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Which statement defines calorimeter?a device that measures the heat that is gained or lost in a chemical change
a device that measures the potential energy of an element
O a device that is used to measure heat capacity
O a device that is used to measure temperature

Answers

Answer 1
Answer:

a device that measures the heat that is gained or lost in a chemical change.

information regarding calorimeter:

  • calorimeter refers to the device for determining the heat creating at the time of  mechanical, electrical, or chemical reaction.
  • Also, it measured the heat capacity of materials.
  • Moreover, the heat should be gained or lost

Find out more information about the  Temperature here:brainly.com/question/7510619?referrer=searchResults
Answer 2
Answer:

Answer:

it's A

Explanation:

I didn't it in a test and got an A I put b but got it wrong so it's A

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What is the potentual energy of 7 kg object, 12 m off the ground?​

Answers

Answer:

823.2 J

Explanation:

PE = mgh

= (7 kg) (9.8 m/s^2) (12 m)

= 823.2 J

Answer:

RequiredAnswer:-

Mass=m=7kg

Height=h=12m

Gravitational force=g=10m/s^2

  • As we know that

{\boxed{\sf Potential\:energy{}_((P.E))=mgh}}

  • Substitute the values

{:}\longmapsto\sf P.E=7×12×10

{:}\longmapsto\sf P.E=84×10

{:}\longmapsto\sf Potential\:energy=840Joule

Select the correct answer.In which research method does a researcher change the value of one variable to determine how the change affects another variable?
O A experiment
OB. observational research
Oc.
survey
OD
case study

Answers

Answer:

experiment is the answer

The enthalpy of combustion of acetylene C2H2 is described by C2H2(g) + (5/2) O2(g) →2CO2(g) + H2O(l), Hrxn= –1299 kJ/mol. Calculate the enthalpy of formation of acetylene, given the following enthalpies of formation Hf[CO2(g)] = –393.5 kJ/mol Hf[H2O(l)] = –285.8 kJ/mo

Answers

Answer:

226.2 kJ/mol

Explanation:

Let's consider the following thermochemical equation for the combustion of acetylene.

C₂H₂(g) + (5/2) O₂(g) → 2 CO₂(g) + H₂O(l), ΔH°rxn = –1299 kJ/mol.

We can also calculate the enthalpy of the reaction per mole of acetylene from the enthalpies of formation.

ΔH°rxn = 2 mol × ΔH°f(CO₂(g)) + 1 mol × ΔH°f(H₂O(l)) - 1 mol × ΔH°f(C₂H₂(g)) - 1 mol × ΔH°f(O₂(g))

1 mol × ΔH°f(C₂H₂(g)) = 2 mol × ΔH°f(CO₂(g)) + 1 mol × ΔH°f(H₂O(l)) - ΔH°rxn - 1 mol × ΔH°f(O₂(g))

1 mol × ΔH°f(C₂H₂(g)) = 2 mol × (-393.5 kJ/mol) + 1 mol × (-285.8 kJ/mol) - (-1299 kJ) - 1 mol × (0 kJ/mol)

ΔH°f(C₂H₂(g)) = 226.2 kJ/mol

Answer:

The enthalpy of formation of acetylene is 226.2 kJ/mol

Explanation:

Step 1: Data given

C2H2 (g) + (5/2)O2 (g)  → 2CO2 (g) + H2O (l)  Heat of Reaction (Rxn) = -1299kJ/mol

Standard formation [CO2 (g)]= -393.5 kJ/mol

Standard formation [H2O (l)] = -285.8 kj/mol

Step 2: The balanced equation

The formation of acetylene is:

2C(s) + H2(g)   → C2H2(g)

Step 3: Calculate the enthalpy of formation of acetylene

It doesn't matter if the process will happen in 1 step or in more steps. What matters is the final result. This is Hess' law of heat summation.

To have the reaction of the formation of acetylene we have to take:

⇒ the reverse equation of the combustion of acetylene

   2CO2 (g) + H2O (l) → C2H2 (g) + (5/2)O2 (g)

⇒  The equation of formation of CO2 (multiplied by 2)

2C(s) + 2O2(g) → 2CO2(g)

⇒ the equation of formation of H2O

H2(g) + 1/2 O2(g) → H2O(l)

2CO2 (g) + H2O (l) + 2C(s) + 2O2(g) + H2(g) + 1/2 O2(g  → C2H2 (g) + (5/2)O2 (g) + 2CO2(g) + H2O(l)

Final reaction = 2C(s) + H2(g)   → C2H2(g)

Calculate the enthalpy of formation of acetylene =

ΔHf = 1299 kJ/mol + (2*-393.5) kJ/mol + (-285.8) kJ/mol

ΔHf = 226.2 kJ/mol

The enthalpy of formation of acetylene is 226.2 kJ/mol

A rock contains 0.37 mg of Pb-206 and 0.95 mg of U-238. Approximately how many U-238 atoms were in the rock when it was formed billions of years ago? (The half-life for 238U  206Pb is 4.5  109 yr.)

Answers

Half-life of a radioactive element

The Half-Life of a radioactive element osbtime taken for half the nucleus of the atom of the element to decay

Calculating the original amount of U-238

  • The number of moles present in each element is first determined:
  • Number of moles = mass/molar mass

For Pb-206:

mass = 0.37 mg = 0.37 * 10⁻³ g

molar mass = 206 g/mol

number of moles = 0.37 * 10⁻³ g/206 g/mol

number of moles of Pb-206 = 1.79 * 10⁻⁶ moles

For U-238:

mass = 0.95 mg = 0.95 * 10⁻³ g

molar mass = 238 g/mol

number of moles = 0.95 * 10⁻³ g/238 g/mol

number of moles = 3.99 * 10⁻⁶ moles

Assuming that all the Pb-206 were formed from U-238

  • Initial moles of U-238 = Moles of present U-238 + molesw of present Pb-208

Initial moles of U-238 = 3.99 * 10⁻⁶ moles + 1.79 * 10⁻⁶ moles

Initial moles of U-238 = 5.78 * 10⁻⁶ moles

One mole of U-238 contains = 6.02 * 10²³ atoms

5.78 * 10⁻⁶ moles of U-238 will contain 6.02 * 10²³ * 5.78 * 10⁻⁶ atoms

Number of atoms of U-238 initially present = 3.48 * 10¹⁸ atoms

Therefore, the number of atoms of U-238 initially present is 3.48 * 10¹⁸ atoms

Learn more about Half-life at: brainly.com/question/4702752

10 POINTS : What are 4 things created from Big Bang?

Answers

Answer:

:0

Explanation:

One second after the Big Bang, the universe was filled with neutrons, protons, electrons, anti-electrons, photons and neutrinos.Jun 17, 2017

Write the balanced reaction and solubility product expression (KSP) for dissolving silver chromate: Ag2CrO4(s). Include all charges, stoichiometric coefficients, and phase subscripts.

Answers

Answer:

2Ag⁺ (aq)  + CrO₄⁻² (aq) ⇄  Ag₂CrO₄ (s) ↓

Ksp = [2s]²  . [s] → 4s³

Explanation:

Ag₂CrO₄ → 2Ag⁺  + CrO₄⁻²

Chromate silver is a ionic salt that can be dissociated. When we have a mixture of both ions, we can produce the salt which is a precipitated.

2Ag⁺ (aq)  + CrO₄⁻² (aq) ⇄  Ag₂CrO₄ (s) ↓ Ksp

That's the expression for the precipitation equilibrium.

To determine the solubility product expression, we work with the Ksp

Ag₂CrO₄ (s)  ⇄ 2Ag⁺ (aq)  + CrO₄⁻² (aq)   Ksp

                          2 s                 s

Look the stoichiometry is 1:2, between the salt and the silver.

Ksp = [2s]²  . [s] → 4s³

 
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