and δs∘ = -319 j/
QveST [7]
First we will calculate free energy change:
ΔG₀ = ΔH₀ - (T * ΔS₀)
        = - 793 kJ - (298 * - 0.319 kJ/K) = - 698 kJ
We know the relation between free energy change and cell potential is:
ΔG₀ = - n F E⁰ where
F = Faraday's constant = 96485 C/mol
n = 2 (given by equation that the electrons involved is 2)
ΔG₀ = - 2 x 96485 x E⁰
- 698 kJ = - 2 x 96485 x E⁰
E⁰ = (698 x 1000) / (2 x 96485) = 3.62 volts   
 
5 0
3 years ago
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