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Ludmilka [50]
3 years ago
13

How long does it take for magnesium citrate to work?.

Chemistry
1 answer:
stepan [7]3 years ago
5 0

Answer:

Magnesium citrate should produce a bowel movement within 30 minutes to 6 hours after you take the medicine. Call your doctor if your symptoms do not improve after 7 days of treatment, or if the medicine produces no results

Explanation:

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the answer is : true ice and snow melts into water in sublimation

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An electron in the hydrogen atom makes a transition from an energy state of principal
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Which of the following contains plasma?
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It is mostly water (up to 95% by volume), and contains dissolved proteins (6–8%) (e.g. serum albumins, globulins, and fibrinogen), glucose, clotting factors, electrolytes (Na+, Ca2+, Mg2+, HCO3−, Cl−, etc.), hormones, carbon dioxide (plasma being the main medium for excretory product transportation) and oxygen.

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Calculate the molarity of H3PO4 when you added 57.3 g into 3,820 mL of water
motikmotik

Answer:

0.153M

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.0585/3.820L=0.153M

5 0
4 years ago
The balanced combustion reaction for C6H6 is 2C6H6(l)+15O2(g)⟶12CO2(g)+6H2O(l)+6542 kJ If 8.600 g C6H6 is burned and the heat pr
jenyasd209 [6]

Answer : The final temperature of the water is, 22.5166^oC

Explanation : Given,

Mass of benzene = 8.600 g

Molar mass of benzene = 78 g/mole

First we have to calculate the moles of benzene.

\text{Moles of benzene}=\frac{\text{Mass of benzene}}{\text{Molar mass of benzene}}=\frac{8.600g}{78g/mol}=0.1103mole

Now we have to calculate the energy of combustion.

The given balanced chemical reaction is:

2C_6H_6(l)+15O_2(g)\rightarrow 12CO_2(g)+6H_2O(l)+6542 kJ

According to reaction,

As, 2 moles of benzene gives 6542 kJ of energy on combustion.

So, 0.1103 mole of benzene gives \frac{6542 kJ}{2}\times 0.1103=360.7913kJ of energy on combustion.

Now we have to calculate the final temperature of the water.

Formula used : q_w=m_w\times c_w\times \Delta T=m_w\times c_w\times (T_{final}-T_{initial})

where,

q_w = heat released = 360.7913 kJ = 36079.13 J

m_w = mass of water = 5691 g

c_w = specific heat of water= 4.18J/g^oC

T_{final} = final temperature = ?

T_{initial} = initial temperature = 21^oC

Now put all the given values in the above formula, we get:

36079.13J=5691g\times 4.18J/g^oC\times (T_{final}-21^oC)

T_{final}=22.5166^oC

Therefore, the final temperature of the water is, 22.5166^oC

8 0
4 years ago
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