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3 years ago

Name 3 elements that are found in the human body

Chemistry

2 answers:

Taya2010 [7]3 years ago

8 0

Answer:

oxygen, carbon, hydrogen, nitrogen, calcium, and phosphorus.

Explanation:

the mass of the human body is made up of six elements: oxygen, carbon, hydrogen, nitrogen, calcium, and phosphorus.

mars1129 [50]3 years ago

4 0

Six elements make up nearly 99 percent of the mass of the human body: oxygen, carbon, hydrogen, nitrogen, calcium, and phosphorus. Just 0.85% of the total is made up of the following five elements: potassium, sulfur, sodium, chlorine, and magnesium. All 11 are needed for survival.
I wrote more than 3 :) you can just choose!
Have a great day! Hope this helps

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Compute the molar enthalpy of combustion of glucose (C6 H12O6 ): C6 H12O6 (s) + 6 O2 (g) → 6 CO2 (g) + 6 H2 O (g) Given that com

lana66690 [7]

Answer:

The molar enthalpy of combustion of glucose is -2819.3 kJ/mol

Explanation:

Step 1: Data given

Mass of glucose = 0.305 grams

Combustion of 0.305 grams causes a raise of 6.30 °C

Calorimeter has a heat capacity of 755 J/°C

Molar mass of glucose = 180.2 g/mol

Step 2: The balanced equation

C6H12O6 (s) + 6O2 (g) → 6CO2 (g) + 6H2O (g)

Step 3:

ΔH = (m * C * ΔT + c(calorimeter) * ΔT)

with m = mass of the solutin = 0.305 grams

with C = heat capacity of water = 4.184 J/g°C

with ΔT = the change in temperature = 6.30 °C

with c(calorimeter) = 755 J/°C

ΔH = 0.305 * 4.184 *6.30 + 755 * 6.30 = 4764.5 J ( negative because it's exothermic)

Step 4: Calculate moles of glucose

Moles glucose = mass glucose / Molar mass glucose

Moles glucose = 0.305 grams / 180.2 g/mol

Moles glucose = 0.00169 moles

Step 5: Calculate molar enthalpy

Molar enthalpy = -4764.5 J / 0.00169 moles

Molar enthalpy = - 2819254.2 J/moles = -2819.3 kJ/moles

The molar enthalpy of combustion of glucose is -2819.3 kJ/mol

5 0

3 years ago

Is water wet? This is a discussion for easy points so I suggest you answer it.

marishachu [46]

Yes , water is wet lol

7 0

3 years ago

Read 2 more answers

An atom of lead has a radius of and the average orbital speed of the electrons in it is about . Calculate the least possible unc

victus00 [196]

The question is incomplete. Here is the complete question.

An atom of lead has a radius of 154 pm and the average orbitalspeed of the electron in it is about 1.8x $10^{8}$ m/s. Calculate the least possible uncertainty in a measurement of the speed of an electron in an atom of lead. Write your answer as a percentage of the average speed, and round it to significant 2 digits.

Answer: v% = 0.21 m/s

Explanation: To calculate the uncertainty, use <u>Heisenberg's Uncertainty Principle</u>, which states that: ΔpΔx≥ $\frac{h}{4\pi }$

where h is <u>Planck's constant</u> and it is equal to 6.626. $10^{-34}$ m²kg/s.

Since p (momentum) is p = m.v:

mΔv.Δx ≥ $\frac{h}{4\pi }$

Δv = $\frac{h}{4\pi.x.m }$

Given that: r = x = 1.54. $10^{-10}$ m and mass of an electron is m=9.1. $10^{-31}$ kg

Δv = $\frac{6.626.10^{-34} }{4.3.14.1.54.10^{-10}.9.1.10^{-31}}$

Δv = 0.0376. $10^{7}$

As percentage of average speed:

Δv. $\frac{1}{v}$ .100% = $\frac{0.0376.10^{7} }{1.8.10^{8} }$ .10² = 0.021.10 = 0.21%

The least possible uncertainty in a speed of an electron is 0.21%.

5 0

4 years ago

A gas occupies 100.0 mL at a pressure of 780 mm ahh. What is the volume of the gas when it’s pressure is increased to 880 mm Hg

vlabodo [156]

Hello!

The volume of the gas when its pressure is increased to 880 mm Hg is 88,64 mL

This question can be easily answered using the Boyle's Law, which states that as pressure increases, the volume is lower. It can be expressed in a mathematical way as follows:

$P_1*V_1=P_2*V_2$

So, from this equation we can clear V₂ to find the volume at 880 mm Hg:

$V_2= \frac{P_1*V_1}{P_2}= \frac{780 mmHg*100 mL}{880 mmHg}=88,64 mL$

Have a nice day!

4 0

4 years ago

Why is creativity important in scientific investigations?

irakobra [83]

I think the answer is option 2

4 0

4 years ago

Read 2 more answers