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Luden [163]
3 years ago
15

If the statement is true, select True. If it is false, select False.

Chemistry
1 answer:
defon3 years ago
5 0

Answer:

False, since boiling is a physical change of state. Although it is at 100°C, it is not a chemical property.

Explanation:

Hope this helped!

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weil es in der Luft schwebt

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What is the particle that is labeled with a question mark (?) in the diagram?
Nat2105 [25]

Answer is: quark.

Quark is a type of elementary particle and a fundamental constituent of matter.

Quarks form composite hadrons (protons and neutrons). Protons and neutrons are in the nucleus of an atom.

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5 0
3 years ago
Read 2 more answers
Need help !
yulyashka [42]
B is the answer loll
6 0
2 years ago
PLEASE HELP ASAP
katrin2010 [14]

Answer: Each ion, or atom, has a particular mass; similarly, each mole of a given pure substance also has a definite mass. The mass of one mole of atoms of a pure element in grams is equivalent to the atomic mass of that element in atomic mass units (amu) or in grams per mole (g/mol).

Explanation:

8 0
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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