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postnew [5]
4 years ago
10

You may experiment alone if you apply which of the following safety precautions?

Chemistry
2 answers:
serg [7]4 years ago
6 0
 The answer is C. Follow all safety protocols
A could have also been the answer, but C is follow all.
ankoles [38]4 years ago
4 0
A and b are both safety protocols, but C says to follow all saftey protocols. your answer is C
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Use standard reduction potentials to calculate the equilibrium constant for the reaction: 2Cr3+(aq) + Pb(s)2Cr2+(aq) + Pb2+(aq)
inn [45]

Answer:

3.47 ×10^-10

Explanation:

The equation of the reaction is 2Cr3+(aq) + Pb(s)------->2Cr2+(aq) + Pb2+(aq)

A total of two moles of electrons were transferred in the process. The chromium was reduced while the lead was oxidized. Hence the lead species will constitute the oxidation half equation and the chromium will constitute the reduction half equation.

E°cell = E°cathode - E°anode

E°cathode = -0.41 V

E°anode = -0.13 V

E°cell = -0.41 -(-0.13) = -0.28 V

From

E°cell = 0.0592/n log K

n= 2, K= the unknown

-0.28 = 0.0592/2 log K

log K = -0.28/0.0296

log K = -9.4595

K = Antilog ( -9.4595)

K= 3.47 ×10^-10

4 0
3 years ago
Sceince problem i need help
Cerrena [4.2K]
D. due to the the water it will bring sand with the water there for us is D.
3 0
2 years ago
Read 2 more answers
How many milliliters of 0.150 M NaOH are required to neutralize 85.0 mL of 0.300 M H2SO4 ? The balanced neutralization reaction
nekit [7.7K]

Answer : The volume of NaOH required to neutralize is, 340 mL

Explanation :

To calculate the volume of base (NaOH), we use the equation given by neutralization reaction:

n_1M_1V_1=n_2M_2V_2

where,

n_1,M_1\text{ and }V_1 are the n-factor, molarity and volume of acid which is H_2SO_4

n_2,M_2\text{ and }V_2 are the n-factor, molarity and volume of base which is NaOH.

We are given:

n_1=2\\M_1=0.300M\\V_1=85.0mL\\n_2=1\\M_2=0.150M\\V_2=?

Putting values in above equation, we get:

2\times 0.300M\times 85.0mL=1\times 0.150M\times V_2\\\\V_2=340mL

Hence, the volume of NaOH required to neutralize is, 340 mL

4 0
3 years ago
Please help! Asap! :)
timama [110]

The first one is true.

The second one is false.

4 0
3 years ago
Read 2 more answers
Many monatomic ions are found in seawater, including the ions formed from the following list of elements. Write the Lewis symbol
77julia77 [94]

Answer:

The Lewis structures are in image attached.

Explanation:

Lewis symbol is a representation of an element symbol along with its valence electrons around it in the form of dot(s).

Mono-atomic ions formed from the following :

(a) Cl

Chlorine's atomic number is 17 in which only 7 electrons are present in its valence shell .So in order to gain noble gas stability it will gain 1 electron to completes its octet

Cl=1s^22s^22p^63s^23p^5

Cl^-=1s^22s^22p^63s^23p^6

(b) Na

Sodium's atomic number is 11 in which only 1 electrons are present in its valence shell .So in order to gain noble gas stability it will loose 1 electron to completes its octet. In the Lewis symbol no dot shown as sodium has lost its 1 electron.

Na=1s^22s^23p^63s^1

Na^+=1s^22s^23p^63s^0

(c) Mg

Magnesium's atomic number is 12 in which only 2 electrons are present in its valence shell .So, in order to gain noble gas stability it will loose 2 electrons to completes its octet.

In the Lewis symbol no dot shown as magnesium has lost its 2 electrons.

Mg=1s^22s^23p^63s^2

Mg^{2+}=1s^22s^23p^63s^0

(d)Ca

Calcium's atomic number is 20 in which only 2 electrons are present in its valence shell .So, in order to gain noble gas stability it will loose 2 electron to completes its octet.

In the Lewis symbol no dot shown as calcium has lost its 2 electron.

Ca= 1s^22s^22p^63s^23p^64s^2

Ca^{2+}=1s^22s^23p^6^23p^64s^0

(e) K

Potassium's atomic number is 19 in which only 1 electrons are present in its valence shell .So, in order to gain noble gas stability it will loose 1 electron to completes its octet.

In the Lewis symbol no dot shown as calcium has lost its 1 electron.

K= 1s^22s^22p^63s^23p^64s^1

K^{+}=1s^22s^23p^6^23p^64s^0

(f) Br

Bromine's atomic number is 35 in which only 7 electrons are present in its valence shell .So in order to gain noble gas stability it will gain 1 electron to completes its octet

Br=1s^22s^22p^63s^23p^63d^{10}4s^24p^5

Br^-= 1s^22s^22p^63s^23p^63d^{10}4s^24p^6

(g) Sr

Strontium's atomic number is 38 in which only 2 electrons are present in its valence shell .So, in order to gain noble gas stability it will loose 2 electron to completes its octet.

In the Lewis symbol no dot shown as calcium has lost its 2 electron.

Sr=1s^22s^22p^63s^23p^63d^{10}4s^24p^65s^2

Sr^{2+}=1s^22s^22p^63s^23p^63d^{10}4s^24p^65s^0

(h) F

Florine's atomic number is 7 in which only 7 electrons are present in its valence shell .So in order to gain noble gas stability it will gain 1 electron to completes its octet.

F=1s^22s^22p^5

F^-=1s^22s^22p^6

5 0
3 years ago
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