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3 years ago

When performing physical exercise, the heart rate, by what percentage can it be increased?

Chemistry

2 answers:

Dafna11 [192]3 years ago

4 0

Answer:

Explanation:

I did that already. You got beo

Lina20 [59]3 years ago

3 0

Answer: 50% to 85% of the person’s maximum heart rate.

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It takes 495.0 kJ of energy to remove 1 mole of electron from an atom on the surface of sodium metal. How much energy does it ta

Zigmanuir [339]

Answer:

$\lambda=241.9\ nm$

Explanation:

The work function of the sodium= 495.0 kJ/mol

It means that

1 mole of electrons can be removed by applying of 495.0 kJ of energy.

Also,

1 mole = $6.023\times 10^{23}\ electrons$

So,

$6.023\times 10^{23}$ electrons can be removed by applying of 495.0 kJ of energy.

1 electron can be removed by applying of $\frac {495.0}{6.023\times 10^{23}}\ kJ$ of energy.

Energy required = $82.18\times 10^{-23}\ kJ$

Also,

1 kJ = 1000 J

So,

Energy required = $82.18\times 10^{-20}\ J$

Also, $E=\frac {h\times c}{\lambda}$

Where,

h is Plank's constant having value $6.626\times 10^{-34}\ Js$

c is the speed of light having value $3\times 10^8\ m/s$

So,

$79.78\times 10^{-20}=\frac {6.626\times 10^{-34}\times 3\times 10^8}{\lambda}$

$\lambda=\frac{6.626\times 10^{-34}\times 3\times 10^8}{82.18\times 10^{-20}}$

$\lambda=\frac{10^{-26}\times \:19.878}{10^{-20}\times \:82.18}$

$\lambda=\frac{19.878}{10^6\times \:82.18}$

$\lambda=2.4188\times 10^{-7}\ m$

Also,

1 m = 10⁻⁹ nm

So,

$\lambda=241.9\ nm$

6 0

2 years ago

How does the A Hreaction relate to the A He of molecules involved in a reaction?

igor_vitrenko [27]

Answer:

B. ΔHreaction = ΔH°f reactants- ΔH°f products

Explanation:

Using Hess's law, it is possible to sum ΔH of several related reactions to find ΔH of a particular reaction.

Having in mind Hess's law, ΔH°f is defined as the change in enthalpy during the formation of 1 mole of substance from its constituent elements (That is, pure elements, mono or diatomics, that have a ΔH° = 0).

For example, in ΔH°f of H₂O, the equation is:

H₂(g) + 1/2O₂(g) → H₂O(g)

The constituent elements with ΔH°f = 0 are H₂(g) and O₂(g).

Now, using Hess's law, you can sum the ΔH°f of substance in a reaction as, for example:

NaOH + HCl → H₂O + NaCl. ΔHr

The ΔH°f for each substance in the reaction is:

NaOH: Na + 1/2H₂ + 1/2O₂ → NaOH (1)

HCl: 1/2H₂ + 1/2Cl₂ → HCl (2)

H₂O: H₂ + 1/2O₂ → H₂O (3)

NaCl: Na + 1/2Cl₂ → NaCl (4)

The algebraic sum of (3) + (4) is -(ΔH°f reactants):

H₂ + 1/2O₂ + Na + 1/2Cl₂ → NaCl + H₂O ΔH°f reactants

This reaction - {(1)+(2)} ΔH°f products

NaOH + HCl → H₂O + NaCl.

ΔHr = ΔH°f reactants- ΔH°f products

In the example, we obtain this relationship that can be expanded for all reactions. Thus, right answer is:

<h3>B. ΔHreaction = ΔH°f reactants- ΔH°f products</h3>

8 0

3 years ago

What is the density of an object that has a mass of 20g and a volume of 10 cc?

Alexxandr [17]

i believe
2 g/mL
good luck!

5 0

3 years ago

Before a Bill can move to "floor consideration" in the General Assembly, it must first pass through

ANEK [815]

Answer:I believe it is the Senate

Explanation:

8 0

2 years ago

Read 2 more answers

How many moles of C are in a diamond that has a mass of 0.80 g? (Atomic mass of C=12.01g)

Alik [6]

Answer:

0.066 moles

Explanation:

0.80/12.01=0.0666

6 0

3 years ago