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ElenaW [278]
2 years ago
15

Giúp mình 2 câu này với ạ

Chemistry
1 answer:
alisha [4.7K]2 years ago
5 0

Answer:it is wrong answer

Explanation:estro man

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Determine the number of grams of hcl that can react with 0.750 g of Al(OH)3 according to the following reaction Al(OH)₃(s)+3HCl(
djyliett [7]

Answer:

1040 g HCl

Explanation:

Al(OH)₃ + 3HCl => AlCl₃ + 3H₂O

moles Al(OH)₃ = 750g/78g·mol⁻¹ = 9.62 mol

moles HCl used = 3 x moles Al(OH)₃ consumed = 3(9.62)mol HCl = 1038.46 g. ≈ 1040 g HCl (3 sig.figs.)

8 0
4 years ago
A helium balloon has a volume of 2.30 L at 23.5 ​o​C and a pressure of 1.00 atm at sea level. The Balloon is released and floats
Paha777 [63]

Answer:

27.3 L

Explanation:

5 0
3 years ago
In what method Copper sulphate solution change blue to colourless ??​
enot [183]

Answer: When an iron nail immersed in the solution of copper sulphate than iron displaces copper from the solution of copper sulphate because iron is more reactive than copper. Therefore copper sulphate solution colour changes from blue to pale green.

7 0
3 years ago
Balloon X is filled with the carbon monoxide. Balloon Y is filled to the same volume with carbon dioxide. The balloons are left
weqwewe [10]

Answer:

The smallest ballon is the ballon X

Explanation:

It is possible to answer this question by using Graham's law:

\frac{v_1}{v_2} =\sqrt{\frac{MW_2}{MW_1}}

Where v is the speed of effusion and MW is molar weight of each compound.

This equation is showing that speed is inversely related to the square root of its molar mass. As carbon dioxide has a bigger MW than carbon monoxide, the speed of effusion of carbon dioxide is lower doing its ballon biggest than carbon monoxide ballon, thus: <em>The smallest ballon is the ballon X</em>

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I hope it helps!

6 0
3 years ago
Consider a galvanic cell based on the reaction Al^3+_(aq) + Mg_(s) rightarrow Al_(s) + Mg^2+ _(aq) The half-reactions are Al^3+
grin007 [14]

<u>Answer:</u> The standard cell potential of the cell is -0.71 V

<u>Explanation:</u>

The half reactions follows:

<u>Oxidation half reaction:</u>  Mg\rightarrow Mg^{2+}+2e^-;E^o_{Mg^{2+}/Mg}=-2.37V  ( × 3)

<u>Reduction half reaction:</u>  Al^{3+}(aq.)+3e^-\rightarrow Al(s);E^o_{Al^{3+}/Al}=-1.66V  ( × 2)

The balanced cell reaction follows:

2Al^{3+}(aq.)+3Mg(s)\rightarrow 2Al(s)+3Mg^{2+}(aq.)

To calculate the E^o_{cell} of the reaction, we use the equation:

E^o_{cell}=E^o_{cathode}-E^o_{anode}

Substance getting oxidized always act as anode and the one getting reduced always act as cathode.

Putting values in above equation, we get:

E^o_{cell}=-2.37-(-1.66)=-0.71V

Hence, the standard cell potential of the cell is -0.71 V

8 0
3 years ago
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