The ground state electron configuration of Fe is what

Calculate the total energy, in kilojoules, that is needed to turn a 46 g block

neonofarm [45]

Answer: The total energy, in kilojoules, that is needed to turn a 46 g block of ice at -25 degrees C into water vapor at 100 degrees C is 11.787 kJ.

Explanation:

Given: Mass = 46 g

Initial temperature = $-25^{o}C$

Final temperature = $100^{o}C$

Specific heat capacity of ice = 2.05 $J/g^{o}C$

Formula used to calculate the energy is as follows.

$q = m \times C \times (T_{2} - T_{1})$

where,

q = heat energy

m = mass

C = specific heat capacity

$T_{1}$ = initial temperature

$T_{2}$ = final temperature

Substitute the values into above formula as follows.

$q = m \times C \times (T_{2} - T_{1})\\= 46 g \times 2.05 J/g^{o}C \times (100 - (-25))^{o}C\\= 11787.5 J (1 J = 0.001 kJ)\\= 11.787 kJ$

Thus, we can conclude that the total energy, in kilojoules, that is needed to turn a 46 g block of ice at -25 degrees C into water vapor at 100 degrees C is 11.787 kJ.

7 0

3 years ago

If a school uses 1200 gallons/day. How many liters/sec is this?

Naddik [55]

Answer:

0.05257 L/s

Explanation:

Step 1: Given data

The school uses 1200 gallons/day

Step 2: Convert "gal/day" to "L/day"

We will use the conversion factor 1 gal = 3.785 L.

1200 gal/day × (3.785L/gal) = 4542 L/day

Step 3: Convert "L/day" to "L/s"

We will use the following conversion factors:

1 day = 24 h
1 h = 3600 s

4542 L/day × (1 day/24 h) × (1 h/3600 s) = 0.05257 L/s

5 0

3 years ago

Calculate the pH of: (a) 0.1M HCl; (b) 0.1M NaOH; (c) 3 X 10% M HNO3; (d) 5 X 10-10 M HCIO.; and (e) 2 x 10-8 M KOH.

Shtirlitz [24]

Answer:

(a) $pH = -Log (0.1M) = 1$

(b) $pH = -Log (10^{-13}M) = 13$

(c) $pH = -Log (3x10^{-3}M) = 2.5$

(d) $pH = -Log (4.93x10^{-10}M) = 9.3$

(e) $pH = -Log (5^{-7}M) = 6.3$

Explanation:

To calculate de pH of an acid solution the formula is:

$pH = -Log ([H^{+}]) = 1$

were [H^{+}] is the concentration of protons of the solution. Therefore it is necessary to know the concentration of the protons for every solution in order to solve the problem.

(a) and (c) are strong acids so they dissociate completely in aqueous solution. Thus, the concentration of the acid is the same as the protons.

(b) and (e) are strong bases so they dissociate completely in aqueous solution too. Thus, the concentration of the base is the same as the oxydriles. But in this case it is necessary to consider the water autoionization to calculate the protons concentration:

$K_{w} =[H^{+} ][OH^{-}]=10^{-14}$