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Zielflug [23.3K]
3 years ago
11

Which statement describes the density of the inner planets?

Chemistry
2 answers:
katrin [286]3 years ago
8 0

Answer:

c

Explanation:

timurjin [86]3 years ago
5 0

Answer:

Answer: The correct statement is : all the inner planets are dense.

Explanation:

i took the text and got 100% ; )

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What’s the difference between internal stimuli and external stimuli
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An external stimuli is a stimulus that comes from outside an organism and causes a reaction.An internal stimuli is a stimulus that comes from inside an organism.

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True or false: Atoms and mass are conserved in nature<br> True<br> False
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It’s true

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If we account for all reactants and products in a chemical reaction, the total mass will be the same at any point in time in any closed system. ... The Law of Conservation of Mass holds true because naturally occurring elements are very stable at the conditions found on the surface of the Earth.

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Consider this reaction: NH + + HPO 4 + NH3 + H2PO4
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What types of atoms will experience radioactivity?
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3 years ago
The enthalpy of solution (∆H) of KOH is -57.6 kJ/mol. If 3.66 g KOH is dissolved in enough water to make a 150.0 mL solution, wh
Wewaii [24]

When 3.66 g of KOH (∆Hsol = -57.6 kJ/mol) is dissolved in 150.0 mL of solution, it causes a temperature change of 5.87 °C.

The enthalpy of solution of KOH is -57.6 kJ/mol. We can calculate the heat released by the solution (Qr) of 3.66 g of KOH considering that the molar mass of KOH is 56.11 g/mol.

3.66g \times \frac{1mol}{56.11g} \times \frac{(-57.6kJ)}{mol} = -3.76 kJ

According to the law of conservation of energy, the sum of the heat released by the solution of KOH (Qr) and the heat absorbed by the solution (Qa) is zero.

Qr+Qa = 0\\\\Qa = -Qr = 3.76 kJ

150.0 mL of solution with a density of 1.02 g/mL were prepared. The mass (m) of the solution is:

150.0 mL \times \frac{1.02g}{mL}  = 153 g

Given the specific heat capacity of the solution (c) is 4.184 J/g・°C, we can calculate the change in the temperature (ΔT) of the solution using the following expression.

Qa = c \times m \times \Delta T\\\\\Delta T = \frac{Qa}{c \times m} = \frac{3.76 \times 10^{3}J  }{\frac{4.184J}{g.\° C }  \times 153g} = 5.87 \° C

When 3.66 g of KOH (∆Hsol = -57.6 kJ/mol) is dissolved in 150.0 mL of solution, it causes a temperature change of 5.87 °C.

Learn more: brainly.com/question/4400908

7 0
2 years ago
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