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fomenos
3 years ago
6

Hi can someone help me with this question! First one to answer it I will make u brainliest

Chemistry
2 answers:
Scrat [10]3 years ago
8 0

Answer:

(B) Test 2

Explanation:

The entire substance changed, and turned into a liquid.

skad [1K]3 years ago
7 0

Answer:

Test 2

Explanation:

............................................

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A mineral that has no planes of weakness in its lattice structure will not break along planes. Which term is appled to the way i
Pie
The answer for this question should be B) Cleavage or the second option.
5 0
3 years ago
Read 2 more answers
Is the reaction to produce zinc from zinc sulfide spontaneous under standard conditions? Coupledreaction: ZnS(s) + H2 (g)  Zn(s
finlep [7]

Answer:

The reaction to produce zinc from zinc sulfide is not spontaneous under standard conditions.

Explanation:

Hess's law states: "When one reaction can be expressed as the algebraic sum of others, its heat of reaction is equal to the same algebraic sum of the partial heats of the partial reactions". So Hess's Law is an indirect method of calculating the heat of reaction or enthalpy of reaction when the chemical reaction occurs in one or more than one stage.

Taking into account that ΔG is a state function, which only depends on the initial and final states, its variation in a reaction is calculated by adding the free energies of the reactants and products involved in it when both are in the normal state. That is, at the pressure of 1natm if it is gases or at the concentration of 1 mol / L for substances in liquid solution.

The sum of the fitted equations should give the problem equation. So, if in a "data" reaction a substance is as a reactant and in the reaction that you must obtain is as a product, you must turn the "data" reaction and ΔG will change its sign. In this case, this happen with the reaction 1 (Rxn1):

Rxn1: ZnS (s) → Zn (s) + S (s)          ΔG1°= 201.3 kJ

Rxn2: S (s) + H₂ (g) → H₂S (g)        ΔG2°= -33.4kJ

Adding both reactions (taking into account that certain substances appear sometimes as a reagent and others as a product, so they are totally eliminated if they appear in the same quantities) you get:

ZnS (s) + H₂ (g) → Zn (s) + H₂S (g)

Adding algebraically ΔG1° and ΔG2° you get:

ΔG°= ΔG1° + ΔG2°

ΔG°= 201.3 kJ - 33.4 kJ

ΔG°= 167.9 kJ

If a chemical reaction proceeds with a ΔG <0 the process is  spontaneous. If, on the other hand, ΔG> 0, the reaction is not spontaneous.

Since ΔG>0, <u><em>the reaction to produce zinc from zinc sulfide is not spontaneous under standard conditions.</em></u>

6 0
3 years ago
People often think bonds store or hold energy. If that is true, should the energy be high or low when a bond forms?
marta [7]

Answer:

The energy should be high.

Explanation:

Bonds do store energy and release it depending if it's endothermic or exothermic. The energy should be low because when a bond forms (endothermic) it releases heat, which helps form bonds. Having a high energy means the bond is absorbing energy, which helps break bonds (endothermic). How this helps!

4 0
3 years ago
Zinc sulfide + magnesium arsenate → zinc arsenate +
Black_prince [1.1K]

The formula equation for the reaction between zinc sulfide and magnesium arsenate is

3ZnS + Mg₃(AsO₄)₂ → Zn₃(AsO₄)₂ + 3MgS

<h3>What is a chemical equation? </h3>

Chemical equations are representations of chemical reactions using symbols and formula of the reactants and products.

The reactants are located on the left side while the products are located on the right side.

Reactants —> Products

The balancing of chemical equations follows the law of conservation of matter which states that matter can neither be created nor destroyed during a chemical reaction but can be transferred from one form to another.

<h3>How to write the formula equation </h3>

Zinc sulfide => ZnS

Magnesium arsenate => Mg₃(AsO₄)₂

Zinc arsenate => Zn₃(AsO₄)₂

Magnesium sulfide => MgS

Zinc sulfide + Magnesium arsenate → Zinc arsenate + Magnesium sulfide

3ZnS + Mg₃(AsO₄)₂ → Zn₃(AsO₄)₂ + 3MgS

Learn more about chemical equation:

brainly.com/question/7181548

3 0
2 years ago
A certain substance X has a normal freezing point of -6.4 C and a molal freezing point depression constant Kf= 3.96 degrees C.kg
Brut [27]

Answer:  1.0\times 10^2g

Explanation:

Depression in freezing point is given by:

\Delta T_f=i\times K_f\times m

\Delta T_f=T_f^0-T_f=(-6.4-(13.6))^0C=7.2^0C = Depression in freezing point

i= vant hoff factor = 1 (for non electrolyte like urea)

K_f = freezing point constant = 3.96^0C/m

m= molality

\Delta T_f=i\times K_f\times \frac{\text{mass of solute}}{\text{molar mass of solute}}\times \text{weight of solvent in kg}}

Weight of solvent (X)= 950 g = 0.95 kg  

Molar mass of non electrolyte (urea) = 60.06 g/mol

Mass of non electrolyte (urea) added = ?

7.2=1\times 3.96\times \frac{xg}{60.06 g/mol\times 0.95kg}

x=1.0\times 10^2g

Thus 1.0\times 10^2g urea was dissolved.

8 0
4 years ago
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