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Tresset [83]
3 years ago
6

What is the empirical formula made of 9.7% aluminum, 38.4% chlorine, and 51.9% oxygen?

Chemistry
1 answer:
lesantik [10]3 years ago
6 0

Answer: AlCl3O9

Explanation:Please see attachment for explanation

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Name some Changes in matter...
frutty [35]

Answer:

Solid, Liquid, Gas.

Explanation:

Quick gg search.

5 0
2 years ago
Read 2 more answers
Suppose that a catalyst lowers the activation barrier of a reaction from 125kJ/mol to 55kJ/mol125⁢kJ/mol to 55kJ/mol. By what fa
sineoko [7]

Answer:

The factor of increasing reaction rate is 1,85x10¹².

Explanation:

Using arrhenius formula:

k = A e^\frac{-E_{a}}{RT}

Where k is rate constant; A is frecuency factor; Eₐ is activation energy; R is gas constant (0,008134 kJ/molK); T is temperature 25°C = 298,15K

Thus, replacing for an activation energy of 125 kJ/mol assuming A as 1:

k = 1,25x10⁻²²

When activation energy is 55kJ/mol:

k = 2,31x10⁻¹⁰

Thus, the factor of increasing reaction rate is:

2,31x10⁻¹⁰/1,25x10⁻²² =<em> 1,85x10¹²</em>

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I hope it helps!

8 0
3 years ago
Does pb(no3)2 + Na3(PO4) = Pb3(PO4)2 + Na(NO3) have a precipitate?
ipn [44]

Answer:

Yes, Pb3(PO4)2.

Explanation:

Hello there!

In this case, according to the given balanced chemical reaction, it is possible to use the attached solubility series, it is possible to see that NaNO3 is soluble for the Na^+ and NO3^- ions intercept but insoluble for the Pb^3+ and PO4^2- when intercepting these two. In such a way, we infer that such reaction forms a precipitate of Pb3(PO4)2, lead (II) phosphate.

Regards!

4 0
3 years ago
Natural gas is almost entirely methane. A container with a volume of 2.65L holds 0.120mol of methane. What will the volume be if
mrs_skeptik [129]

The final volume of the methane gas in the container is 6.67 L.

The given parameters;

  • <em>initial volume of gas in the container, V₁ = 2.65 L</em>
  • <em>initial number of moles of gas, n₁ = 0.12 mol</em>
  • <em>additional concentration, n = 0.182 mol</em>

The total number of moles of gas in the container is calculated as follows;

n_t = 0.12 + 0.182 = 0.302 \ mol

The final volume of gas in the container is calculated as follows;

PV = nRT\\\\\frac{V}{n} = \frac{RT}{P} \\\\\frac{V_1}{n_1} = \frac{V_2}{n_2} \\\\V_2 = \frac{V_1 n_2}{n_1} \\\\V_2 = \frac{2.65 \times 0.302}{0.12} \\\\V_2 = 6.67 \ L

Thus, the final volume of the methane gas in the container is 6.67 L.

Learn more here:brainly.com/question/21912477

5 0
3 years ago
What is the mass of 19.5 moles of carbon dioxide?
Olenka [21]

Explanation:

mass = moles * molar mass

molar mass of carbon dioxide (CO2) = 12+32 = 44

= 19.5* 44 = 858

proof

moles = mass /molar mass

= 858/44 = 19.5 moles

6 0
3 years ago
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