All categories

3 years ago

Label the parts of the electric circuit that best correspond to the heart, arteries, veins, and cells.

Chemistry

1 answer:

katovenus [111]3 years ago

3 0

Answer:

1 ➡️ Cells

2 ➡️ Arteries

3 ➡️ Veins

4 ➡️ Heart

Explanation:

The parts of the electric circuit that best correspond to the heart, arteries, veins, and cells have been properly labeled.

The circulatory system involves the transportation of nutrients, oxygen and water by blood to other the parts of the body.

From the electric circuit, we see that arteries transport blood away from the heart to the other cells in the body. The veins actually return the blood back to the heart from the cells. The heart pumps the blood

You might be interested in

A plastic bin is found to hold 3.1x10^24 molecules of water.

motikmotik

Answer:

$\boxed {\boxed {\sf 5.1 \ mol \ H_2O}}$

Explanation:

To convert from representative particles to moles, Avogadro's Number: 6.02*10²³, which tells us the number of particles (atoms, molecules, etc.) in 1 mole of a substance.

We can use it in a ratio.

$\frac {6.02*10^{23} \ molecules \ H_2O}{1 \ mol \ H_2O}$

Multiply by the given number of molecules.

$3.1*10^{24} \ molecules \ H_2O*\frac {6.02*10^{23} \ molecules \ H_2O}{1 \ mol \ H_2O}$

Flip the ratio so the molecules of water cancel out.

$3.1*10^{24} \ molecules \ H_2O*\frac {1 \ mol \ H_2O}{6.02*10^{23} \ molecules \ H_2O}$

$3.1*10^{24} *\frac {1 \ mol \ H_2O}{6.02*10^{23} }$

$\frac {3.1*10^{24} \ mol \ H_2O}{6.02*10^{23} }$

Divide.

$5.14950166113 \ mol \ H_2O$

The original number of molecules has 2 significant figures: 3 and 1, so our answer must have the same. For the number we calculated, that is the tenth place. The 4 in the hundredth place tells us to leave the 1.

$5.1 \ mol \ H_2O$

There are about 5.1 moles of water in 3.1*10²⁴ molecules of water.

5 0

2 years ago

How does gravity affect the moon?

Shalnov [3]

A, The moon pulls on the earth making the tides rise and fall

8 0

3 years ago

Read 2 more answers

The volume of 7.91 M HCl needed to make 196.1 mL of 2.13 M HCl is ____.

Alona [7]

Answer:

a. 52.8

Explanation:

To find the number of moles of HCl we use the relation M₁V₁=M₂V₂

where M₁ is the initial molarity, M₂ the new molarity, V₁ the initial volume used, and V₂ the final volume obtained.

M₁=7.91 M

M₂=2.13 M

V₁=?

V₂=196.1 mL

Replacing these values in the relationship.

M₁V₁=M₂V₂

7.91 M× V₁=2.13 M×196.1 mL

V₁=(2.13 M×196.1 mL)/7.91 M

=52.8 mL

3 0

3 years ago

Explain how the structure and bonding in bromine account for it's relatively low melting point. [3]

Sever21 [200]

Melting point is dependent on the intermolecular forces which means the bonds between the molecules of bromine as it is a simple molecular structure the intermolecular bonds of bromine are weak bcz they are weak vandervaal forces thats why Bromines melting point is low..In short when intermolecular bomds are weak the M.P is lower

6 0

3 years ago

Read 2 more answers

Consider the following reaction where Kc = 1.80×10-2 at 698 K:

Klio2033 [76]

Answer:

The system is not in equilibrium and the reaction must run in the forward direction to reach equilibrium.

Explanation:

The reaction quotient Qc is a measure of the relative amount of products and reagents present in a reaction at any given time, which is calculated in a reaction that may not yet have reached equilibrium.

For the reversible reaction aA + bB⇔ cC + dD, where a, b, c and d are the stoichiometric coefficients of the balanced equation, Qc is calculated by:

$Qc=\frac{[C]^{c}*[D]^{d} } {[A]^{a}*[B]^{b}}$

In this case:

$Qc=\frac{[H_{2} ]*[I_{2} ] } {[HI]^{2}}$

Since molarity is the concentration of a solution expressed in the number of moles dissolved per liter of solution, you have:

$[H_{2} ]=\frac{2.09*10^{-2} moles}{1 Liter}$ =2.09*10⁻² $\frac{moles}{liter}$

$[I_{2} ]=\frac{4.14*10^{-2} moles}{1 Liter}$ =4.14*10⁻² $\frac{moles}{liter}$

$[I_{2} ]=\frac{0.280 moles}{1 Liter}$ = 0.280 $\frac{moles}{liter}$

So,

$Qc=\frac{2.09*10^{-2} *4.14*10^{-2} } {0.280^{2} }$

Qc= 0.011

Comparing Qc with Kc allows to find out the status and evolution of the system:

If the reaction quotient is equal to the equilibrium constant, Qc = Kc, the system has reached chemical equilibrium.

If the reaction quotient is greater than the equilibrium constant, Qc> Kc, the system is not in equilibrium. In this case the direct reaction predominates and there will be more product present than what is obtained at equilibrium. Therefore, this product is used to promote the reverse reaction and reach equilibrium. The system will then evolve to the left to increase the reagent concentration.

If the reaction quotient is less than the equilibrium constant, Qc <Kc, the system is not in equilibrium. The concentration of the reagents is higher than it would be at equilibrium, so the direct reaction predominates. Thus, the system will evolve to the right to increase the concentration of products.

Being Qc=0.011 and Kc=1.80⁻²=0.018, then Qc<Kc. <u><em>The system is not in equilibrium and the reaction must run in the forward direction to reach equilibrium.</em></u>

8 0

3 years ago