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Tamiku [17]
3 years ago
8

How many grams are in 5.2 grams of K

Chemistry
2 answers:
balandron [24]3 years ago
5 0
The answer would be 00.052.
AleksAgata [21]3 years ago
4 0
I think you meant K as in kilograms?
If so, the answer should be 0.0052.
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How many moles of water are produced from 40.0 g of C2H6?
inn [45]

Answer:

4 moles of water

Explanation:

this is a combustion reaction, so the balanced equation is: 2C2H6 + 7O2 → 4CO2 + 6H2O.

the molar mass of C2H6 is 30.07g, so 40.0 g of C2H6 is 1.33 moles of C2H6.

mole ratio of H2O to C2H6 is 6/2, or 3.

1.33 moles C2H6 * 3 moles H2O/1 mole C2H6 = 4 moles H2O

4 0
2 years ago
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Nat2105 [25]

A good reason for a desert fox to show this pattern of behavior because hunting at night allows the fox to use its night vision.

<h3>What is Hunting?</h3>

Thi9s is commonly practised by predators such as fox in which they capture and kill other animals for food.

The fox has a good night vision which makes it able to hunt for animals during the night also. This is why option C is chosen as the most appropriate choice.

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6 0
2 years ago
In which pair of elements are both metalloids?
Aleksandr [31]

Answer:    

Sb and Te

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there are on the stair step

3 0
3 years ago
Read 2 more answers
N204(0) + 2NO2(g)
user100 [1]

setup 1 : to the right

setup 2 : equilibrium

setup 3 : to the left

<h3>Further explanation</h3>

The reaction quotient (Q) : determine a reaction has reached equilibrium

For reaction :

aA+bB⇔cC+dD

\tt Q=\dfrac{C]^c[D]^d}{[A]^a[B]^b}

Comparing Q with K( the equilibrium constant) :

K is the product of ions in an equilibrium saturated state  

Q is the product of the ion ions from the reacting substance  

Q <K = solution has not occurred precipitation, the ratio of the products to reactants is less than the ratio at equilibrium. The reaction moved to the right (products)

Q = Ksp = saturated solution, exactly the precipitate will occur, the system at equilibrium

Q> K = sediment solution, the ratio of the products to reactants is greater than the ratio at equilibrium. The reaction moved to the left (reactants)

Keq = 6.16 x 10⁻³

Q for reaction N₂O₄(0) ⇒ 2NO₂(g)

\tt Q=\dfrac{[NO_2]^2}{[N_2O_4]}

Setup 1 :

\tt Q=\dfrac{0.0064^2}{0.098}=0.000418=4.18\times 10^{-4}

Q<K⇒The reaction moved to the right (products)

Setup 2 :

\tt Q=\dfrac{0.0304^2}{0.15}=0.00616=6.16\times 10^{-3}

Q=K⇒the system at equilibrium

Setup 3 :

\tt Q=\dfrac{0.230^2}{0.420}=0.126

Q>K⇒The reaction moved to the left (reactants)

8 0
2 years ago
Read 2 more answers
P4 (s) + Cl2 (g) → 4PCl3 (l)
schepotkina [342]
If you are wanting the equation to be balanced it would be: P4 + 6Cl2 → 4PCl3
3 0
3 years ago
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