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vodka [1.7K]
3 years ago
11

What is the chemical formula for mercury(I) nitrate? Hgmc021-1.jpg(NOmc021-2.jpg) Hg(NOmc021-3.jpg)mc021-4.jpg Hgmc021-5.jpg(NOm

c021-6.jpg)mc021-7.jpg Hgmc021-8.jpg(NOmc021-9.jpg)mc021-10.jpg
Physics
2 answers:
Anit [1.1K]3 years ago
8 0
If you just type "<span>What is the chemical formula for mercury(I) nitrate?" into google you get the answer but HG(NO3)2 is the correct one.
sorry no one helped you in time hope you passed anyway</span>
Anton [14]3 years ago
3 0

Answer: The symbol of Mercury(I) nitrate: HgNO_3

Explanation:

  • Symbol of the mercury metal is Hg.
  • The symbol of nitrate ion is NO_3^-.

The chemical formula of Mercury(I) nitrate: HgNO_3

It is formed by the combination of mercury ion and nitrate ion. Nitrate ion is a polyatomic ion.

The reaction for the combination of the ions to form mercury (I) nitrate follows the equation:

Hg^++NO_{3}^-\rightarrow HgNO_3

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tia_tia [17]

The distance travelled by the ball that is thrown horizontally from a window that is 15.4 meters high at a speed of 3.01 m/s is 5.34 m

s = ut + 1 / 2 at²

s = Distance

u = Initial velocity

t = Time

a = Acceleration

Vertically,

s = 15.4 m

u = 0

a = 9.8 m / s²

15.4 = 0 + ( 1 / 2 * 9.8 * t² )

t² = 3.14

t = 1.77 s

Horizontally,

u = 3.01 m / s

a = 0 ( Since there is no external force )

s = ( 3.01 * 1.77 ) + 0

s = 5.34 m

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How is Coulomb’s law similar to newton’s law of gravitational force? How is it different
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The similarities and the differences between gravitational and electric force are listed below

Explanation:

- The magnitude of the gravitational force between two objects is given by Newton's law of gravitation:

F=G\frac{m_1 m_2}{r^2}

where

G=6.67\cdot 10^{-11} m^3 kg^{-1}s^{-2} is the gravitational constant

m_1, m_2 are the masses of the two objects

r is the separation between them

- Coloumb's law gives instead the strength of the electrostatic force between two charged objects, which is

F=k\frac{q_1 q_2}{r^2}

where:

k=8.99\cdot 10^9 Nm^{-2}C^{-2} is the Coulomb's constant

q_1, q_2 are the two charges

r is the separation between the two charges

By comparing the two equations, we find the following similarities:

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Instead, we have the following differences:

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5 0
3 years ago
A piston-cylinder device initially contains 1.4 kg saturated liquid water at 200oC. Now heat is transferred to the water until t
postnew [5]

Answer:

Explanation:

Given

mass of saturated liquid water m=1.4\ kg

at 200^{\circ} specific volume is \nu =0.001157\ m^3\kg(From Table A-4,Saturated water Temperature table)

V_1=m\nu _1

V_1=1.4\times 0.001157

V_1=1.6198\times 10^{-3}\ m^3

Final Volume V_2=4V_1

V_2=4\times (1.6198\times 10^{-3})

V_2=6.4792\times 10^{-3}\ m^3

Specific volume at this stage

\nu _2=\frac{V_2}{m}

\nu _2=\frac{6.4792\times 10^{-3}}{1.4}

\nu _2=0.004628\ m^3/kg

Now we see the value and find the temperature it corresponds to specific volume at vapor stage in the table.

T_2=T_1^{*}+\frac{T_2^{*}-T_1^{*}}{\alpha _2^{*}-\alpha _1^{*}}\times (\alpha _2-\alpha _1^{*})

T_2=370^{\circ}+\frac{373.95-370}{0.003106-0.004953}\times (0.004628-0.004953)

T_2=370.7^{\circ} C

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