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3 years ago

Acid-base neutralization means having a pH of 7. true or false ?

Chemistry

2 answers:

Aleks04 [339]3 years ago

8 0

Answer:

True

Explanation:

A pH of 7 is greater than basic but lesser than acidic. Therefore, it's neutral

Firdavs [7]3 years ago

6 0

True, When a strong acid is neutralized by a strong base there are no excess hydrogen ions left in the solution.

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A year has 365 days, and a day has 24 hours. If an hour has 60 minutes and a minute has 60 seconds, how many seconds are there i

Cloud [144]

Answer:

31,536,000 seconds

Explanation:

off google

3 0

3 years ago

Read 2 more answers

A 27 kg iron block initially at 375 C is quenched in an insulated tank that contains 130kg of water at 26 C. Assume the water th

Bess [88]

Solution :

a). Applying the energy balance,

$\Delta E_{sys}=E_{in}-E_{out}$

$0=\Delta U$

$0=(\Delta U)_{iron} + (\Delta U)_{water}$

$0=[mc(T_f-T_i)_{iron}] + [mc(T_f-T_i)_{water}]$

$0 = 27 \times 0.45 \times (T_f - 375) + 130 \times 4.18 \times (T_f-26)$

$t_f=33.63^\circ C$

b). The entropy change of iron.

$\Delta s_{iron} = mc \ln\left(\frac{T_f}{T_i} \right)$

$= 27 \times 0.45\ \ln\left(\frac{33.63 + 273}{375 + 273} \right)$

= -9.09 kJ-K

Entropy change of water :

$\Delta s_{water} = mc \ \ln\left(\frac{T_f}{T_i} \right)$

$= 130 \times 4.18\ \ln\left(\frac{33.63 + 273}{26 + 273} \right)$

= 10.76 kJ-K

So, the total entropy change during the process is :

$\Delta s_{tot} = \Delta s_{iron} + \Delta s_{water}$

= -9.09 + 10.76

= 1.67 kJ-K

c). Exergy of the combined system at initial state,

$X=(U-U_{0}) - T_0(S-S_0)+P_0(V-V_0)$

$X=mc (T-T_0) - T_0 \ mc \ \ln \left(\frac{T}{T_0} \right)+0$

$X=mc\left((T-T_0)-T_0 \ ln \left(\frac{T}{T_0} \right)\right)$

$X_{iron, i} = 27 \times 0.45\left(((375+273)-(12+273))-(12+273) \ln \frac{375+273}{12+273}\right)$

$X_{iron, i} =63.94 \ kJ$

$X_{water, i} = 130 \times 4.18\left(((26+273)-(12+273))-(12+273) \ln \frac{26+273}{12+273}\right)$

$X_{water, i} =-13.22 \ kJ$

Therefore, energy of the combined system at the initial state is

$X_{initial}=X_{iron,i} +X_{water, i}$

= 63.94 -13.22

= 50.72 kJ

Similarly, Exergy of the combined system at initial state,

$X=(U_f-U_{0}) - T_0(S_f-S_0)+P_0(V_f-V_0)$

$X=mc\left((T_f-T_0)-T_0 \ ln \left(\frac{T_f}{T_0} \right)\right)$

$X_{iron, f} = 27 \times 0.45\left(((33.63+273)-(12+273))-(12+273) \ln \frac{33.63+273}{12+273}\right)$

$X_{iron, f} = 216.39 \ kJ$

$X_{water, f} = 130 \times 4.18\left(((33.63+273)-(12+273))-(12+273) \ln \frac{33.63+273}{12+273}\right)$

$X_{water, f} =-9677.95\ kJ$

Thus, energy or the combined system at the final state is :

$X_{final}=X_{iron,f} +X_{water, f$

= 216.39 - 9677.95

= -9461.56 kJ

d). The wasted work

$X_{in} - X_{out}-X_{destroyed} = \Delta X_{sys}$

$0-X_{destroyed} =$

$X_{destroyed} = X_{initial} - X_{final}$

= 50.72 + 9461.56

= 9512.22 kJ

6 0

3 years ago

What is the mass (in mg) of 2.63 moles of nickel?

zaharov [31]

Data:
Molar Mass of Nickel = 58,7 g/mol

Solving:
58,7 g → 1 mol
y -------→ 2.63 mol

Solving: (They are proportional measures, the rule of three is made (directly proportional)

$\frac{58.7}{y} = \frac{1}{2.63}$
multiply cross
$1*y = 58.7*2.63$
$y = 154.381\:g \stackrel{converting}{\longrightarrow}\:\boxed{y = 154381\:mg}$

8 0

3 years ago

What is changed when matter undergoes a physical change?

Furkat [3]

I think its the mass that changes but im not sure

3 0

3 years ago

A government laboratory wants to determine whether water in a certain city has any traces of fluoride and whether the concentrat

nexus9112 [7]

<u>Answer:</u> The given sample of water is not safe for drinking.

<u>Explanation:</u>

We are given:

Concentration of fluorine in water recommended = 4.00 ppm

ppm is the amount of solute (in milligrams) present in kilogram of a solvent. It is also known as parts-per million.

To calculate the ppm of fluorine in water, we use the equation:

$\text{ppm}=\frac{\text{Mass of solute}}{\text{Mass of solution}}\times 10^6$

Both the masses are in grams.

We are given:

Mass of fluorine = $0.152mg=0.152\times 10^{-3}g$ (Conversion factor: 1 g = 1000 mg)

Mass of water = 5.00 g

Putting values in above equation, we get:

$\text{ppm of fluorine in water}=\frac{0.152\times 10^{-3}}{5}\times 10^6\\\\\text{ppm of fluorine in water}=30.4$

As, the calculated concentration is greater than the recommended concentration. So, the given sample of water is not safe for drinking.

Hence, the given sample of water is not safe for drinking.

6 0

3 years ago