Which two structures would provide a positive identification of an plant cell under a microscope?

Cindy made tea. She started with 300 grams of water at 20 degrees Celsius. She transferred 18,000 calories to the water. What wa

Serjik [45]

Answer:

T final = 80°C

Explanation:

Q = mCpΔT

∴ Q = 18000 cal

∴ m H2O = 300 g

∴ Cp H2O (15°C) = 0.99795 cal/g.K ≅ 1 cal/g.K

∴ T1 = 20°C = 293 K

∴ T2 = ?

⇒ 18000 cal = (300 g)(1 cal/g.K)(T2 - 293 K)

⇒ (18000 cal)/(300 cal/K) = T2 - 293 K

⇒ T2 = 293 K + 60 K

⇒ T2 = 353 K (80°C)

8 0

3 years ago

Consider a generic redox reaction?

vesna_86 [32]

The Nernst equation allows us to predict the cell potential for voltaic cells under conditions other than the standard conditions of 1M, 1 atm, 25°C. The effects of different temperatures and concentrations may be tracked in terms of the Gibbs energy change ΔG. This free energy change depends upon the temperature & concentrations according to ΔG = ΔG° + RTInQ where ΔG° is the free energy change under conditions and Q is the thermodynamic reaction quotient. The free energy change is related to the cell potential Ecell by ΔG= nFEcell

so for non-standard conditions
-nFEcell = -nFE°cell + RT InQ

or

Ecell = E°cell - RT/nF (InQ)

which is called Nernst equation.

5 0

3 years ago

write equations to show the chemical processes which occur when the first ionization and the second ionization energies of lithi

diamong [38]

Answer:

First ionization of lithium:

$\text{Li}\;(g)\to \text{Li}^{+} \; (g) + \text{e}^{-}$ .

Second ionization of lithium:

$\text{Li}^{+}\;(g) \to\text{Li}^{2+} \;(g) + \text{e}^{-}$ .

Explanation:

The ionization energy of an element is the energy required to remove the outermost electron from an atom or ion of the element in gaseous state. (Refer to your textbook for a more precise definition.) Some features of the equation:

Start with a gaseous atom (for the first ionization energy only) or a gaseous ion. Write the gaseous state symbol $(g)$ next to any atom or ion in the equation.
The product shall contain one gaseous ion and one electron. The charge on the ion shall be the same as the order of the ionization energy. For the second ionization energy, the ion shall carry a charge of +2.
Charge shall balance on the two sides of the equation.

First Ionization Energy of Li:

The products shall contain a gaseous ion with charge +1 $\text{Li}^{+}\;(g)$ as well as an electron $\text{e}^{-}$ .
Charge shall balance on the two sides. There's no net charge on the product side. Neither shall there be a charge on the reactant side. The only reactant shall be a lithium atom which is both gaseous and neutral: $\text{Li}\;(g)$ .

Hence the equation: $\text{Li}\;(g) \to \text{Li}^{+}\;(g) + \text{e}^{-}$ .

Second Ionization Energy of Li:

The product shall contain a gaseous ion with charge +2: $\text{Li}^{2+}\;(g)$ as well as an electron $\text{e}^{-}$ .
Charge shall balance on the two sides. What's the net charge on the product side? That shall also be the charge on the reactant side. What will be the reactant?

The equation for this process is $\text{Li}^{+} \; (g) \to \text{Li}^{2+}\;(g) + \text{e}^{-}$ .

5 0

3 years ago

Friction is generated when _____ interfere with each other on sliding surfaces. solids molecules forces fluids

9966 [12]

Friction is generated when sliding

5 0

3 years ago

How much energy, in joules, does 150.0 g of water with an initial temperature of 25 C need to absorb be raised to a final temper

satela [25.4K]

Answer:

31395 J

Explanation:

Given data:

mass of water = 150 g

Initial temperature = 25 °C

Final temperature = 75 °C

Energy absorbed = ?

Solution:

Formula:

q = m . c . ΔT

we know that specific heat of water is 4.186 J/g.°C

ΔT = final temperature - initial temperature

ΔT = 75 °C - 25 °C

ΔT = 50 °C

now we will put the values in formula

q = m . c . ΔT

q = 150 g × 4.186 J/g.°C × 50 °C

q = 31395 J

so, 150 g of water need to absorb 31395 J of energy to raise the temperature from 25°C to 75 °C .

5 0

3 years ago