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frozen [14]
3 years ago
15

Metallic materials are becoming more and more widely used. They have

Chemistry
2 answers:
weqwewe [10]3 years ago
7 0

Answer:

Its B

Explanation:

hope its help by the way iam dream

Tema [17]3 years ago
6 0
It’s B hehehehhdhdhfgfgfhrhthhrhdhr
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The dumping of wastewater from this pipe is an example of what type of pollution?
Veseljchak [2.6K]

Answer:

dumping of wastewater is an example of water polution.

Explanation:

water polution is when an alienated foreign contaminent is put into a water source either intentionally or accidentally

3 0
3 years ago
Consider an experimental setup with two compartments separated by a phospholipid bilayer membrane containing ion channels select
Schach [20]

This question is incomplete, the complete question is;

Consider an experimental setup with two compartments separated by a phospholipid bilayer membrane containing ion channels selectively permeable only to chloride ions. The left compartment (also called inside) contains 1 mM Cl- and the right compartment (also called outside) contains 100 mM Cl-. What will the electrical potential be when the system attains equilibrium? [ assume body temperature; log 100 = 2, log 10 = 1, log 1 = 0, log 0.1 = -1, log 0.01 = -2]

Options;

a) -62 mV

b) -124 mV

c) +62 mV

d) 0 mV

e) +124 mV

Answer:

the electrical potential be when the system attains equilibrium is  –124mV

Option b) –124mV is the correct answer

Explanation:

Given the data in the question;

Two compartments are divided by lipid bilayer;

In inside compartment Cl- ion concentration- 1mM and out side of the cell concentration is 100mM

now we apply the Nernst equilibrium potential equation;

Chlorine ion valency is z = –1

So

E_{eq} = 62/z × log(ion outside/ ion inside) [for Cl‐ ions]

E_{eq} = (62 / –1) x log(  100 / 1 )

 E_{eq} = -62 x 2 =

E_{eq}  = –124mV

Therefore, the electrical potential be when the system attains equilibrium is  –124mV

Option b) –124mV is the correct answer

4 0
3 years ago
What is the new volume of a 61 L sample at STP that is moved to 183 K and 0.60 atm?
Volgvan

Answer: The new volume of a 61 L sample at STP that is moved to 183 K and 0.60 atm is 54.63 L.

Explanation:

Given: V_{1} = 61 L,      T_{1} = 183 K,      P_{1} = 0.60 atm

At STP, the value of pressure is 1 atm and temperature is 273.15 K.

Now, formula used to calculate the new volume is as follows.

\frac{P_{1}V_{1}}{T_{1}} = \frac{P_{2}V_{2}}{T_{2}}

Substitute the values into above formula as follows.

\frac{P_{1}V_{1}}{T_{1}} = \frac{P_{2}V_{2}}{T_{2}}\\\frac{0.60 atm \times 61 L}{183 K} = \frac{1 atm \times V_{2}}{273.15 K}\\V_{2} = 54.63 L

Thus, we can conclude that the new volume of a 61 L sample at STP that is moved to 183 K and 0.60 atm is 54.63 L.

4 0
3 years ago
Copper is used for electrical wirin g and cooking pans however iron is much more common on earth than copper why might copper be
pishuonlain [190]
Because copper has a higher electrical and thermal conductivity than iron and is not susceptible to oxidation as easily.
8 0
3 years ago
Given that ΔH = −571.6 kJ/mol for the reaction 2 H2(g) + O2(g) → 2 H2O(l), calculate ΔH for these reactions. (a) 2 H2O(l) → 2 H2
pashok25 [27]

Answer : The value of \Delta H for the reaction is +571.6 kJ/mole.

Explanation :

According to Hess’s law of constant heat summation, the heat absorbed or evolved in a given chemical equation is the same whether the process occurs in one step or several steps.

According to this law, the chemical equation can be treated as ordinary algebraic expression and can be added or subtracted to yield the required equation. That means the enthalpy change of the overall reaction is the sum of the enthalpy changes of the intermediate reactions.

The given chemical reaction is,

2H_2(g)+O_2(g)\rightarrow 2H_2O(l)     \Delta H_1=-571.6kJ/mole

Now we have to determine the value of \Delta H for the following reaction i.e,

2H_2O(l)\rightarrow 2H_2(g)+O_2(g)    \Delta H_2=?

According to the Hess’s law, if we reverse the reaction then the sign of \Delta H change.

So, the value \Delta H_2 for the reaction will be:

\Delta H_2=-(-571.6kJ/mole)

\Delta H_2=+571.6kJ/mole

Hence, the value of \Delta H for the reaction is +571.6 kJ/mole.

6 0
3 years ago
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