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Virty [35]
3 years ago
8

Which one of the following statements is correct?

Chemistry
1 answer:
adell [148]3 years ago
7 0

Answer:

Tissues form organs, and organs form systems. Hope this helped!

Explanation:

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Write the balanced molecular and net ionic equations for the reaction between aluminum metal and silver nitrate. identify the ox
asambeis [7]

Well in this case, silver nitrate is reduced:

Ag<span>+  </span><span>+  </span>e<span>−  </span>→ Ag(s) ↓

 

Meanwhile, the aluminum is oxidized forming a positive ion:

Al(s<span>)  →  </span>Al<span>3+  </span><span>+  3</span>e−

 

To get the overall reaction,  we add the half equations so that the electrons are eliminated:

Al(s<span>)  +   3</span>Ag<span>+  </span><span>→  </span>Al<span>3+  </span><span>+  3</span>Ag(s)

 

And similarly:

Al(s<span>)  +  3</span>AgNO3(aq<span>)  →  </span>Al(NO3)3(aq<span>)  +  3</span>Ag(s<span>)</span>

4 0
2 years ago
Read 2 more answers
What would be the result on the Insect population if the Bats and Mice were removed?
Ivanshal [37]

Answer:

We wouldn't have Coronavirus

Explanation:

5 0
2 years ago
Suppose 50.0g of silver nitrate is reacted with 50g of hydrochloric acid producing silver chloride and a mixture of other produc
docker41 [41]

Answer:

53.6 grams of silver chloride was produced.

Explanation:

AgNO_3+HCl+\rightarrow AgCl+HNO_3

Law of conservation of mass states that mass can neither be created nor be destroyed but it can only be transformed from one form to another form.

This also means that total mass on the reactant side must be equal to the total mass on the product side.

Mass of silver nitrate = 50.0 g

Mass of hydrogen chloride = 50.0 g

Mass of silver chloride = x

Mass of  nitric acid = 46.4 g

Mass of silver nitrate + Mass of hydrogen chloride =

                             Mass of silver chloride + Mass of  nitric acid

[te]50.0 g+50.0 g=x+46.4 g[/tex]

x=50.0 g+50.0 g - 46.4 g = 53.6 g

53.6 grams of silver chloride was produced.

8 0
3 years ago
The average propane cylinder for a residential grill holds approximately 18 kg of propane. How much energy (in kJ) is released b
brilliants [131]
Combustion is a chemical reaction between a fuel and an oxidant, oxygen, to give off combustion products and heat. Complete combustion results when all of the fuel is consumed to form carbon dioxide and water, as in the case of a hydrocarbon fuel. Incomplete combustion results when insufficient oxygen reacts with the fuel, forming soot and carbon monoxide. 

The complete combustion of propane proceeds through the following reaction:
C_{3} H_{8} + 5O_{2} --> 3CO_{2} + 4H_{2}O

Combustion is an exothermic reaction, which means that it gives off heat as the reaction proceeds. For the complete combustion of propane, the heat of combustion is (-)2220 kJ/mole, where the minus sign indicates that the reaction is exothermic. 

The molar mass of propane is 44.1 grams/mole. Using this value, the number of moles propane to be burned can be determined from the mass of propane given. Afterwards, this number of moles is multiplied by the heat of combustion to give the total heat produced from the reaction of the given mass of propane.

 14.50 kg propane  x <u> 1000 g </u> x <u>  1 mole propane   </u>  x <u>  2220 kJ  </u>   
                                     1 kg              44.1 g                     1 mole

= 729,931.97 kJ
8 0
2 years ago
Calculate the amount of oxygen gas collected by the displacement of water at 14◦C if the atmospheric pressure is 790 Torr and th
zhannawk [14.2K]

Answer : The amount of oxygen gas collected are, 0.217 mol

Explanation :

Using ideal gas equation :

PV=nRT

where,

P = pressure of gas = (790-12)torr=778torr=1.02atm     (1 atm = 760 torr)

V = volume of gas = 5 L

T = temperature of gas = 14^oC=273+14=287K

n = number of moles of gas = ?

R = gas constant  = 0.0821 L.atm/mol.K

Now put all the given values in the ideal gas equation, we get:

(1.02atm)\times (5L)=n\times (0.0821L.atm/mol.K)\times (287K)

n=0.217mole

Thus, the amount of oxygen gas collected are, 0.217 mol

8 0
3 years ago
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