Can someone help me with these. I don’t understand how to answer them or do the work

Why blowing at the ping ping ball did not change its shape and size?

goldfiish [28.3K]

Answer:

Because it does not have to

Explanation:

Why because when you blow at it it might move it does not change it's shape because it is a solid

4 0

2 years ago

We know that organisms pass traits onto their offspring through reproduction. So how is it

Tpy6a [65]

Answer:

There are recessive traits and dominant traits

Explanation:

recessive traits are always hidden

dominant traits are the ones that are present in an organism

some recessive traits can be passed on to offsprings to become dominant

and some dominant traits can be passed on to become recessive traits

that's life

6 0

3 years ago

Read 2 more answers

B) calculate the ratio of moles of h2o to moles of anhydrous kal(so4)2. note: report the ratio to the closest whole number. (sho

Sindrei [870]

Answer: -

Mass of Hydrated KAl(SO₄)₂ = 2.0 g

Molar mass of anhydrous KAl(SO₄)₂ = 258.20 g/ mol

Mass of of anhydrous KAl(SO₄)₂ = mass of the 2nd heating = A

The mass of water released = mass of the Aluminum Cup + 2.0 grams of KAl(SO₄)₂ - mass of the 2nd heating

= H g

Moles of water released = $\frac{H g}{18 g/mol}$

Moles of anhydrous KAl(SO₄)₂ = $\frac{A g}{258.20 g/mol}$

Required ratio = $\frac{moles of water}{moles of anhydrous KAl(SO4)2}$

5 0

3 years ago

How to solve for K when given your anode and cathode equations and voltage

Aleks04 [339]

Answer:

See Explanation

Explanation:

In thermodynamics theory the Free Energy (ΔG) of a chemical system is described by the expression ΔG = ΔG° + RTlnQ. When chemical system is at equilibrium ΔG = 0. Substituting into the system expression gives ...

0 = ΔG° + RTlnKc, which rearranges to ΔG° = - RTlnKc. ΔG° in electrochemical terms gives ΔG° = - nFE°, where n = charge transfer, F = Faraday Constant = 96,500 amp·sec and E° = Standard Reduction Potential of the electrochemical system of interest.

Substituting into the ΔG° expression above gives

-nFE°(cell) = -RTlnKc => E°(cell) = (-RT/-nF)lnKc = (2.303·R·T/n·F)logKc

=> E°(cell) = (0.0592/n)logKc = E°(Reduction) - E°(Oxidation)

Application example:

Calculate the Kc value for a Zinc/Copper electrochemical cell.

Zn° => Zn⁺² + 2e⁻ ; E°(Zn) = -0.76 volt

Cu° => Cu⁺² + 2e⁻ ; E°(Cu) = 0.34 volt

By natural process, charge transfer occurs from the more negative reduction potential to the more positive reduction potential.

That is,

Zn° => Zn⁺² + 2e⁻ (Oxidation Rxn)

Cu⁺² + 2e⁻ => Cu° (Reduction Rxn)

E°(Zn/Cu) = (0.0592/n)logKc

= (0.0592/2)logKc = E°(Cu) - E°(Zn) = 0.34v - (-0.76v) = 1.10v

=> logKc = 2(1.10)/0.0592 = 37.2

=> Kc = 10³⁷°² = 1.45 x 10³⁷

3 0

3 years ago

Calculate the volume of air in liters that you might inhale (and exhale) in 12.0 hours. Assume that each breath has a volume of

zloy xaker [14]

Taking into account the rule of three, the volume of air that you inhale (and exhale) in 12.0 hours is 4,384.8 liters.

<h3>Rule of three</h3>

In first place, the rule of three is a way of solving problems of proportionality between three known values and an unknown value, establishing a relationship of proportionality between all of them.

That is, what is intended with it is to find the fourth term of a proportion knowing the other three.

If the relationship between the magnitudes is direct, that is, when one magnitude increases, so does the other (or when one magnitude decreases, so does the other) , the direct rule of three must be applied.

To solve a direct rule of three, the following formula must be followed, being a, b and c known data and x the variable to be calculated:

a ⇒ b

c ⇒ x

So: $x=\frac{cxb}{a}$

<h3 /><h3>Volume of air inhaled and exhaled</h3>

To calculated the volume of air inhaled and exhaled in 12 hours, you know that each breath has a volume of 0.435 liters, and that you are breathing 14 times a minute.

The you can apply the following rule of three: If each breath has a volume of 0.435 liters, 14 breaths have how much volume?

$volume=\frac{14 breathsx0.435 liters}{1 breath}$

<u><em>volume= 6.09 liters</em></u>

So, 14 breaths a minute have a volume of 6.09 liters, this means that in 1 minute you inhale and exhale 6.09 liters.

Now, knowing that 1 hour has 60 minutes, 12 hours has 720 minutes. So you can apply the following rule of three: in 1 minute you inhale and exhale 6.09 liters, in 720 minutes you inhale and exhale how much volume?

$volume=\frac{720 minutesx6.09 liters}{1 minute}$

<u><em>volume= 4,384.8 liters</em></u>

In summary, the volume of air that you inhale (and exhale) in 12.0 hours is 4,384.8 liters.

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3 0

2 years ago