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IceJOKER [234]
3 years ago
6

The body loses water and salts in sweat . Explain why drinking large volumes of plain water after exercising may affect the salt

balance in the body.
Chemistry
1 answer:
algol133 years ago
8 0
They balance out how much salt is in your body
You might be interested in
The bonds in the compound MgSO4 can be described asA. ionic, onlyB. covalent, onlyC. both ionic and covalentD. neither ionic nor
Galina-37 [17]
I believe the answer is C. The bonds in the compound magnesium sulfate is ionic and covalent. Magnesium sulfate is soluble in water. When the said compound is dissolved in water, it dissociates into magnesium ions and sulfate ions. However, the bonds that held together the sulfate ions is covalent.
6 0
4 years ago
Mr. Chem S. Tree added water to 250. ML of a 2.50 M NaOH solution, until the final volume was 500. ML. What is the new molarity
tresset_1 [31]

Answer:

molarity of diluted solution = 1.25 M

Explanation:

Using,          

C1V1 (Stock solution) = C2V2 (dilute solution)

given that

C1 = 2.50M

V1 = 250ML

C2 = ?

V2 = 500ML

2.50 M x 250 mL = C2 x 500 mL

C2 = (2.50 M x 250 mL) / 500 mL

C2 = 1.25 M

Hence, molarity of diluted solution = 1.25 M

4 0
3 years ago
A 825 g iron block is heated to 352 degrees C and is placed in an insulated container (of negligible heat capacity) containing 4
Stella [2.4K]

Answer : The final equilibrium temperature of the water and iron is, 537.12 K

Explanation :

In this problem we assumed that heat given by the hot body is equal to the heat taken by the cold body.

q_1=-q_2

m_1\times c_1\times (T_f-T_1)=-m_2\times c_2\times (T_f-T_2)

where,

c_1 = specific heat of iron =  560 J/(kg.K)

c_1 = specific heat of water = 4186 J/(kg.K)

m_1 = mass of iron = 825 g

m_2 = mass of water = 40 g

T_f = final temperature of water and iron = ?

T_1 = initial temperature of iron = 352^oC=273+352=625K

T_2 = initial temperature of water = 20^oC=273+20=293K

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

(825\times 10^{-3}kg)\times 560J/(kg.K)\times (T_f-625K)=-(40\times 10^{-3}kg)\times 4186J/(kg.K)\times (T_f-293K)

T_f=537.12K

Therefore, the final equilibrium temperature of the water and iron is, 537.12 K

8 0
3 years ago
(Thermodynamics)
frutty [35]

Answer:

3853 g

Step-by-step explanation:

M_r: 107.87

         16Ag + S₈ ⟶ 8Ag₂S; ΔH°f =  -31.8 kJ·mol⁻¹

1. Calculate the moles of Ag₂S

Moles of Ag₂S = 567.9 kJ × 1 mol Ag₂S/31.8kJ = 17.858 mol Ag₂S

2. Calculate the moles of Ag

Moles of Ag = 17.86 mol Ag₂S × (16 mol Ag/8 mol Ag₂S) = 35.717 mol Ag

3. Calculate the mass of Ag

Mass of g = 35.717 mol Ag × (107.87 g Ag/1 mol Ag) = 3853 g Ag

You must react 3853 g of Ag to produce 567.9 kJ of heat

3 0
3 years ago
A particular radioactive nuclide has a half-life of 1000 years. What percentage of an initial population of this nuclide has dec
Delicious77 [7]

Answer:

91.16% has decayed & 8.84% remains

Explanation:

A = A₀e⁻ᵏᵗ => ln(A/A₀) = ln(e⁻ᵏᵗ) => lnA - lnA₀ = -kt => lnA = lnA₀ - kt

Rate Constant (k) = 0.693/half-life = 0.693/10³yrs = 6.93 x 10ˉ⁴yrsˉ¹

Time (t) = 1000yrs  

A = fraction of nuclide remaining after 1000yrs

A₀ = original amount of nuclide = 1.00 (= 100%)  

lnA = lnA₀ - kt

lnA = ln(1) – (6.93 x 10ˉ⁴yrsˉ¹)(3500yrs) = -2.426

A = eˉ²∙⁴²⁶ = 0.0884 = fraction of nuclide remaining after 3500 years

Amount of nuclide decayed = 1 – 0.0884 = 0.9116 or 91.16% has decayed.

3 0
4 years ago
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