All categories

2 years ago

Explain any differences in the pulse rate at rest and after exercise ( in own words btw^^

Chemistry

1 answer:

timurjin [86]2 years ago

5 0

Answer:

Pulse rate depends on the activity we do, while exercising theres need of more oxygen so it increases but while at rest body requires less oxygen so low pulse rate.

You might be interested in

A student mixes some sugar into a cup of hot water. The sugar dissolves in the liquid. What are the BEST ways for the student to

polet [3.4K]

You would have to evaporate the water to get just the sugar

6 0

3 years ago

Practice entering numbers that include a power of 10 by entering the diameter of a hydrogen atom in its ground state, dH=1.06×10

kkurt [141]

Answer:

The diameter of the hydrogen $\mathbf{d =1.0605 \times 10^{-10}\ m}$

Explanation:

From the given information:

Using the concept of Bohr's Model, the equation for the angular momentum can be expressed as:

$L = \dfrac{nh}{2 \pi}$

Where the generic expression for angular momentum is:

L = mvr.

replacing the value of L into the previous equation, we have:

$mvr= \dfrac{nh}{2 \pi}$

$v= \dfrac{nh}{2 \pi mr}$ ----- (1)

The electron in the hydrogen atom posses an electrostatic force which gives a centripetal force.

$\dfrac{ke^2}{r^2} = \dfrac{mv^2}{r}$ ----- (2)

replacing the value of v in equation (1) into (2), and taking r as the subject of the formula, we have:

$\dfrac{ke^2}{r} = m (\dfrac{nh}{2 \pi mr})^2$

$ke^2=\dfrac{n^2h^2}{4 \pi^2 mr}$

$r =\dfrac{n^2h^2}{4 \pi^2 mke^2}$

For ground-state n = 1

$h = (6.625 \times 10^{-34} \ J.s)^2$

$m =( 9.1 \times 10^{-31} \ kg)(9 \times 10^9 \ N .m^2/C^2)$

$Ke = (1.6 \times 10^{-19} \ C)^2$

$r =\dfrac{(1)^2(6.625 \times 10^{-34})^2}{4 \pi^2 (9.1 \times 10^{-31} )(9 \times 10^9 ) (1.6 \times 10^{-19})^2}$

$r =\dfrac{4.3890625 \times 10^{-67}}{8.27720295 \times 10^{-57}}$

$\mathbf{r = 5.3025 \times 10^{-11} \ m}$

Therefore, the diameter of hydrogen d = 2r

$\mathbf{d = ( 2 \times 5.3025 \times 10^{-11} \ m})}$

$\mathbf{d =1.0605 \times 10^{-10}\ m}}$

4 0

3 years ago

Determine the concentrations of mgcl2, mg2 , and cl– in a solution prepared by dissolving 2.75 × 10–4 g mgcl2 in 1.75 l of water

inysia [295]

Answer:

M of MgCl₂ = 1.65 × 10⁻⁶ M

M of Mg²⁺ = 1.65 × 10⁻⁶ M

M of Cl⁻ = 3.30 × 10⁻⁶ M

Explanation:

1) MgCl₂

Molarity = number of moles of solute / volume of solution in liters, M = n / V

n = mass in grams / molar mass

molar mass of MgCl₂ = 24.305 g/mol + 2(35.543 g/mol) = 95.211 g/mol

n = 2.75 × 10⁻⁴ g / 95.211 g/mol = 2.89×10⁻³ moles

⇒ M = n / V = 2.89×10⁻³ moles / 1.75 l = 1.65 × 10⁻⁶ M

2) Mg²⁺ and Cl⁻

Those are the ions in solution.

You assume 100% dissociation of the ionic compound (strong electrolyte).

Then the equation is: MgCl₂ → Mg²⁺ + 2Cl⁻

That means that 1 mol of MgCl₂ produces 1 mol of Mg²⁺ and 2 moles of Cl⁻.

That yields the same molarity concentration of Mg²⁺ , while the molarity concentration of Cl⁻ is the double.

So, the results are:

M of MgCl₂ = 1.65 × 10⁻⁶ M

M of Mg²⁺ = 1.65 × 10⁻⁶ M

M of Cl⁻ = 3.30 × 10⁻⁶ M

8 0

3 years ago

Read 2 more answers

A 0.24g sample of compound of oxygen and boron was found by analysis to contain 0.144g of Oxygen.calculate the percentage compos

oksano4ka [1.4K]

Answer:

Mass of compound = 0.24 g and, mass of boron = 0.096 g percentage of boron in the compound = mass of boron / mass of compound * 100 = 0.096/0.24 * 100 = 40% mass of oxygen = 0.144 g again, mass of compound = 0.24 g percentage of oxygen in compound = mass of oxygen/mass of of compound * 100 = 0.144/0.24 * 100 = 60%

<h2>Hope it's Helpful!!✌️</h2>

5 0

2 years ago

When this happens the inward force of gravity is exactly balanced with the outward force created by _______ and _____________.

4vir4ik [10]

Sw ocjhohz hug tú the funny ohhzubutgrrfuY

8 0

3 years ago