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aev [14]
3 years ago
14

3) Plasma occurs in electrical sparks. These sparks contain

Chemistry
2 answers:
Rudiy273 years ago
7 0

Answer:

C)  ions and electrons.

Explanation:

They are correct took the test

blsea [12.9K]3 years ago
4 0

Answer:

Explanation:

Plasma is the 4rth state of matter

Plasma is essentially composed of mixtures of ions and electrons .the ions and electrons are mixed containing equal amounts of positive and negative charges .

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Slkjflsdfjsfjlsm d dd dd dld dld dd;s dpd<br><br>.
MrMuchimi
Do you need any help ?
6 0
3 years ago
Onsider this representation of an original DNA sequence: THE TALL MAN WAS SAD.
kenny6666 [7]

Answer:

Substitution

Explanation:

just did it on edge

7 0
4 years ago
Read 2 more answers
C. Use Hess's law and the following equations to calculate ΔH for the reaction 4NH3 (g) + 5O2 (g) 4NO(g) + 6H2 O(g). Show your w
Monica [59]

Considering the Hess's Law, the enthalpy change for the reaction is -906.4 kJ/mol.

<h3>Hess's Law</h3>

Hess's Law indicates that the enthalpy change in a chemical reaction will be the same whether it occurs in a single stage or in several stages. That is, the sum of the ∆H of each stage of the reaction will give us a value equal to the ∆H of the reaction when it occurs in a single stage.

<h3>ΔH in this case</h3>

In this case you want to calculate the enthalpy change of:

4 NH₃ + 5 O₂ → 4 NO + 6 H₂O

which occurs in three stages.

You know the following reactions, with their corresponding enthalpies:

Equation 1: 2 N₂ + 6 H₂ → 4 NH₃   ΔH = –183.6 kJ/mol

Equation 2:  2 N₂ + 2 O₂ → 4 NO     ΔH = 361.1 kJ/mol

Equation 3: 2 H₂ + O₂→ 2 H₂O     ΔH = -483.7 kJ/mol

Because of the way formation reactions are defined, any chemical reaction can be written as a combination of formation reactions, some going forward and some going back.

In this case, first, to obtain the enthalpy of the desired chemical reaction you need 4 moles of NH₃ on reactant side and it is present in first equation on product side. So you need to invert the reaction, and when an equation is inverted, the sign of delta H also changes.

Now, 4 moles of NO must be a product and is present in the second equation, so let's write this as such.

Finally, you need 6 moles of H₂O on the product side, so you need to multiply by 3 the third equation to obtain the amount of water that you need. Since enthalpy is an extensive property, that is, it depends on the amount of matter present, since the equation is multiply by 3, the variation of enthalpy also.

In summary, you know that three equations with their corresponding enthalpies are:

Equation 1: 2 4 NH₃ → N₂ + 6 H₂  ΔH = 183.6 kJ/mol

Equation 2:  2 N₂ + 2 O₂ → 4 NO     ΔH = 361.1 kJ/mol

Equation 3: 6 H₂ + 3 O₂→ 6 H₂O     ΔH = -1,451.1 kJ/mol

Adding or canceling the reactants and products as appropriate, and adding the enthalpies algebraically, you obtain:

4 NH₃ + 5 O₂ → 4 NO + 6 H₂O ΔH= -906.4 kJ/mol

Finally, the enthalpy change for the reaction is -906.4 kJ/mol.

Learn more about Hess's law:

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#SPJ1

7 0
2 years ago
Describe two ways you think a sound wave could be stopped or could change change direction.
Alla [95]

Answer:

Like any wave, a sound wave doesn't just stop when it reaches the end of the medium or when it encounters an obstacle in its path. Rather, a sound wave will undergo certain behaviors when it encounters the end of the medium or an obstacle. Possible behaviors include reflection off the obstacle, diffraction around the obstacle, and transmission (accompanied by refraction) into the obstacle or new medium

8 0
3 years ago
Read 2 more answers
Find the final equilibrium temperature when 15.0 g of milk at 13.0 degrees c is added to 148 g of coffee with a temperature of 8
user100 [1]

According to zeroth law of thermodynamics, when two objects are kept in contact, heat (energy) is transferred from one to the other until they reach the same temperature (are in thermal equilibrium). When the objects are at the same temperature there is no heat transfer.

So, at equilibrium, q_{lost}=q_{gain},  q_{lost}= q_{milk} + q_{coffee}

q=m×c×T, where q = heat energy, m = mass of a substance, c = specific heat (units J/kg∙K), T is temperature

q_{lost}= (15X13X4.19)+(148X88.3X4.19), (15+148)X4.19XT_{final}=(15X13X4.19)+(148X88.3X4.19)

T_{final}= 81.37 ° C

6 0
3 years ago
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