Gold at 100.0°C is placed in 2.00×10^2 g of water at 20.0°C. The mixture reaches equilibrium at 21.2°C. the specific heat of gold is 0.129 (J/g°C) What is the mass of the gold? Specific heat of water is 4.18 (J/g°C) .

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Answer 1

Answer:

Answer:

there it is fella atleast i tried

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An electron in a hydrogen atom undergoes a transition from the n = 3 level to the n = 6 level. To accomplish this, energy, in the form of light, must be absorbed by the hydrogen atom. Calculate the energy of the light (in kJ/photon) associated with this transition.

Answers

Answer: $1.816* 10^(-19) J$

Explanation:

Given

$E=(hc)/(\lambda )$

$E=2.18* 10^(-18)((1)/(n_1^2)-(1)/(n_2^2))$

where h=Planck constant

c=speed of light

$E=2.18* 10^(-18)((1)/(3^2)-(1)/(6^2))$

$E=2.18* 10^(-18)* (1)/(12)$

$E=1.816* 10^(-19) J$

Answer:

1.82 × 10⁻¹⁹ J

Explanation:

The Bohr model of the atom states that the energy required to transition between two energy levels is equal to the difference between the inverse squares of the energy levels multiplied by the Rydberg constant:

A speedy rabbit is hopping to the right with a velocity of 4.0 \,\dfrac{\text m}{\text s}4.0 s m 4, point, 0, start fraction, start text, m, end text, divided by, start text, s, end text, end fraction when it sees a carrot in the distance. The rabbit speeds up to its maximum velocity of 13 \,\dfrac{\text m}{\text s}13 s m 13, start fraction, start text, m, end text, divided by, start text, s, end text, end fraction with a constant acceleration of 2.0 \,\dfrac{\text m}{\text s^2}2.0 s 2 m 2, point, 0, start fraction, start text, m, end text, divided by, start text, s, end text, squared, end fraction rightward.

Answers

Answer: 38.25 m

Explanation:

In this situation we need to find the distance $d$ between the rabbit and the carrot, and we can use the following equation, since the rabbit's acceleration is constant:

$V^(2)=V_(o)^(2) + 2ad$ (1)

Where:

$V=13 m/s$ is the rabbit's maximum velocity (final velocity)

$V_(o)=4 m/s$ is the rabbit's initial velocity

$a=2 m/s^(2)$ is the rabbit's acceleration

$d$ is the distance between the rabbit and the carrot

Isolating $d$ :

$d=(V^(2)-V_(o)^(2))/(2a)$ (2)

$d=((13 m/s)^(2)-(4 m/s)^(2))/(2(2 m/s^(2)))$ (3)

Finally:

$d=38.25 m$

Answer:

4.5s

Explanation:

Cause that's what it says on my test hints

Extremely high temperatures and pressures are necessary for fusion reactions to take place in stars. true or false.

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That statement stands unique in its simple veracity and truthiness.

Which is an IUPAC name for a covalent compound

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In naming covalent compound (binary) based in IUPAC naming, we have 4 rules to be followed:

1. The first element of the formula will use the normal name of the given element. for example: CO2 ( Carbon Dioxide), Carbon is the element name of the first element of the formula.

2. The second element is named as if they are treated like an anion but put in mind that these are no ions in a covalent compound but we put -ide on the second element as if it is an anion.

3. Prefixes are used to indicate the number of atom of the elements in the compound. for example: mono- 1 atom, di- 2atoms, tri- 3 atoms and etc

4. Prefix "mono"is never used in naming the first element. For example: Carbon dioxide, there should be no monocarbon dioxide.

An IUPAC name for a covalent compound is ethane. For covalent compounds, IUPAC names are based on the composition and structure of the molecules. Covalent compounds typically consist of nonmetals or a combination of nonmetals and metalloids.

Ethane (C₂H₆) is a covalent compound that consists of two carbon atoms bonded to each other with single bonds, and each carbon atom is also bonded to three hydrogen atoms.

Other examples of IUPAC names for covalent compounds include:

Methane (CH₄)

Propane (C₃H₈)

Ethene (C₂H₄)

Nitrogen dioxide (NO₂)

These names are derived based on the IUPAC rules for naming covalent compounds, which consider the number and types of atoms present in the molecule.

To know more about covalent compounds:

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You travel in a circle, whose circumference is 8 kilometers, at an average speed of 8 kilometers/hour. If you stop at the same point you started from, what is your average velocity?

Answers

Answer: The average velocity will be 0 m/s.

Explanation:

Average velocity is defined as the fraction of displacement and total time.

$\text{Average velocity}=\frac{Displacement}{\text{total time}}$

In the given question, it states that the person stops at the same point from where it started, which means that the displacement of this person is 0 meters.

Therefore, the average velocity becomes:

$\text{Average velocity}=\frac{0}{\text{total time}}=0m/s$

Hence, the average velocity will be 0 m/s.

the answer is 8 kilometers per hour

At the local grocery store, you push a 14.5-kg shopping cart. You stop for a moment to add a bag of dog food to your cart. With a force of 12.0 N you now accelerate the cart from rest through a distance of 2.29 m in 3.00 s. What was the mass of the dog food?