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larisa [96]
3 years ago
5

Why chemistry sucks so much...!!!:( it is so tuff help me ​

Chemistry
1 answer:
Flura [38]3 years ago
7 0

Answer:

hahaha fighting! you can do it

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What is the volume in liters of 5.25 moles of He gas?​
serious [3.7K]

Answer:

\boxed {\boxed {\sf 117.6 \ L \ He}}

Explanation:

Regardless of the type of gas, 1 mole at standard temperature and pressure (STP) occupies a volume of 22.4 liters. In this case the gas is helium (He).

We can set up a ratio.

\frac { 22.4 \ L \ He}{ 1 \ mol \ He}

Multiply by the given number of moles.

5.25 \ mol \ He *\frac { 22.4 \ L \ He}{ 1 \ mol \ He}

The moles of helium will cancel.

5.25 *\frac { 22.4 \ L \ He}{ 1 }

5.25 * { 22.4 \ L \ He}

Multiply.

117.6 \ L \ He

5.25 moles of helium gas at STP is 117.6 liters of helium.

5 0
3 years ago
According to reference table adv-10, which is the strongest reducing agent?
Mumz [18]
Li(s)    (answer  A)   
    Li  is  strongest  reducing  agent  because  of  the   lowest  standard reduction  potential.  when  something  is  oxidized, it  reduces  another  substance,  becoming a  reducing.Hence  Lithium  is  strongest  reducing  agent. Reducing  agent   is  stronger  when  it  has  a  more  positive  oxidation potential.
3 0
3 years ago
A solution of KCIO3 is prepared using 75 grams of the solute in enough water to make 0.250 liters of solution. The gram-formula
charle [14.2K]
Using the mass/volume percentage method for percentages of the solution, you simply divide the grams of solute by the volume of the solution and multiply by 100 to get your percentage.
(75.0g/250mL)•100 = 30.0% solute
3 0
3 years ago
Butane (C4 H10(g), Hf = –125.6 kJ/mol) reacts with oxygen to produce carbon dioxide (CO2 , Hf = –393.5 kJ/mol ) and water (H2 O,
WITCHER [35]

Answer: Enthalpy of combustion (per mole) of C_4H_{10} (g) is -2657.5 kJ

Explanation:

The chemical equation for the combustion of butane follows:

2C_4H_{10}(g)+4O_2(g)\rightarrow 8CO_2(g)+10H_2O(g)

The equation for the enthalpy change of the above reaction is:

\Delta H^o_{rxn}=[(8\times \Delta H^o_f_{CO_2(g)})+(10\times \Delta H^o_f_{H_2O(g)})]-[(1\times \Delta H^o_f_{C_4H_{10}(g)})+(4\times \Delta H^o_f_{O_2(g)})]

We are given:

\Delta H^o_f_{(C_4H_{10}(g))}=-125.6kJ/mol\\\Delta H^o_f_{(H_2O(g))}=-241.82kJ/mol\\\Delta H^o_f_{(O_2(g))}=0kJ/mol\\\Delta H^o_f_{(CO_2(g))}=-393.5kJ/mol\\\Delta H^o_{rxn}=?

Putting values in above equation, we get:

\Delta H^o_{rxn}=[(8\times -393.5)+(10\times -241.82)]-[(2\times -125.6)+(4\times 0)]\\\\\Delta H^o_{rxn}=-5315kJ

Enthalpy of combustion (per mole) of C_4H_{10} (g) is -2657.5 kJ

6 0
2 years ago
What is the uses of plastic?
Alex Ar [27]

Answer:

Plastic is used across almost every sector, including to produce packaging, in building and construction, in textiles, consumer products, transportation, electrical and electronics and industrial machinery.

7 0
2 years ago
Read 2 more answers
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