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deff fn [24]
3 years ago
12

Is the dissolution reaction a spontaneous or non spontaneous?

Chemistry
1 answer:
slavikrds [6]3 years ago
6 0
Bottom line Your dissolution reaction<span> is endothermic, but because </span>dissolution reactions<span> lead to states of higher entropy, your particular </span>reaction is spontaneous<span> (it occurs) at your specific temperature nonetheless</span><span />
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PLEASE HELP!!!
hammer [34]

B. Magnesium + Hydrogen Sulfide (Reactors) ---->  Magnesium Sulfide + Hydrogen (Products)

8 0
3 years ago
Read 2 more answers
5. What is the mass of 9.80 x 1023 formula units of zinc chlorate, Zn(CO3)2?
Zielflug [23.3K]

Answer:

\boxed{\text{378 g}}

Explanation:

We must convert formula units of Zn(ClO₃)₂ to moles and then to grams of Zn(ClO₃)₂.

Step 1. Convert formula units to moles

\text{Moles of Zn(ClO$_{3}$)$_{2}$}\\\\= 9.80 \times10^{23}\text{ formula units Zn(ClO$_{3}$)$_{2}$} \times \dfrac{\text{1 mol Zn(ClO$_{3}$)$_{2}$}}{6.022 \times\ 10^{23} \text{ formula units Zn(ClO$_{3}$)$_{2}$}}\\\\= \text{1.627 mol Zn(ClO$_{3}$)$_{2}$}

Step 2. Convert moles to grams

\text{Mass of Zn(ClO$_{3}$)$_{2}$}\\\\= \text{1.627 mol Zn(ClO$_{3}$)$_{2}$} \times \dfrac{\text{232.29 g Zn(ClO$_{3}$)$_{2}$}}{\text{1 mol Zn(ClO$_{3}$)$_{2}$}}\\\\= \text{378 g Zn(ClO$_{3}$)$_{2}$}\\\\\text{The mass of Zn(ClO$_{3}$)$_{2}$ is } \boxed{\textbf{378 g}}

5 0
3 years ago
What is the empirical formula of a molecule containing 65.5% carbon, 5.5% hydrogen, and 29.0% oxygen
natta225 [31]
Carbon(C):
number of moles= mass/molar mass(Mr)
=65.5/12
=5.5 moles

Hydrogen(H):
number of moles=mass/molar mass (Mr)
=5.5/1
=5.5 moles

Oxygen (O):
number of moles = mass/molar mass (Mr)
=29.0/16
=1.8 moles

EF= lowest number of moles over each of the elements

So,
C= 5.5/1.8 = 3
H= 5.5/1.8 = 3
O= 1.8/1.8 = 1

Therefore Emperical formula= C3H3O
6 0
2 years ago
Tengo que resolver con procedimiento
zzz [600]

Answer:

Distance = 200 km

Distance = 204 km

Speed = 77 km/h

Time = 21.42 h

Explanation:

Given:

A.

Speed = 100 km/h , Time = 2 h

Find:

Distance

B.

Speed = 68 km/h , Time = 3 h

Find:

Distance

C.

Distance = 154 km , Time = 2 h

Find:

Speed

D.

Distance = 1500 km speed = 70 km/h

Find:

Time

Computation:

Speed = distance / time

A.

Distance = 100 x 2

Distance = 200 km

B.

Distance = 68 x 3

Distance = 204 km

C.

Speed = 154 / 2

Speed = 77 km/h

D.

Time = 1500 / 70

Time = 21.42 h

6 0
2 years ago
Can a pressure change shift the equilibrium position in every reversible reaction? Explain
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3 years ago
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