Answer: A typical hydrogen fuel cell produces 0.5 V to 0.8 V per cell. To increase the voltage individual cells can be connected in series.
Answer:
-1 Coulomb meter = -2.997 × 10²⁹ Debye
Explanation:
Given:
Coulomb meter = -1 CM
Find:
In debye
Computation:
We know that,
1 Coulomb meter = 299,792,458,178,090,000,000,000,000,000 Debye
So,
-1 Coulomb meter = -299,792,458,178,090,000,000,000,000,000 Debye
-1 Coulomb meter = -2.997 × 10²⁹ Debye
Answer:
x(t) = −39e
−0.03t + 40.
Explanation:
Let V (t) be the volume of solution (water and
nitric acid) measured in liters after t minutes. Let x(t) be the volume of nitric acid
measured in liters after t minutes, and let c(t) be the concentration (by volume) of
nitric acid in solution after t minutes.
The volume of solution V (t) doesn’t change over time since the inflow and outflow
of solution is equal. Thus V = 200 L. The concentration of nitric acid c(t) is
c(t) = x(t)
V (t)
=
x(t)
200
.
We model this problem as
dx
dt = I(t) − O(t),
where I(t) is the input rate of nitric acid and O(t) is the output rate of nitric acid,
both measured in liters of nitric acid per minute. The input rate is
I(t) = 6 Lsol.
1 min
·
20 Lnit.
100 Lsol.
=
120 Lnit.
100 min
= 1.2 Lnit./min.
The output rate is
O(t) = (6 Lsol./min)c(t) = 6 Lsol.
1 min
·
x(t) Lnit.
200 Lsol.
=
3x(t) Lnit.
100 min
= 0.03 x(t) Lnit./min.
The equation is then
dx
dt = 1.2 − 0.03x,
or
dx
dt + 0.03x = 1.2, (1)
which is a linear equation. The initial condition condition is found in the following
way:
c(0) = 0.5% = 5 Lnit.
1000 Lsol.
=
x(0) Lnit.
200 Lsol.
.
Thus x(0) = 1.
In Eq. (1) we let P(t) = 0.03 and Q(t) = 1.2. The integrating factor for Eq. (1) is
µ(t) = exp Z
P(t) dt
= exp
0.03 Z
dt
= e
0.03t
.
The solution is
x(t) = 1
µ(t)
Z
µ(t)Q(t) dt + C
= Ce−0.03t + 1.2e
−0.03t
Z
e
0.03t
dt
= Ce−0.03t +
1.2
0.03
e
−0.03t
e
0.03t
= Ce−0.03t +
1.2
0.03
= Ce−0.03t + 40.
The constant is found using x(t) = 1:
x(0) = Ce−0.03(0) + 40 = C + 40 = 1.
Thus C = −39, and the solution is
x(t) = −39e
−0.03t + 40.
0.0340 g O2
Step 1. Write the balanced chemical equation
4Fe(OH)^(+) + 4OH^(-) + O2 + 2H2O → 4Fe(OH)3
Step 2. Calculate the moles of Fe^(2+)
Moles of Fe^(2+) = 50.0 mL Fe^(2+) × [0.0850 mmol Fe^(2+)/1 mL Fe^(2+)]
= 4.250 mmol Fe^(2+)
Step 3. Calculate the moles of O2
Moles of O2 = 4.250 mmol Fe^(2+) × [1 mmol O2/4 mmol Fe^(2+)]
= 1.062 mmol O2
Step 4. Calculate the mass of O2
Mass of O2 = 1.062 mmol O2 × (32.00 mg O2/1 mmol O2) = 34.0 mg O2
= 0.0340 g O2
0.0342 grams of O2 are consumed to precipitate all of the iron in 50.0 mL of 0.0850 M Fe(II) solution.
To solve this problem, we need to first calculate the number of moles of Fe(II) in 50.0 mL of 0.0850 M Fe(II) solution.
Moles of Fe(II) = (0.0850 mol/L) * (50.0 mL) = 0.00425 mol
According to the balanced chemical equation, 4 moles of Fe(II) react with 1 mole of O2. Therefore, the number of moles of O2 required to precipitate all of the iron in 50.0 mL of 0.0850 M Fe(II) solution is:
Moles of O2 = (0.00425 mol Fe(II)) * (1 mol O2 / 4 mol Fe(II)) = 0.00106 mol O2
Now we can convert the moles of O2 to grams using the molar mass of O2 (32.00 g/mol):
Grams of O2 = (0.00106 mol O2) * (32.00 g/mol) = 0.0342 g O2
Therefore, 0.0342 grams of O2 are consumed to precipitate all of the iron in 50.0 mL of 0.0850 M Fe(II) solution.
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Answer:
This question is incomplete
Explanation:
This question is incomplete but some general explanation provides a clear answer to what is been asked in the question.
An ionic/electrovalent compound is a compound whose constituent atoms are joined together by ionic bond. Ionic bond is a bond involving the transfer of valence electron(s) from an atom (to form a positively charged cation) to another atom (to form a negatively charged anion). The atom transferring is usually a metal while the atom receiving is usually a non-metal.
For example (as shown in the attachment), in the formation of NaCl salt, the sodium (Na) transfers the single electron (valence) on it's outermost shell to chlorine (Cl) which ordinarily has 7 electrons on it's outermost shell but becomes 8 after receiving the valence electron from sodium. It should also be noted that Na is a metal while Cl is a non-metal.
There are two types of chemical compound one is covalent compound and other is ionic compound, covalent compound formed by sharing of electron and ionic compound formed by complete transfer of electron. Therefore, the name and formula of the compound made of magnesium and fluorine is Magnesium fluoride and MgF₂ respectively.
Chemical Compound is a combination of molecule, Molecule forms by combination of element and element forms by combination of atoms in fixed proportion.
An ionic compound is a metal and nonmetal combined compound. Ionic compound are very hard. They have high melting and boiling point because of strong ion bond.
The name and formula of the compound made of magnesium and fluorine is Magnesium fluoride and MgF₂ respectively. Magnesium fluoride is an ionic compound.
Therefore, the name and formula of the compound made of magnesium and fluorine is Magnesium fluoride and MgF₂ respectively.
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Answer:Magnesium fluorine
MgF2
Explanation:
God knows
When you fill a basin with liquid water, you can see that the water takes the shape of the container in which it is contained. This is because in the liquid state, water has molecules farther apart than in the solid state.
You can notice this property when performing an experiment with liquid and solid water.
When filling a glass, liquid water takes on the shape of a glass, and solid water, such as an ice cube, remains the same shape when placed in a glass.
Therefore, when filling a basin with water we perceive a property of the physicalstate of water, in liquid form. Water is one of the few substances that can be found naturally in liquid, solid and gaseous states.
Learn more here:
Answer:
Cautiously and avoiding filling in the central area so that it does not overflow when filling, since being very beach makes filling difficult.
Explanation:
The basins are shallow, that is why filling is difficult, the filling must be slow, low intensity and at the edges not placing the water filling in the center of the basin.