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ArbitrLikvidat [17]
2 years ago
14

What does not represent a compound

Chemistry
1 answer:
Ainat [17]2 years ago
5 0

Answer:

<h2>Every combination of atoms is a molecule. A compound is a molecule made of atoms from different elements. All compounds are molecules, but not all molecules are compounds. Hydrogen gas (H2) is a molecule, but not a compound because it is made of only one element.</h2>

Explanation:

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The chemical properties of a substance can only be observed when it is changed into a different substance. True or False
HACTEHA [7]
I am pretty sure that the answer is true
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3 years ago
Read 2 more answers
ILL GIVE BRAILNEST!!!!! PLZ HELP ME FAST
NemiM [27]
Multiple states of matter - mixture
Same state of matter - pure
Separated physically- mixture
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Can’t be a single element - mixture
3 0
3 years ago
Select the correct answer.
Norma-Jean [14]

Answer:

a t o m i c w e i h g t

B

Explanation:

  1. a to m i c w e i g h t
7 0
3 years ago
10 pts) A student titrates a 20.00 mL sample of an aqueous borax solution with 1.03 M H2SO4. If 2.07 mL of acid are needed to re
Natasha2012 [34]

Answer: The molarity of the borax solution is 0.107 M

Explanation:

The neutralization reaction is:

Na_2B_4O_7.10H_2O+H_2SO_4(aq)\rightarrow Na_2SO_4+4H_3BO_3+5H_2O

According to neutralization law:

n_1M_1V_1=n_2M_2V_2

where,

n_1 = basicity of H_2SO_4 = 2

n_2 = acidity of borax = 2

M_1 = concentration of H_2SO_4 = 1.03 M

M_2 = concentration of borax =?

V_1 = volume of H_2SO_4  = 2.07ml

V_2 = volume of borax = 20.0 ml

Now put all the given values in the above law, we get the molarity of borax:

(2\times 1.03\times 2.07)=(2\times M_2\times 20.0)

By solving the terms, we get :

M_2=0.107M

Thus the molarity of the borax solution is 0.107 M

7 0
3 years ago
How do I do this question?The aluminum cup inside your calorimeter weighs 39.78 g. You add 50.01 g of ice cold water to the calo
konstantin123 [22]

Answer:

Cp_{metal}=0.922\frac{J}{g\°C}

Explanation:

Hello.

In this problem we must realize that there is heat flow that moves from the hot metal object and the hot water to the cold water and the cold aluminum cup, which allows us to write:

Q_{cup}+Q_{cold,w}=-(Q_{metal}+Q_{hot,w})

Which means that the heat lost be the hot metal object and the hot water is gained by both the cold water and the cold aluminum cup, which can be written in terms of mass, specific heats and change in temperature towards the equilibrium temperature (35.9 °C):

m_{cup}Cp_{cup}(T_{eq}-T_{cup})+m_{cold,w}Cp_{cold,w}(T_{eq}-T_{cold,w})=-(m_{metal}Cp_{metal}(T_{eq}-T_{metal})+m_{hot,w}Cp_{hot,w}(T_{eq}-T_{hot,w})

We need to solve for the specific heat of the metal as shown below:

Cp_{metal}=\frac{m_{cup}Cp_{cup}(T_{eq}-T_{cup})+m_{cold,w}Cp_{cold,w}(T_{eq}-T_{cold,w})+m_{hot,w}Cp_{hot,w}(T_{eq}-T_{hot,w})}{-m_{metal}(T_{eq}-T_{metal})} \\\\Cp_{metal}=\frac{39.78g*0.903\frac{J}{g\°C}(35.9-0.5)\°C+50.01g*4.184\frac{J}{g\°C}(35.9-0.5)\°C +50.72g*4.184\frac{J}{g\°C}(35.9-69.5)\°C  }{-49.98g(35.9-69.5)\°C } \\\\Cp_{metal}=\frac{1271.6J+7407.2J-7130.3J}{-1679.3g\°C} \\\\Cp_{metal}=0.922\frac{J}{g\°C}Best regards.

3 0
3 years ago
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