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Serjik [45]
3 years ago
15

Which domain consists of single-celled organisms, some of which can be treated with anitibiotics

Chemistry
2 answers:
Aneli [31]3 years ago
3 0
I got bacteria for this one................
My name is Ann [436]3 years ago
3 0
Yes the correct answer is Bacteria
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Which chemical equation represents a precipitation reaction? A. Mg(ClO3)2(aq) + 2HCl(aq) → MgCl2(aq) + 2HClO3(aq) B. AlBr3(aq) +
Rama09 [41]

Answer is D.

Among the given options, the products of equation A, B and C are aqueous solutions. Those are indicated by using (aq).

The balanced equation for the reaction between Li₂CO₃(aq) and FeCl₂(aq) is Li₂CO₃(aq) + FeCl₂(aq) → 2LiCl(aq) + FeCO₃(s)

LiCl is a salt while FeCO₃ is a Grey/green color precipitate. Hence, option D represents a precipitation reaction.

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Would copper exist in a monatomic form, <br> a lattice or as molecules?
Elan Coil [88]

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Calculate the total amount of energy required to change 10.0 g of water from 35.0 degrees Celsius to 110. degrees Celsius.
Makovka662 [10]

Answer:

The total amount of energy required is 25,515.2 J.

Explanation:

Calorimetry is the measurement and calculation of the amounts of heat exchanged by a body or a system.

When a system absorbs (or gives up) a certain amount of heat, it can happen that:

  • experience a change in its temperature, which involves sensible heat,
  • undergoes a phase change at constant temperature, or latent heat.

To calculate the latent heat the formula is used:

Q = m. L

Where

  • Q: amount of heat
  • m: mass
  • L: latent heat

To calculate sensible heat the following formula is used:

Q = m. c. ΔT

where:

  • Q: amount of sensible heat  
  • m: body mass
  • c: specific heat of the substance
  • ΔT: temperature range

In this case, you have in the first place a heat to raise the temp of the water from 35.0 C to 100 C, where the specific heat value for water is  4.184 \frac{J}{g*C}:

q1 = m*c*(Tfinal-Tinitial)

q1 = 10.0 g *(4.184 \frac{J}{g*C})* (100 - 35.0 C) = 2719.6 J

Now you have the heat to vaporize the water, where the heat of vaporization is 2259.36 \frac{J}{g}:

q2 = m*(heat of vaporization)

q2 = 10.0 g*(2259.36 \frac{J}{g}) = 22593.6 J

Finally, you have the heat to raise temp of steam to 110 C, where the specific heat value for steam is  2.02 \frac{J}{g*C} :

q3 = m*c*(Tfinal-Tinitial)

q3 = 10.0 g*(2.02 \frac{J}{g*C})*(110-100 C) = 202 J

The total amount of energy can be calculated as:

Q= q1 + q2 + q3

Q= 2719.6 J + 22593.6 J + 202 J

Q=25,515.2 J

<u><em>The total amount of energy required is 25,515.2 J.</em></u>

5 0
3 years ago
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