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3 years ago

How many grams are in 88.1 moles of magnesium?

1 answer:

4 0

For getting the result of this problem, the knowledge of periodic table is very important. From the periodic table we come to know that. The knowledge of atomic mass of magnesium is also required to solve the problem.
1 mole of magnesium = 24.3 gm
88.1 moles of magnesium = (24.3 * 88.1) gms
= 2140.83 gms
So 2140.83 grams are there in 88.1 moles of magnesium.

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What is the mass of 1.24 x 10^23 molecules of N2O2

Scorpion4ik [409]

<h3>Answer:</h3>

12.4 g N₂O₂

<h3>General Formulas and Concepts:</h3>

<u>Chemistry</u>

<u>Atomic Structure</u>

Reading a Periodic Table
Moles
Avogadro's Number - 6.022 × 10²³ atoms, molecules, formula units, etc.

<u>Stoichiometry</u>

Using Dimensional Analysis

<h3>Explanation:</h3>

<u>Step 1: Define</u>

[Given] 1.24 × 10²³ molecules N₂O₂

[Solve] grams N₂O₂

<u>Step 2: Identify Conversions</u>

Avogadro's Number

[PT] Molar Mass of N - 14.01 g/mol

[PT] Molar Mass of O - 16.00 g/mol

Molar Mass of N₂O₂ - 2(14.01) + 2(16.00) = 60.02 g/mol

<u>Step 3: Convert</u>

[DA] Set up: $\displaystyle 1.24 \cdot 10^{23} \ molecules \ N_2O_2(\frac{1 \ mol \ N_2O_2}{6.022 \cdot 10^{23} \ molecules \ N_2O_2})(\frac{60.02 \ g \ N_2O_2}{1 \ mol \ N_2O_2})$
[DA] Divide/Multiply [Cancel out units]: $\displaystyle 12.3588 \ g \ N_2O_2$

<u>Step 4: Check</u>

<em>Follow sig fig rules and round. We are given 3 sig figs.</em>

12.3588 g N₂O₂ ≈ 12.4 g N₂O₂

7 0

2 years ago

I need help with this i'm really dumb.<br> Balance<br> p4+o2 ----> p4o10

kogti [31]

Answer:

P4 + 5O2 → P4O10

Explanation:

White Phosphorus - P4

Dioxygen - O2

Tetraphosphorus Decaoxide - P4O10

4 0

3 years ago

An element has two isotopes. 90% of the isotopes have a mass number of 20 amu, while 10% have a mass number of 22 amu. Calculate

jek_recluse [69]

Explanation:

aam= 90%•20amu+10%•22amu/100

aam= 2020/100

aam= 20.2

8 0

2 years ago

Which of the following in the definition of control?

Advocard [28]

D The Part That Stays The Same

5 0

2 years ago

Hydrazine (N2H4) is a fuel used by some spacecraft. It is normally oxidized by N2O4 according to the following equation: N2H4(l)

vitfil [10]

Answer:

The enthalpy of the reaction is coming out to be -380.16 kJ.

Explanation:

Enthalpy change is defined as the difference in enthalpies of all the product and the reactants each multiplied with their respective number of moles. It is represented as $\Delta H$

The equation used to calculate enthalpy change is of a reaction is:

$\Delta H_{rxn}=\sum [n\times \Delta H_f(product)]-\sum [n\times \Delta H_f(reactant)]$

For the given chemical reaction:

$N_2H_4(l)+N_2O_4(g)\rightarrow 2N_2O(g)+2H_2O(g)$

The equation for the enthalpy change of the above reaction is:

$\Delta H_{rxn}=[(2 mol\times \Delta H_f_{(N_2O)})+(2 mol\times\Delta H_f_{(H_2O)} )]-[(1 mol\times \Delta H_f_{(N_2H_4)})+(1 mol\times \Delta H_f_{(N_2O_4)})]$

We are given:

$\Delta H_f_{(N_2O)}=81.6 kJ/mol\\\Delta H_f_{(H_2O)}=-241.8 kJ/mol\\\Delta H_f_{(N_2H_4)}= 50.6 kJ/mol\\\Delta H_f_{(N_2O_4)}=9.16 kJ/mo$

Putting values in above equation, we get:

$\Delta H_{rxn}=[(2 mol\times 81.6 kJ/mol)+2 mol\times -241.8 kJ/mol)]-[(1 mol\times (50.6 kJ/mol))+(1 mol\times (9.16))]\\\\\Delta H_{rxn}=-380.16 kJ$

Hence, the enthalpy of the reaction is coming out to be -380.16 kJ.

6 0

2 years ago