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Lesechka [4]
3 years ago
14

If the strength of the magnetic field at B is 10 units, the strength of the magnetic field at A is _____.

Chemistry
2 answers:
Sergeeva-Olga [200]3 years ago
8 0
The answer is 40 units.

Nostrana [21]3 years ago
5 0

Answer:

40 units

Explanation:

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According to the graph, which of the following substances is least soluble in 100 grams of water at 50°C?
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Answer:2) NaCl

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Many children are uncomfortable around white lab coats. Specifically explain why this occurs in what a doctor or dentist could d
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Sanitize

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Each 5-ml teaspoon of Extra Strength Maalox Plus contains 450 mg of magnesium hydroxide and 500 mg of aluminum hydroxide. How ma
Art [367]

Answer:

0.0347 moles of hydronium ions

Explanation:

The equation of the neutralization reaction between hydroxide and hydronium ions is given below:

H₃O+ (aq) + OH- (aq) ----> 2 H₂O (l)

From the equation above, 1 mole of hydroxide ions will neutralize one mole hydronium ions.

The moles of hydroxide ions present in 1 teaspoon or 5 mL of antacid product is calculated as follows:

Number of moles = mass / molar mass

Molar mass of Magnesium hydroxide, Mg(OH)₂ = 58 g/mol

Molar mass of aluminium hydroxide, Al(OH)₃ = 78 g/mol

Mass of magnesium hydroxide = 450 g = 0.45 g

Mass of aluminium hydroxide = 500 mg = 0.5 g

Moles of magnesium hydroxide = (0.45/58) moles

Moles of aluminium hydroxide = (0.5/78) moles

Equation of the ionization of magnesium hydroxide and aluminium hydroxide is given below:

Mg(OH)₂ (aq) ----> Mg²+ (aq) + 2 OH- (aq)

Al(OH)₃ (aq) ---> Al³+ (aq) + 3 OH- (aq)

Number of moles of hydroxide ions present in (0.45/58) moles of magnesium hydroxide = 2 × (0.45/58) moles = 0.0155 moles

Number of moles of hydroxide ions present in (0.5/78) moles of aluminium hydroxide = 3 × (0.5/78) moles = 0.0192 moles

Total moles of hydroxide ions = 0.0155 + 0.0192 = 0.0347 moles hydroxide ions

Therefore, 0.0347 moles of hydroxide ions will neutralize 0.0347 moles of hydronium ions.

3 0
3 years ago
Aluminum chloride can be formed from its elements:
saul85 [17]

<u>Answer:</u> The \Deltas H^o_{formation} for the reaction is -1406.8 kJ.

<u>Explanation:</u>

Hess’s law of constant heat summation states that the amount of heat absorbed or evolved in a given chemical equation remains the same whether the process occurs in one step or several steps.

According to this law, the chemical equation is treated as ordinary algebraic expressions and can be added or subtracted to yield the required equation. This means that the enthalpy change of the overall reaction is equal to the sum of the enthalpy changes of the intermediate reactions.

The chemical reaction for the formation reaction of AlCl_3 is:

2Al(s)+3Cl_2(g)\rightarrow 2AlCl_3 (s)    \Delta H^o_{formation}=?

The intermediate balanced chemical reaction are:

(1) HCl(g)\rightarrow HCl(aq.)    \Delta H_1=-74.8kJ    ( ×  6)

(2) H_2(g)+Cl_2(g)\rightarrow 2HCl(g)    \Delta H_2=-185kJ     ( ×  3)

(3) AlCl_3(aq.)\rightarrow AlCl_3(s)    \Delta H_3=+323kJ     ( ×  2)

(4) 2Al(s)+6HCl(aq.)\rightarrow 2AlCl_3(aq.)+3H_2(g)    \Delta H_4=-1049kJ

The expression for enthalpy of formation of AlCl_3 is,

\Delta H^o_{formation}=[6\times \Delta H_1]+[3\times \Delta H_2]+[2\times \Delta H_3]+[1\times \Delta H_4]

Putting values in above equation, we get:

\Delta H^o_{formation}=[(-74.8\times 6)+(-185\times 3)+(323\times 2)+(-1049\times 1)]=-1406.8kJ

Hence, the \Deltas H^o_{formation} for the reaction is -1406.8 kJ.

6 0
3 years ago
When metals react with acid, bubbles of what gas are produced?​
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hydrogen is the answer

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