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jasenka [17]
3 years ago
5

During fission, _____. protons collide to create uranium-235 new elements are created when uranium breaks apart hydrogen atoms a

re fused together to create helium helium atoms are fused together to create hydrogen
Chemistry
2 answers:
tresset_1 [31]3 years ago
7 0
The correct answer is the second option. During fission, new elements are created when uranium breaks apart. This takes place in any of the heavy nuclei after capture of a neutron. This is the opposite of nuclear fusion. In this case, nuclei are broken down into two.
masya89 [10]3 years ago
5 0

Answer:

B. new elements are created when uranium breaks apart

Explanation:

You might be interested in
Argon, which comprises almost 1% of the atmosphere, is approximately 27 times more abundant than CO 2, but does not contribute t
kotegsom [21]

Answer:

The gas argon does not reach a state of vibrational excitation when infrared radiation strikes this gas.

Explanation:

The dry atmosphere is composed almost entirely of nitrogen (in a volumetric mixing ratio of 78.1%) and oxygen (20.9%), plus a series of oligogases such as argon (0.93%), helium and gases of greenhouse effect such as carbon dioxide (0.035%) and ozone. In addition, the atmosphere contains water vapor in very variable amounts (about 1%) and aerosols.

Greenhouse gases or greenhouse gases are the gaseous components of the atmosphere, both natural and anthropogenic, that absorb and emit radiation at certain wavelengths of the infrared radiation spectrum emitted by the Earth's surface, the atmosphere and clouds . In the Earth's atmosphere, the main greenhouse gases (GHG) are water vapor (H2O), carbon dioxide (CO2), nitrous oxide (N2O), methane (CH4) and ozone (O3 ). There is also in the atmosphere a series of greenhouse gases (GHG) created entirely by humans, such as halocarbons (compounds containing chlorine, bromine or fluorine and carbon, these compounds can act as potent greenhouse gases in the atmosphere and they are also one of the causes of the depletion of the ozone layer in the atmosphere) regulated by the Montreal Protocol. In addition to CO2, N2O and CH4, the Kyoto Protocol sets standards regarding sulfur hexafluoride (SF6), hydrofluorocarbons (HFCs) and perfluorocarbons (PFCs).

The difference between argon and greenhouse gases such as CO2 is that the individual atoms in the argon do not have free bonds and therefore do not vibrate. As a consequence, it does not reach a state of vibrational excitation when infrared radiation strikes this gas.

6 0
3 years ago
Put the list in chronological order (1–5).
Leokris [45]

Explanation:

Filtration is a separation technique in which solid particles suspended in liquid medium are separated by allowing the mixture through the pores of the filter paper. By this solid particles get collect on filter paper and liquid drains out from the pores of the filter paper.

The chronological order for given steps will be:

  1. Weigh and fold the filter paper.
  2. Place the filter paper in the funnel, then place the funnel in the Erlenmeyer flask.
  3. Allow the solid/liquid mixture to drain through the filter.
  4. Use water to rinse the filter paper containing the mixture.
  5. Weigh the dried filter paper and copper.
4 0
3 years ago
Read 2 more answers
Which energy profile best shows that the enthalpy of formation of CS2 is 89.4 KJ/mol?
Anna11 [10]

Answer:

Option C. Energy Profile D

Explanation:

Data obtained from the question include:

Enthalpy change ΔH = 89.4 KJ/mol.

Enthalpy change (ΔH) is simply defined as the difference between the heat of product (Hp) and the heat of reactant (Hr). Mathematically, it is expressed as:

Enthalpy change (ΔH) = Heat of product (Hp) – Heat of reactant (Hr)

ΔH = Hp – Hr

Note: If the enthalpy change (ΔH) is positive, it means that the product has a higher heat content than the reactant.

If the enthalpy change (ΔH) is negative, it means that the reactant has a higher heat content than the product.

Now, considering the question given, the enthalpy change (ΔH) is 89.4 KJ/mol and it is a positive number indicating that the heat content of the product is higher than the heat content of the reactant.

Therefore, Energy Profile D satisfy the enthalpy change (ΔH) for the formation of CS2 as it indicates that the heat content of product is higher than the heat content of the reactant.

7 0
3 years ago
Read 2 more answers
Consider the reaction below for which K = 78.2 atm-1. A(g) + B(g) ↔ C(g) Assume that 0.386 mol C(g) is placed in the cylinder re
borishaifa [10]

Answer:

1.65 L

Explanation:

The equation for the reaction is given as:

                        A            +            B           ⇄        C

where;

numbers of moles = 0.386 mol C  (g)

Volume =  7.29 L

Molar concentration of C = \frac{0.386}{7.29}

= 0.053 M

                        A            +            B           ⇄        C

Initial               0                           0                      0.530    

Change          +x                          +x                       - x

Equilibrium      x                           x                      (0.0530 - x)

K = \frac{[C]}{[A][B]}

where

K is given as ; 78.2 atm-1.

So, we have:

78.2=\frac{[0.0530-x]}{[x][x]}

78.2= \frac{(0.0530-x)}{(x^2)}

78.2x^2= 0.0530-x

78.2x^2+x-0.0530=0  

Using quadratic formula;

\frac{-b+/-\sqrt{b^2-4ac} }{2a}

where; a = 78.2 ; b = 1 ; c= - 0.0530

= \frac{-b+\sqrt{b^2-4ac} }{2a}   or \frac{-b-\sqrt{b^2-4ac} }{2a}

= \frac{-(1)+\sqrt{(1)^2-4(78.2)(-0.0530)} }{2(78.2)}  or \frac{-(1)-\sqrt{(1)^2-4(78.2)(-0.0530)} }{2(78.2)}

= 0.0204  or -0.0332

Going by the positive value; we have:

x = 0.0204

[A] = 0.0204

[B] = 0.0204

[C] = 0.0530 - x

     = 0.0530 - 0.0204

     = 0.0326

Total number of moles at equilibrium = 0.0204 +  0.0204 + 0.0326

= 0.0734

Finally, we can calculate the volume of the cylinder at equilibrium using the ideal gas; PV =nRT

if we make V the subject of the formula; we have:

V = \frac{nRT}{P}

where;

P (pressure) = 1 atm

n (number of moles) = 0.0734 mole

R (rate constant) = 0.0821 L-atm/mol-K

T = 273.15 K  (fixed constant temperature )

V (volume) = ???

V=\frac{(0.0734*0.0821*273.15)}{(1.00)}

V = 1.64604

V ≅ 1.65 L

3 0
3 years ago
2-phosphoglycerate(2PG) is converted to phosphoenolpyruvate (PEP) by the enzyme enolase. The standard free energy change(deltaGo
pogonyaev

Answer:

The correct option is: (D) -2.4 kJ/mol

Explanation:

<u>Chemical reaction involved</u>: 2PG ↔ PEP

Given: The standard Gibb's free energy change: ΔG° = +1.7 kJ/mol

Temperature: T = 37° C = 37 + 273.15 = 310.15 K    (∵ 0°C = 273.15K)

Gas constant: R = 8.314 J/(K·mol) = 8.314 × 10⁻³ kJ/(K·mol)     (∵ 1 kJ = 1000 J)

Reactant concentration: 2PG = 0.5 mM

Product concentration: PEP = 0.1 mM

Reaction quotient: Q_{r} =\frac{\left [ PEP \right ]}{\left [ 2PG \right ]} = \frac{0.1 mM}{0.5 mM} = 0.2

<u>To find out the Gibb's free energy change at 37° C (310.15 K), we use the equation:</u>

\Delta G = \Delta G^{\circ } + 2.303 R T log Q_{r}

\Delta G = 1.7 kJ/mol + [2.303 \times (8.314 \times 10^{-3} kJ/(K.mol))\times (310.15 K)] log (0.2)

\Delta G = 1.7 + [5.938] \times (-0.699) = 1.7 - 4.15 = (-2.45 kJ/mol)

<u>Therefore, the Gibb's free energy change at 37° C (310.15 K): </u><u>ΔG = (-2.45 kJ/mol)</u>

4 0
3 years ago
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