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2 years ago

One of the essential minerals in the human body is salt. How much salt (NaCl) is in the average adult human body?

1 answer:

8 0

The human body contains many salts, of which sodium chloride NaCl is the major one, making up around 0.4 per cent of the body's weight. So a 50kgperson would contain around 200g of sodium chloride - around 40 teaspoons. Since we lose salt whenever we sweat, it has to be continually replaced. Answer: C.

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(h) The student made observations related to the contents of the Erlenmeyer flask during the titration. Identify an observation

katen-ka-za [31]

Answer:

Explanation:

During titration indicators are often used to identify chemical changes between reacting species.

For colorless solutions in which no noticeable changes can easily be seen, indicators are the best bet. Most titration processes involves a combination of acids and bases to an end point.

Indicators are substances whose color changes to signal the end of an acid-base reaction. Examples are methyl orange, methyl red, phenolphthalein, litmus, cresol red, cresol green, alizarin R3, bromothymol blue and congo red.

Most of these indicators have various colors when chemical changes occur.

Also, there are heat changes that accompanies most of these reactions. These are also indicators of chemical changes.

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3 years ago

Which is an example of a set point for the body? A. hair color B. type of blood C. thumb crossing style D. amount of blood gluco

Mama L [17]

The Answer Is B Type Of Blood

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2 years ago

Identify the base in this acid-base reaction:

yarga [219]

Answer:

$\boxed{\text{NaOH}}$

Explanation:

$\rm \underbrace{\hbox{NaOH }}_{\hbox{base}} + \underbrace{\hbox{HCl}}_{\hbox{acid}} \longrightarrow \, NaCl + H$_{2}$O$

A hydroxide of a metal is a base.

$\text{The base is }\boxed{\textbf{NaOH}}$

B is wrong. HCl is an acid.

C is wrong. NaCl is a salt.

D is wrong. Water is neutral.

8 0

3 years ago

Read 2 more answers

Calculate the solubilities of the following compounds in a 0.02 M solution of barium nitrate using molar concentrations, first i

Law Incorporation [45]

Answer:

a. 1.7 × 10⁻⁴ mol·L⁻¹; b. 5.5 × 10⁻⁹ mol·L⁻¹

c. 2.3 × 10⁻⁴ mol·L⁻¹; 5.5 × 10⁻⁸ mol·L⁻¹

Explanation:

a. Silver iodate

Let s = the molar solubility.

AgIO₃(s) ⇌ Ag⁺(aq) + IO₃⁻(aq); Ksp = 3.0 × 10⁻⁸

E/mol·L⁻¹: s s

$K_{sp} =\text{[Ag$^{+}$][IO$_{3}$$^{-}$]} = s\times s = s^{2} = 3.0\times 10^{-8}\\s = \sqrt{3.0\times 10^{-8}} \text{ mol/L} = 1.7 \times 10^{-4} \text{ mol/L}$

b. Barium sulfate

BaSO₄(s) ⇌ Ba²⁺(aq) + SO₄²⁻(aq); Ksp = 1.1 × 10⁻¹⁰

I/mol·L⁻¹: 0.02 0

C/mol·L⁻¹: +s +s

E/mol·L⁻¹: 0.02 + s s

$K_{sp} =\text{[Ba$^{2+}$][SO$_{4}$$^{2-}$]} = (0.02 + s) \times s \approx 0.02s = 1.1\times 10^{-10}\\s = \dfrac{1.1\times 10^{-10}}{0.02} \text{ mol/L} = 5.5 \times 10^{-9} \text{ mol/L}$

c. Using ionic strength and activities

(i) Calculate the ionic strength of 0.02 mol·L⁻¹ Ba(NO₃)₂

The formula for ionic strength is

$\mu = \dfrac{1}{2} \sum_{i} {c_{i}z_{i}^{2}}\\\\\mu = \dfrac{1}{2} (\text{[Ba$^{2+}$]}\cdot (2+)^{2} + \text{[NO$_{3}$$^{-}$]}\times(-1)^{2}) = \dfrac{1}{2} (\text{0.02}\times 4 + \text{0.04}\times1)= \dfrac{1}{2} (0.08 + 0.04)\\\\= \dfrac{1}{2} \times0.12 = 0.06$

(ii) Silver iodate

a. Calculate the activity coefficients of the ions

$\log \gamma = -0.51z^{2}\sqrt{I} = -0.051(1)^{2}\sqrt{0.06} = -0.51\times 0.24 = -0.12\\\gamma = 10^{-0.12} = 0.75$

b. Calculate the solubility

AgIO₃(s) ⇌ Ag⁺(aq) + IO₃⁻(aq)

$K_{sp} =\text{[Ag$^{+}$]$\gamma_{Ag^{+}}$[IO$_{3}$$^{-}$]$\gamma_{IO_{3}^{-}}$} = s\times0.75\times s \times 0.75 =0.56s^{2}= 3.0 \times 10^{-8}\\s^{2} = \dfrac{3.0 \times 10^{-8}}{0.56} = 5.3 \times 10^{-8}\\\\s =2.3 \times 10^{-4}\text{ mol/L}$

(iii) Barium sulfate

a. Calculate the activity coefficients of the ions

$\log \gamma = -0.51z^{2}\sqrt{I} = -0.051(2)^{2}\sqrt{0.06} = -0.51\times16\times 0.24 = -0.50\\\gamma = 10^{-0.50} = 0.32$

b. Calculate the solubility

BaSO₄(s) ⇌ Ba²⁺(aq) + SO₄²⁻(aq

$K_{sp} =\text{[Ba$^{2+}$]$\gamma_{ Ba^{2+}}$[SO$_{4}$$^{2-}$]$\gamma_{ SO_{4}^{2-}}$} = (0.02 + s) \times 0.32\times s\times 0.32 \approx 0.02\times0.10s\\2.0\times 10^{-3}s = 1.1 \times 10^{-10}\\s = \dfrac{1.1\times 10^{-10}}{2.0 \times 10^{-3}} \text{ mol/L} = 5.5 \times 10^{-8} \text{ mol/L}$

7 0

3 years ago

Explain what happens when two light waves traveling from the same direction of displacement meet.

Aneli [31]

When two waves come in contact with each other while traveling on the same medium, its called constructive interference. Constructive interference happens when two waves come together.

7 0

3 years ago