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lisabon 2012 [21]
2 years ago
14

If 8.25 grams of lead II nitrate reacts, how many grans of sodium nitrate is produced?

Chemistry
1 answer:
NeTakaya2 years ago
5 0
Hope this helps, I found the paper key too your question.

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Use sedimentary in a sentence
VMariaS [17]
'The Sedimentary rock formed from years of sediments piling on top of it and being compressed.'
5 0
2 years ago
Read 2 more answers
How many grams of Fe3O4 are required to react completely with 300 grams of H2?
vekshin1

Answer:

2023.04 g

Explanation:

Magnetite reacts with hydrogen to produce Iron metal and steam. Steam instead of water is produced as the reaction occurs at temperatures above the boiling point of water.

Fe₃O₄ + 4 H₂ → 3 Fe +4 H₂O

From the equation, 1 mole of Fe₃O₄ reacts with 4 moles of H₂.

69.76 grams of H₂ has the following number of moles.

Number of moles= mass/RAM

=69.76/2

=34.88 moles.

The reaction ratio of Fe₃O₄ to H₂ is 1:4

Thus number of moles of magnetite= (1×34.88)/4

=8.72 moles.

Mass= moles × molecular weight

=8.72 moles × (56×3+16×4)

=2023.04 grams

8 0
3 years ago
A. Calculate the empirical formula of a molecule with percent compositions: 55.3% potassium (K), 14.6% phosphorus (P), and 30.1%
Otrada [13]
The way you calculate the empirical formula is to firstly assume 100g. To find each elements moles you take each elements percentage listed, times it by one mole and divide it by its atomic mass. (ex: moles of K =55.3g x 1 mole/39.1g, therefore there is 1.41432225 moles of Potassium) Once you’ve completed this for every element you list each elements symbol beside it’s number of moles and divide by the smallest number because it can only go into its self once. After you’ve done this, you’ve found your empirical formula, which is the simplest whole number ratio of atoms in a compound. I’ve added an example of a empirical question I completed last semester :)

6 0
3 years ago
Read 2 more answers
Dinitrogen tetraoxide, N2O4, decomposes to nitrogen dioxide, NO2, in a first-order process. If k = 1.5 x 103 s-1 at 5 ºC and k =
Arada [10]

Answer:

The activation energy for the decomposition = 33813.28 J/mol

Explanation:

Using the expression,

\ln \dfrac{k_{1}}{k_{2}} =-\dfrac{E_{a}}{R} \left (\dfrac{1}{T_1}-\dfrac{1}{T_2} \right )

Wherem  

k_1\ is\ the\ rate\ constant\ at\ T_1

k_2\ is\ the\ rate\ constant\ at\ T_2

E_a is the activation energy

R is Gas constant having value = 8.314 J / K mol  

Thus, given that, E_a = ?

k_2=4.0\times 10^3s^{-1}

k_1=1.5\times 10^3s^{-1}  

T_1=5\ ^0C  

T_2=25\ ^0C  

The conversion of T( °C) to T(K) is shown below:

T(K) = T( °C) + 273.15  

So,  

T = (5 + 273.15) K = 278.15 K  

T = (25 + 273.15) K = 298.15 K  

T_1=278.15\ K

T_2=298.15\ K

So,

\ln \frac{1.5\times 10^3}{4.0\times 10^3}\:=-\frac{E_{a}}{8.314}\times \left(\frac{1}{278.15}-\frac{1}{298.15}\right)

E_a=-\ln \frac{1.5\times \:10^3}{4.0\times \:10^3}\:\times \frac{8.314}{\left(\frac{1}{278.15}-\frac{1}{298.15}\right)}

E_a=-\frac{8.314\ln \left(\frac{1.5\times \:10^3}{4\times \:10^3}\right)}{\frac{1}{278.15}-\frac{1}{298.15}}

E_a=-\frac{689483.53266 \ln \left(\frac{1.5}{4}\right)}{20}

E_a=33813.28\ J/mol

<u>The activation energy for the decomposition = 33813.28 J/mol</u>

8 0
3 years ago
How does an atom become an atom of another element is
ivann1987 [24]
Can you give me the anwsers

3 0
3 years ago
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