How many grams of C3H8 are in a 7 L tank at 293 K and 5.45 atm?

Balance the following equation in acidic conditions Phases are optional. S2O3 2- + Cu 2+ ---> S4O6 2- + Cu+

jonny [76]

To balance the the chemical reaction, the number of moles per element is balance is both side of the reaction and also the charge in both sides of the reation. to balnce the reaction:

S2O3 2- + Cu 2+ ---> S4O6 2- + Cu+

2S2O3 2- + Cu 2+ ---> S4O6 2- + Cu+ + e

3 0

3 years ago

Does a precipitate form when a solution of calcium chloride and a solution of mercury(I) nitrate are mixed together? Write the n

olganol [36]

Answer : Yes, a precipitate form when a solution of calcium chloride and a solution of mercury(I) nitrate are mixed together.

The net ionic equation will be,

$2Hg^{+}(aq)+2Cl^{-}(aq)\rightarrow Hg_2Cl_2(s)$

Explanation :

In the net ionic equations, we are not include the spectator ions in the equations.

Spectator ions : The ions present on reactant and product side which do not participate in a reactions. The same ions present on both the sides.

The given balanced ionic equation will be,

$CaCl_2(aq)+2HgNO_3(aq)\rightarrow Ca(NO_3)_2(aq)+Hg_2Cl_2(s)$

The ionic equation in separated aqueous solution will be,

$Ca^{2+}(aq)+2Cl^{-}(aq)+2Hg^{+}(aq)+2NO_3^{-}(aq)\rightarrow Hg_2Cl_2(s)+Ca^{2+}(aq)+2NO_3^{-}(aq)$

In this equation, $Ca^{2+}\text{ and }NO_3^-$ are the spectator ions.

By removing the spectator ions from the balanced ionic equation, we get the net ionic equation.

The net ionic equation will be,

$2Hg^{+}(aq)+2Cl^{-}(aq)\rightarrow Hg_2Cl_2(s)$

7 0

3 years ago

This element is a liquid at room temperature. (A) Hg (B) Th (C) Na (D) Cl (E) Co

gulaghasi [49]

A) Hg, or Mercury, is a liquid at room temperature. Hope this helps!!

7 0

3 years ago

Read 2 more answers

What is the atomic number of hydrogen??

Nataly_w [17]

The number is is surely 1

5 0

3 years ago

A 1.8 g sample of octane C8H18 was burned in a bomb calorimeter and the temperature of 100 g of water increased from 21.36 C to

melomori [17]

Answer:

HEAT OF COMBUSTION PER GRAM OF OCTANE IS 1723.08 J OR 1.72 KJ/G OF HEAT

HEAT OFF COMBUSTION PER MOLE OF OCTANE IS 196.4 KJ/ MOL OF HEAT

Explanation:

Mass of water = 100 g

Change in temperature = 28.78 °C - 21.36°C = 7.42 °C

Heat capcacity of water = 4.18 J/g°C

Mass of octane = 1.8 g

Molar mass of octane = C8H18 = (12 * 8 + 1 * 18) g/mol= 96 + 18 = 114 g/mol

First is to calculate the heat evolved when 100 g of water is used:

Heat = mass * specific heat capacity * change in temperature

Heat = 100 * 4.18 * 7.42

Heat = 3101.56 J

In other words, 3101.56 J of heat was evolved from the reaction of 1.8 g octane with water.

Heat of combustion of octane per gram:

1.8 g of octane produces 3101.56 J of heat

1 g of octane will produce ( 3101.56 * 1 / 1.8)

= 1723.08 J of heat

So, heat of combustion of octane per gram is 1723.08 J

Heat of combustion per mole:

1.8 g of octane produces 3101.56 J of heat

1 mole of octane will produce X J of heat

1 mole of octane = 114 g/ mol of octane

So we have:

1.8 g of octane = 3101.56 J

114 g of octane = (3101.56 * 114 / 1.8) J of heat

= 196 432.13 J

= 196. 4 kJ of heat

The heat of combustion of octane per mole is 196.4 kJ /mol.

Mass of water = 100 g

Change in temperature = 28.78 °C - 21.36°C = 7.42 °C

Heat capcacity of water = 4.18 J/g°C

Mass of octane = 1.8 g

Molar mass of octane = C8H18 = (12 * 8 + 1 * 18) g/mol= 96 + 18 = 114 g/mol

First is to calculate the heat evolved when 100 g of water is used:

Heat = mass * specific heat capacity * change in temperature

Heat = 100 * 4.18 * 7.42

Heat = 3101.56 J

8 0

2 years ago