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2 years ago

Classify the following as a type of potential energy or kinetic energy (use the letters K or P)

2 answers:

5 0

K, P, K, K, P, K, K, P, K, P. If it is moving, it is kinetic, if it isn't, it's potential. the sugar one is a little tricky using that method though, because we generally consider this in terms of spacial movement, but sugar holds energy which is later released by your body to allow you to move.the chemical bonds have potential energy because they release energy when broken.

katen-ka-za [31]2 years ago

5 0

1. k
2. p
3. k
4.k
5.
6.k
7.k
8.p
9.k
10.p

You might be interested in

Cobalt-60 is produced by a three reaction process involving neutron capture, beta-emission, and neutron capture. The initial rea

Dima020 [189]

Answer : The correct option is, (C) $^{58}\textrm{Fe}$

Explanation :

Neutron capture : In this decay process, an atomic nucleus and one or more number of neutrons collide and combine to form a heavier nucleus. The mass number changes in this process.

The neutron capture equation is represented as,

$_Z^A\textrm{X}+_{0}^1\textrm{n}\rightarrow _{Z}^{A+1}\textrm{X}+\gamma$

(A is the atomic mass number and Z is the atomic number)

Beta emission or beta minus decay : It is a type of decay process, in which a neutrons gets converted to proton, an electron and anti-neutrino. In this the atomic mass number remains same.

The beta minus decay equation is represented as,

$_Z^A\textrm{X}\rightarrow _{Z+1}^A\textrm{Y}+_{-1}^0e$

(A is the atomic mass number and Z is the atomic number)

As per question, the cobalt-60 is produced by a three reaction process involving neutron capture, beta-emission, and neutron capture.

Process 1 : Neutron capture.

$_{26}^{58}\textrm{Fe}+_{0}^1\textrm{n}\rightarrow _{26}^{59}\textrm{Fe}+\gamma$

Process 2 : Beta emission.

$_{26}^{59}\textrm{Fe}\rightarrow _{27}^{59}\textrm{Co}+_{-1}^0e$

Process 3 : Neutron capture.

$_{27}^{59}\textrm{Co}+_{0}^1\textrm{n}\rightarrow _{26}^{60}\textrm{Co}+\gamma$

From this we conclude that, the initial reactant in the production of cobalt-60 is $_{26}^{58}\textrm{Fe}$

Hence, the correct option is, (C) $^{58}\textrm{Fe}$

7 0

2 years ago

For scuba dives below 150 ft, helium is often used to replace nitrogen in the scuba tank. If 15.2 g of He(g) and 30.6 g of O2(g)

abruzzese [7]

Answer:

see explanation below

Explanation:

To do this exercise, we need to use the following expression:

P = nRT/V

This is the equation for an ideal gas. so, we have the temperature of 22 °C, R is the gas constant which is 0.082 L atm / mol K, V is the volume in this case, 5 L, and n is the moles, which we do not have, but we can calculate it.

For the case of the oxygen (AW = 16 g/mol):

n = 30.6 / 32 = 0.956 moles

For the case of helium (AW = 4 g/mol)_

n = 15.2 / 4 = 3.8 moles

Now that we have the moles, let's calculate the pressures:

P1 = 0.956 * 0.082 * 295 / 5

P1 = 4.63 atm

P2 = 3.8 * 0.082 * 295 / 5

P2 = 18.38 atm

Finally the total pressure:

Pt = 4.63 + 18.38

Pt = 23.01 atm

7 0

2 years ago

Methane, CH4, reacts with I2 according to the reaction CH4(g)+I2(g)⇌CH3I(g)+HI(g)

gtnhenbr [62]

Answer:

pCH₄ = 105.1 - 0.42 = 104.68 torr

pI₂ = 7.96 -0.42 = 7.54 torr

pCH₃I = 0.42 torr

pHI = 0.42 torr

Explanation:

Kp is the equilibrium constant for the partial pressure of the gases in the reaction, and it is calculated for a general equation:

aA(g) + bB(g) ⇄ cC(g) + dD(g)

$Kp = \frac{(pC)^cx(pD)^d}{(pA)^ax(pB)^b}$ , where p is the partial pressure in the equilibrium. By the reaction given:

CH₄(g) + I₂(g) ⇄ CH₃I(g) + HI(g)

105.1 torr 7.96 torr 0 0 <em> initial partial pressure</em>

-x -x +x +x <em> react</em>

105.1-x 7.96-x x x <em>equilibrium</em>

Then:

$Kp = \frac{pCH3IxpHI}{pCH4xpI2} = \frac{x^2}{(105.1-x)(7.96-x)}$

$2.26x10^{-4} = \frac{x^2}{836.596 - 113.06x -x^2}$

x² = 0.1891 - 0.0255x -2.26x10⁻⁴x²

0.9997x² + 0.0255x - 0.1891 = 0

Using Bhaskara's rule:

Δ = (0.0255)² - 4x(0.9997)x(-0.1891)

Δ = 0.7568

$x = \frac{-b+/-\sqrt{0.7568} }{2a} = \frac{-0.0255 +/-0.8699}{1.9994}$

Using only the positive term, x = 0.42 torr.

So,

pCH₄ = 105.1 - 0.42 = 104.68 torr

pI₂ = 7.96 -0.42 = 7.54 torr

pCH₃I = 0.42 torr

pHI = 0.42 torr

8 0

2 years ago

Read 2 more answers

Which of the compounds above are strong enough acids to react almost completely with a hydroxide ion (pka of h2o = 15.74) or wit

luda_lava [24]

The compounds can react with OH⁻ and HCO₃⁻ only C₅H₆N pyridinium

<h3><em>Further explanation </em></h3>

In an acid-base reaction, it can be determined whether or not a reaction occurs by knowing the value of pKa or Ka from acid and conjugate acid (acid from the reaction)

Acids and bases according to Bronsted-Lowry

Acid = donor (donor) proton (H + ion)

Base = proton (receiver) acceptor (H + ion)

If the acid gives (H +), then the remaining acid is a conjugate base because it accepts protons. Conversely, if a base receives (H +), then the base formed can release protons and is called the conjugate acid from the original base.

From this, it can be seen whether the acid in the product can give its proton to a base (or acid which has a lower Ka value) so that the reaction can go to the right to produce the product.

The step that needs to be done is to know the pKa value of the two acids (one on the left side and one on the right side of the arrow), then just determine the value of the equilibrium constant

Can be formulated:

K acid-base reaction = Ka acid on the left : K acid on the right.

or:

pK = acid pKa on the left - pKa acid on the right

K = equilibrium constant for acid-base reactions

pK = -log K;

$K~=~10^{-pK}$

K value> 1 indicates the reaction can take place, or the position of equilibrium to the right.

There is some data that we need to complete from the problem above, which is the pKa value of some compounds that will react, namely:

pyridinium pKa = 5.25

acetone pKa = 19.3

butan-2-one pKa = 19

Let's look at the K value of each possible reaction:

pka H₂O = 15.74, pka of H₂CO₃ = 6.37)

1. C₅H₆N pyridinium

* with OH⁻

C₅H₆N + OH- ---> C₅H₅N- + H₂O

pK = pKa pyridinium - pKa H₂O

pK = 5.25 - 15.74

pK = -10.49

$K~=~10^{4.9}$

K values> 1 indicate the reaction can take place

* with HCO3⁻

C₅H₆N + HCO₃⁻-- ---> C₅H₅N⁻ + H₂CO₃

pK = 5.25 - 6.37

pK = -1.12

$K`=~10^{1.12]$

Reaction can take place

2. Acetone C₃H₆O

* with OH-

C₃H₆O + OH⁻ ---> C₃H₅O- + H₂O

pK = 19.3 - 15.74

pK = 3.56

$K~=~10^{ -3.56}$

Reaction does not happen

* with HCO₃-

C₃H₆O + HCO₃⁻ ----> C₃H₅O⁻ + H₂CO₃

pK = 19.3 - 6.37

pK = 12.93

$K`=~10 ^{-12.93}$

Reaction does not happen

3. butan-2-one C₄H₇O

* with OH-

C₄H₇O + OH- ---> C₄H₆O- + H₂O

pK = 19 - 15.74

pK = 3.26

$K~=~10^{-3.26}$

Reaction does not happen

* with HCO₃⁻

C₄H₇O + HCO₃⁻ ---> C₄H₆O⁻ + H₂CO₃

pK = 19 - 6.37

pK = 12.63

$K~=~ 10^{-12.63}$

Reaction does not happen

So that can react with OH⁻ and HCO₃⁻ only C₅H₆N pyridinium

<h3><em>Learn more </em></h3>

the lowest ph

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the concentrations at equilibrium.

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the ph of a solution

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Keywords : acid base reaction, the equilibrium constant

5 0

2 years ago

Read 2 more answers

Determine the normality of the following solutions note the species of interest is H 95 g of PO4 3- in 100mL solution

Aliun [14]

Answer : The normality of the solution is, 30.006 N

Explanation :

Normality : It is defined as the number of gram equivalent of solute present in one liter of the solution.

Mathematical expression of normality is:

$\text{Normality}=\frac{\text{Gram equivalent of solute}}{\text{Volume of solution in liter}}$

or,

$\text{Normality}=\frac{\text{Weight of solute}}{\text{Equivalent weight of solute}\times \text{Volume of solution in liter}}$

First we have to calculate the equivalent weight of solute.

Molar mass of solute $PO_4^{3-}$ = 94.97 g/mole

$\text{Equivalent weight of solute}=\frac{\text{Molar mass of solute}}{\text{charge of the ion}}=\frac{94.97}{3}=31.66g.eq$

Now we have to calculate the normality of solution.

$\text{Normality}=\frac{95g}{31.66g.eq\times 0.1L}=30.006eq/L$

Therefore, the normality of the solution is, 30.006 N

5 0

2 years ago