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larisa86 [58]
3 years ago
13

Calcium hydroxide reacts with Hydrogen nitrate to form Calcium nitrate and Dihydrogen monoxide.

Chemistry
1 answer:
ra1l [238]3 years ago
5 0

Answer:

Ca(OH)2 + HNO3 = Ca(NO3)2 + H2O | Chemical reaction and equation.

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Use the standard reduction potentials located in the 'Tables' linked above to calculate the equilibrium constant for the reactio
Paladinen [302]

Answer:

Check the explanation

Explanation:

cell CuE Ecell 0.337 (-0.14) Ecl0.477 V

Since E^o_{ cell } > 0 , the value of \Delta G^o will be negative.

\Delta G^o < 0

\Delta G^o =-nFE^o_{ cell }......(1)

But

\Delta G^o =-RT ln K......(2)

From (1) and (2)

\Delta G^o =-RT ln K=-nFE^o_{ cell }

ln K =\frac{nFE^o_{ cell } }{RT }

ln K =\frac{ 2 \times 96500 \times 0.477 }{8.314 \times \left ( 25+273.15 \right ) }

ln K =37.139

K =1.3468 \times 10^{16}

Hence, the value of the equilibrium constant is 1.35 \times 10^{16}

6 0
3 years ago
1.
xenn [34]
The answer is OH.

Hope this helps!
4 0
3 years ago
Whats the very small dense center of atom
Gemiola [76]
The Nucleus is the very small dense center of an atom.
7 0
4 years ago
Which of the following objects would require the use of a large-scale model A. An atom B. A car engine C. A frog's heart D. The
levacccp [35]

Answer:

A. An atom

Explanation:

A large scale model is one that is bigger than the object itself. These are needed to visualise structures and behaviours that we cannot normally see. A large scale model is needed for small objects, such as an atom. Larger objects do not need<em> even bigger</em> models

4 0
3 years ago
What volume does 1.70 ×10–3 mol of chlorine gas occupy if its temperature is 20.2 °C and its pressure is 795 mm Hg?
Cerrena [4.2K]

Explanation:

The given data is as follows.

     No. of moles = 1.70 \times 10^{-3},           V = ?

     T = 20.2 + 273 K = 293.2 K,             P = \frac{795 mm Hg}{760.0 mm Hg/atm} = 1.046 atm,                      R = 0.0821 L atm K^{-1}mol ^{-1}

Calculate the volume using ideal gas equation as follows.

                                    P V = n R T

    1.046 atm \times V = 1.70 \times 10^{-3} \times 0.0821 L atm K^{-1}mol ^{-1} \times 293.2 K

                                  V = \frac{1.70 \times 10^{-3} \times 0.0821 L atm K^{-1}mol ^{-1} \times 293.2 K}{1.046 atm}

                                   = \frac{40.921 L atm}{1.046 atm}

                                   = 39.122 \times 10^{-3} L

Thus, we can conclude that volume of the gas is 39.122 \times 10^{-3} L.


3 0
3 years ago
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