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ludmilkaskok [199]

3 years ago

7

Acid indigestion is sometimes neutralized with an antacid such as magnesium hydroxide (Mg(OH)2). What products will be released

when the antacid is mixed with the hydrochloric acid in the stomach

1 answer:

NemiM [27]3 years ago

7 0

Answer:

Magnesium chloride and water

Explanation:

Mg(OH)₂ + 2HCl ⟶ MgCl₂ + 2H₂O

magnesium chloride water

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What is the identity of an ion having 46 protons, 42 electrons and 60 neutrons?

Anon25 [30]

Answer:

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Explanation:

repeat this process till the dong bleeds!

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2 years ago

Landslides and slumps are forms of ?

masha68 [24]

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5 0

3 years ago

I2(g) + Cl2(g)2ICl(g) Using standard thermodynamic data at 298K, calculate the entropy change for the surroundings when 1.62 mol

galina1969 [7]

Answer:

The change in entropy of the surrounding is -146.11 J/K.

Explanation:

Enthalpy of formation of iodine gas = $\Delta H_f_{(I_2)}=62.438 kJ/mol$

Enthalpy of formation of chlorine gas = $\Delta H_f_{(Cl_2)}=0 kJ/mol$

Enthalpy of formation of ICl gas = $\Delta H_f_{(ICl)}=17.78 kJ/mol$

The equation used to calculate enthalpy change is of a reaction is:

$\Delta H_{rxn}=\sum [n\times \Delta H_f(product)]-\sum [n\times \Delta H_f(reactant)]$

For the given chemical reaction:

$I_2(g)+Cl_2(g)\rightarrow 2ICl(g),\Delta H_{rxn}=?$

The equation for the enthalpy change of the above reaction is:

$\Delta H_{rxn}=[(2\times \Delta H_f_{(ICl)})]-[(1\times \Delta H_f_{(I_2)})+(1\times \Delta H_f_{(Cl_2)})]$

$=[2\times 17.78 kJ/mol]-[1\times 0 kJ/mol+1\times 62.436 kJ/mol]=-26.878 kJ/mol$

Enthaply change when 1.62 moles of iodine gas recast:

$\Delta H= \Delta H_{rxn}\times 1.62 mol=(-26.878 kJ/mol)\times 1.62 mol=-43.542 kJ$

Entropy of the surrounding = $\Delta S^o_{surr}=\frac{\Delta H}{T}$

$=\frac{-43.542 kJ}{298 K}=\frac{-43,542 J}{298 K}=-146.11 J/K$

1 kJ = 1000 J

The change in entropy of the surrounding is -146.11 J/K.

4 0

3 years ago

How do you think a redox reaction works?

11Alexandr11 [23.1K]

Any chemical reaction in which the oxidation number of an atom,molecule,or ion changes by gaining or losing an electron

3 0

3 years ago

Write this number in standard notation.<br> 1.986 x 10negative exponent 6 or 1.986 x 10^6

Brums [2.3K]

Answer:

$0.000001986$

Explanation:

We have the number $1.986*10^{-6}$ . Standard notation would basically be this number without the $10^{-6}$ part.

To get rid of this part, we need to move the decimal point 6 places to the left (we go left because it's negative 6, indicating division).

So, when we move the decimal point 6 places to the left, the resulting number is: $0.000001986$ , with 5 zeros after the decimal point before a nonzero number.

Thus, the answer is $0.000001986$ .

7 0

3 years ago