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Rudiy27
2 years ago
8

Spattering does occur and some of the sample is lost onto the laboratory bench. Will the reported mole ratio of tin to oxygen be

too high, too low, or unaffected ? Explain
Chemistry
1 answer:
expeople1 [14]2 years ago
7 0

Answer:

too high

Explanation:

The mole ratio is obtained from number of moles. The number of moles of a substance is obtained as its mass divided by its molar mass. This implies that the mass seriously affects the number of moles.

In this present case, the mole ratio of oxygen to tin will be too high because of the spattering. Spattering means that less mass is reported.

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For the reaction A+B↽−−⇀C+D, assume that the standard change in free energy has a positive value. Changing the conditions of the
77julia77 [94]

Explanation:

a. Adding a catalyst

no effect .( Catalyst can only change the activation energy but not the free energy).

b. increasing [C] and [D]

Increase the free energy .

c. Coupling with ATP hydrolysis

decrease the free energy value .

d.Increasing [A] and [B]

decrease the free energy.

7 0
2 years ago
A 25.0 g sample of an alloy was heated to 100.0 oC and dropped into a beaker containing 90 grams of water at 25.32 oC. The tempe
elena55 [62]

Answer:

The specific heat of the alloy C_{a} = 0.37 \frac{KJ}{Kg K}

Explanation:

Mass of an alloy m_{a} = 25 gm

Initial temperature T_{a} = 100°c = 373 K

Mass of water m_{w} = 90 gm

Initial temperature of water T_{w} = 25.32 °c = 298.32 K

Final temperature T_{f} = 27.18 °c = 300.18 K

From energy balance equation

Heat lost by alloy = Heat gain by water

m_{a} C_{a}  [T_{a} - T_{f}] = m_{w} C_w (T_{f} -T_{w} )

25 × C_{a} × ( 373 - 300.18 ) = 90 × 4.2 (300.18 - 298.32)

C_{a} = 0.37 \frac{KJ}{Kg K}

This is the specific heat of the alloy.

4 0
2 years ago
Imagine that a new polyatomic anion called "platoate" is invented. What would an acid of "platoate" be named?
o-na [289]
<h3>Answer:</h3>

                   Platoic Acid

<h3>Explanation:</h3>

                        While naming Carboxylic Acids we know that when the Carboxylic Acid looses proton it is converted into corresponding conjugate base called as Carboxylate.

Examples:

                        HCOOH           →           HCOO⁻  +  H⁺

                     Formic acid                     Formate Ion

                    H₃CCOOH           →           H₃CCOO⁻  +  H⁺

                     Acetic acid                       Acetate Ion

                    H₅C₂COOH           →           H₅C₂COO⁻  +  H⁺

                 Propanoic acid                       Propanoate Ion

Therefore, if the conjugate base is Platoate then the corresponding acid will be Platoic Acid means we will replace the -ate by -ic acid <em>i.e.</em>

                   RCOO⁻  +  H⁺        →            RCOOH

                   Platoate Ion                         Platoic Acid

8 0
2 years ago
Read 2 more answers
When an athlete gets hurt they apply sometimes an instant cold pack to the injury. The instant cold pack works by mixing two che
NeX [460]

Answer:

The evidence showing that there is a chemical reaction taking place is the instantaneous temperature drop once the cold pack is shaken.

Explanation:

When an athlete applies a cold pack to the injury, they shake it before, mixing the water and <em>ammonium-nitrate fertilizer</em> inside the cold pack. This mixing is an endothermic reaction, which means it absorbs heat. In turn, the temperature falls to 35 F for around 10 minutes.

3 0
2 years ago
Explain the effects of nh3 and hcl on the cuso4 solution in terms of le chatelier's principle
Fittoniya [83]

The Principle of Le Chatelier states that if a system in equilibrium is subjected to a disturbance, the system will react in such a way that it will diminish the effect of that disturbance. Thus, when the concentration of one of the substances in an equilibrium system is changed, the equilibrium varies in such a way that it can compensate for this change.

For example, if the concentration of one of the reactants is increased, the equilibrium shifts to the right or to the side of the products. Also, if you add more reagents, the reaction will move even more to the right until the balance is re-established again, increasing the quantity of products.

In this way, adding HCl to a solution of CuSO4 will produce the following reaction:

CuSO4 (aq) + 2HCl (aq) ⇔ CuCl2 (aq) + H2SO4 (aq)

Initially the solution of CuSO4 in water will be blue, but when adding HCl the solution will change color to green, since the aqueous solutions of CuCl2 are green. By adding more HCl this color will intensify as the balance shifts to the right, producing more CuCl2 and H2SO4.

On the other hand, adding NH3 to a solution of CuSO4 will produce the following reaction

CuSO4 (aq) + 4NH3 (aq) ⇔ [Cu(NH3)4] SO4 (s)

Thus, by adding NH3 to the CuSO4 solution we will observe the formation of a precipitate corresponding to [Cu(NH3) 4] SO4. <u>When adding more NH3, the formation of more precipitate will be observed as the equilibrium moves to the right, producing a greater quantity of [Cu (NH3) 4] SO4.</u>

6 0
2 years ago
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