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Mazyrski [523]
3 years ago
13

If Mary likes Johnny, but Johnny likes Katie, and Katie likes Mary. What do they do?

Chemistry
2 answers:
RUDIKE [14]3 years ago
8 0

Answer:

love triangle?- orr they can figure something out

elena-s [515]3 years ago
4 0

Answer:

Russian Roulette?

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Kp/Kc for reaction for the equilibrium, A(g) ⇌ C(g)+B(g), is _______.
Sophie [7]

Kp/Kc = RT

<h3>Further explanation</h3>

Given

Reaction

A(g) ⇌ C(g)+B(g)

Required

Kp/Kc

Solution

For reaction :

pA + qB ⇒ mC + nD  

\large {\boxed {\bold {Kc ~ = ~ \frac {[C] ^ m [D] ^ n} {[A] ^ p [B] ^ q}}}}

While the equilibrium constant Kp is based on the partial pressure  

\large {\boxed {\bold {Kp ~ = ~ \frac {[pC] ^ m [pD] ^ n} {[pA] ^ p [pB] ^ q}}}}

The value of Kp and Kc can be linked to the formula '  

\large {\boxed {\bold {Kp ~ = ~ Kc. (R.T) ^ {\Delta n}}}}

R = gas constant = 0.0821 L.atm / mol.K  

Δn=moles products - moles reactants or

number of product coefficients-number of reactant coefficients  

For reaction :

A(g) ⇌ C(g)+B(g)

number of product coefficients = 1+1=2

number of reactant coefficients   = 1

Δn= 2 - 1 =1

So Kp/Kc = RT

5 0
3 years ago
How much heat is required to decompose 25.5 grams of NaHCO3? 2NaHCO3(s) + 129 kJ 2Na2CO3(s) + H2O(g) + CO2(g)
Scrat [10]
The chemical equation is missbalaced.


The right balanced chemical equation is:


<span>2NaHCO3(s) + 129 kJ ---> Na2CO3(s) + H2O(g) + CO2(g)



Then, now you know that 2 moles of NaHCO3 requires 129 kJ heat, and you just need to convert 25.5 grams of NaHCO3 into number of moles to calculare the amount of heat needed.


A) Number of moles in 25.5 g of NaHCO3, n:

n = grams / molar mass


molar mass NaHCO3 = 23 g/mol + 1g/mol + 12g/mol + 3*16g/mol = 84 g/mol


n = 25.5 g / 84 g/mol = 0.304 mol


B) Heat required


Make the proportion with the theoretical ratio:


2mol / 129 kJ = 0.304 mol / x


=> x = 0.304 mol * 129 kJ / 2 mol =  19.608 kJ


Answer: 19.6 kJ
</span>
7 0
3 years ago
Given 4.80g of ammonium carbonate, find:
V125BC [204]

Answer:

1) 0.05 mol.

2) 0.1 mol.

3) 0.05 mol.

4) 0.4 mol.

5) 2.4 x 10²³ molecules.

Explanation:

<em>1) Number of moles of the compound:</em>

no. of moles of ammonium carbonate = mass/molar mass = (4.80 g)/(96.09 g/mol) = 0.05 mol.

<em>2) Number of moles of ammonium ions :</em>

  • Ammonium carbonate is dissociated according to the balanced equation:

<em>(NH₄)₂CO₃ → 2NH₄⁺ + CO₃²⁻.</em>

It is clear that every 1.0 mole of (NH₄)₂CO₃ is dissociated to produce 2.0 moles of NH₄⁺ ions and 1.0 mole of CO₃²⁻ ions.

<em>∴ The no. of moles of NH₄⁺ ions in 0.05 mol of (NH₄)₂CO₃ </em>= (2.0)(0.05 mol) =  <em>0.1 mol.</em>

<em>3) Number of moles of carbonate ions :</em>

  • Ammonium carbonate is dissociated according to the balanced equation:

<em>(NH₄)₂CO₃ → 2NH₄⁺ + CO₃²⁻.</em>

It is clear that every 1.0 mole of (NH₄)₂CO₃ is dissociated to produce 2.0 moles of NH₄⁺ ions and 1.0 mole of CO₃²⁻ ions.

∴ The no. of moles of CO₃²⁻ ions in 0.05 mol of (NH₄)₂CO₃ = (1.0)(0.05 mol) = 0.05 mol.

<em>4) Number of moles of hydrogen atoms:</em>

  • Every 1.0 mol of (NH₄)₂CO₃  contains:

2.0 moles of N atoms, 8.0 moles of H atoms, 1.0 mole of C atoms, and 3.0 moles of O atoms.

<em>∴ The no. of moles of H atoms in 0.05 mol of (NH₄)₂CO</em>₃ = (8.0)(0.05 mol) = <em>0.4 mol.</em>

<em>5) Number of hydrogen atoms:</em>

  • It is known that every mole of a molecule or element contains Avogadro's number (6.022 x 10²³) of molecules or atoms.

<u><em>Using cross multiplication:</em></u>

1.0 mole of H atoms contains → 6.022 x 10²³ atoms.

0.4 mole of H atoms contains → ??? atoms.

<em>∴ The no. of atoms in  0.4 mol of H atoms</em> = (6.022 x 10²³ molecules)(0.4 mole)/(1.0 mole) = <em>2.4 x 10²³ molecules.</em>

8 0
3 years ago
An object has________when it has been stretched or compressed
Alina [70]

Answer:

Elasticity or Flexibility

Explanation:

The ability of an object to resume its normal shape after being stretched or compressed is called elasticity

4 0
3 years ago
CAN SOMEONE PLEASE HELP ME ASAP?!!!!
krok68 [10]
A and d is physical, b and c is chemical
4 0
3 years ago
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