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ArbitrLikvidat [17]
3 years ago
5

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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Calculate the number of moles in the following: 2.8 X 10^24 atoms of Cl2
vova2212 [387]

Answer:

<h3>The answer is 4.65 moles</h3>

Explanation:

To find the number of moles given it's number of entities we use the formula

n =  \frac{N}{L} \\

where n is the number of moles

N is the number of entities

L is the Avogadro's constant which is

6.02 × 10²³ entities

From the question

N = 2.8 × 10²⁴ atoms of Cl2

So we have

n =  \frac{2.8 \times  {10}^{24} }{6.02 \times  {10}^{23}  }  \\  = 4.65116279069...

We have the final answer as

<h3>4.65 moles</h3>

Hope this helps you

7 0
3 years ago
PLS HELP!!!!
Mashcka [7]

Answer:

the 3rd answer

7 0
2 years ago
Read 2 more answers
2. Which equation represents a transmutation?
antiseptic1488 [7]

8935982560 plz contact me

5 0
2 years ago
Calculate the change in temperature (ΔT) that occurs when 8000 J of energy (q) is used to heat up a mass (m) of 75 g of water. (
pickupchik [31]

Answer:

The change in temperature that occurs when 8000 J of heat is used by a mass 75 g of water is 25.4 °C

Explanation:

H = mc ΔT

m = 75 g

c = 4. 200 J/ g °C

H = 8000 J

ΔT =?

Rearranging the formula, making ΔT the subject of formula;

ΔT = H / m c

ΔT = 8000 / 75 * 4.200

ΔT = 8000 / 315

ΔT = 25.4 °C

8 0
3 years ago
Read 2 more answers
Argon (Ar) and helium (He) are initially in separate compartments of a container at 25°C. The
love history [14]

Answer:

(a) V_B=11.68L

(b) x_{He}=0.533

Explanation:

Hello,

In this case, since the both gases behave ideally, with the given information we can compute the moles of He in A:

n_A=\frac{0.082\frac{atm*L}{mol*K}*298K}{1.974 atm*6.00L}=2.063mol

Thus, since the final pressure is 3.60 bar, we can write:

P=x_{Ar}P_A+x_{He}P_B\\\\P=\frac{n_{Ar}}{n_{Ar}+n_{He}} P_A+\frac{n_{He}}{n_{Ar}+n_{He}} P_B\\\\3.60bar=\frac{2.063mol}{2.063mol+n_{He}} *2.00bar+\frac{n_{He}}{2.063mol+n_{He}} *5.00bar

The moles of helium could be computed via solver as:

n_{He}=2.358mol

Or algebraically:

3.60bar=\frac{1}{2.063mol+n_{He}} *(4.0126+5.00*n_{He})\\\\7.314+3.60n_{He}=4.013+5.00*n_{He}\\\\7.314-4.013=5.00*n_{He}-3.60n_{He}\\\\n_{He}=\frac{3.3}{1.4}=2.358mol

In such a way, the volume of the compartment B is:

V_B=\frac{n_{He}RT}{P_B}=\frac{2.358mol*0.082\frac{atm*L}{mol*K}*298.15K}{4.935atm}\\  \\V_B=11.68L

Finally, he mole fraction of He is:

x_{He}=\frac{2.358}{2.358+2.063}\\ \\x_{He}=0.533

Regards.

8 0
3 years ago
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