2.46 grams of Cu(NO3)2 would contain how many formula units

You want to minimize an objects thermal energy loss on a cold day. How does heat energy transfer affect thermal energy loss?

san4es73 [151]

Answer:

4- A material that transfers heat energy more easily than another material will experience a greater rate of thermal energy loss than an object that does not transfer heat energy easily.

Explanation:

Thermal energy loss has to do with loss of heat energy by a body to another body or its environment. The aim of the process is usually the attainment of thermal equilibrium between the body and its environment.

On a cold day, a material that transfers thermal energy more easily will loose thermal energy faster than an object that does not transfer thermal energy. The rate of heat transfer of a body determines its rate of loss of thermal energy.

6 0

3 years ago

The color of light emitted during a flame test depends on

tiny-mole [99]

option of d is the write answer

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3 years ago

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Chili can be made using ground meat or chunks of meat Which would you choose if you were in a hurry? Explain.

goblinko [34]

I would choose Ground meat if i was in a hurry Hope this helps:)

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3 years ago

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Naturally occurring silicon has an atomic mass of 28.086 and consists of three isotopes. The major isotope is 28Si, natural abun

Elden [556K]

Answer:

29Si has a natural abundance of 4.68%.

30Si has a relative atomic mass of 29.99288 and a natural abundance of 3.09%.

Explanation:

The atomic mass of silicon is given by:

Si=Si²⁸×A₁+Si²⁹×A₂+Si³⁰×A₃

Where:

Si: atomic mass of silicon (28.086)

Si²⁸: relative atomic mass of 28Si (27.97693)

A₁: natural abundance of 28Si (92.23%)

Si²⁹: relative atomic mass of 29Si (28.97649)

A₂: natural abundance of 29Si

Si³⁰: relative atomic mass of 30Si

A₃: natural abundance of 30Si

We also know that 30Si natural abundance is in the ratio of 0.6592 to that of 29Si.

We have to set up a system of three equations in three unknowns:

Si=Si²⁸×A₁+Si²⁹×A₂+Si³⁰×A₃

A₃=0.6592×A₂

A₁+A₂+A₃=1

First, we find substitute the value of A₃ in the third equation and solv teh value of A₂:

A₁+A₂+0.6592×A₂=1

A₁+1.6592×A₂=1

1.6592×A₂=1-A₁

A₂= $\frac{1-A₁}{1.6592}=\frac{1-0.9223}{1.6592}$ =0.0468

Then, we find the value of A₃:

A₃=0.6592×A₂

A₃=0.6592×0.0468=0.0309

Finally, we find the value of Si³⁰ in the first equation:

Si=Si²⁸×A₁+Si²⁹×A₂+Si³⁰×A₃

28.086=27.97693×0.9223+28.97649×0.0468+Si³⁰×0.0309

28.086=27.15922+Si³⁰×0.0309

28.086-27.15922=Si³⁰×0.0309

$\frac{0.92678}{0.0309}$ =Si³⁰

Si³⁰=29.99288

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3 years ago

One way the U.S. Environmental Protection Agency (EPA) tests for chloride contaminants in water is by titrating a sample of silv

Nastasia [14]

The balanced chemical reaction is expressed as follows:

<span>CuCl2 (aq) + 2AgNO3 (aq) → 2AgCl (s) + CuNO32 (aq)

To determine the </span><span>concentration of copper(II) chloride contaminant in the original groundwater sample, we use the final amount of silver chloride that was produced from the reaction and the relation of the substances from the chemical reaction. We calculate as follows:

mmol AgCl = 6.1 mg AgCl ( 1 mmol / 143.35 mg ) = 0.0426 mmol
mmol CuCl2 = </span>0.0426 mmol AgCl ( 1 mmol CuCl2 / 2 mmol AgCl ) = 0.0213 mmol CuCl2
concentration of CuCl2 in the original water sample = 0.0213 mmol CuCl2 / 200.0 mL = 1.0638 x 10^-4 mmol / mL or 1.0638 x 10^-4 mol/L

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3 years ago