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stellarik [79]
3 years ago
5

Question 6 (3 points)

Chemistry
1 answer:
Natasha_Volkova [10]3 years ago
7 0

Answer:

c) 12

Explanation:

electron = 23-11 = 12

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The reaction is proceeding at a rate of 0.0080 Ms-1 in 50.0 mL of solution in a system with unknown concentrations of A and B. W
Leokris [45]

Answer:

0.0010 mol·L⁻¹s⁻¹  

Explanation:

Assume the rate law is  

rate = k[A][B]²

If you are comparing two rates,

\dfrac{\text{rate}_{2}}{\text{rate}_{1}} = \dfrac{k_{2}\text{[A]}_2[\text{B]}_{2}^{2}}{k_{1}\text{[A]}_1[\text{B]}_{1}^{2}}= \left (\dfrac{\text{[A]}_{2}}{\text{[A]}_{1}}\right ) \left (\dfrac{\text{[B]}_{2}}{\text{[B]}_{1}}\right )^{2}

You are cutting each concentration in half, so

\dfrac{\text{[A]}_{2}}{\text{[A]}_{1}} = \dfrac{1}{2}\text{ and }\dfrac{\text{[B]}_{2}}{\text{[B]}_{1}}= \dfrac{1}{2}

Then,

\dfrac{\text{rate}_{2}}{\text{rate}_{1}} = \left (\dfrac{1}{2}\right ) \left (\dfrac{1}{2}\right )^{2} = \dfrac{1}{2}\times\dfrac{1}{4} = \dfrac{1}{8}\\\\\text{rate}_{2} = \dfrac{1}{8}\times \text{rate}_{1}= \dfrac{1}{8}\times \text{0.0080 mol$\cdot$L$^{-1}$s$^{-1}$} = \textbf{0.0010 mol$\cdot$L$^{-1}$s$^{-1}$}\\\\\text{The new rate is $\large \boxed{\textbf{0.0010 mol$\cdot$L$^\mathbf{{-1}}$s$^{\mathbf{-1}}$}}$}

8 0
3 years ago
If measuring experimental results, what can you predict about the work output of a 1200 watt hair dryer? the work output of the
eimsori [14]

Answer: first option, the work output of the hairdryer will be less than the work input.


Explanation:


1) The work output measured in watts is the power of hair dryer measured in joules per second.


2) The hair dryer converts electrical energy from the wall outlet to mechanical and thermal energy: hot wind.


3) Nevertheless, you can never expect a 100% efficiency of the machines: due to friction, some energy is converted into useless energy.


So, efiiviency = power output / power input< 1 ⇒


power output = work output / time


input power = work input / time


⇒ work output / work input < 1


⇒ work output < work input.


Which is the first option: the work output of the hairdryer will be less than the work input

6 0
3 years ago
Read 2 more answers
Identify the major ionic species present in an aqueous solution of C12H22O11 (sucrose).
Marina86 [1]
<h3><u>Answer;</u></h3>

No ions present

<h3><u>Explanation;</u></h3>
  • Ionic compounds are compounds made up of ions. These ions are atoms that gain or lose electrons, giving them a net positive or negative charge.
  • Atoms that gain electrons and therefore have a net negative charge are known as anions. Conversely, atoms that lose electrons have a net positive charge are called cations.
  • C12H22O11 (sucrose) is not an ionic compound, and therefore does not have any ions. Sucrose is a molecular compound.
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3 years ago
HOW DOES DENSITY AFFECT THE MELTING POINT OF WATER?
ddd [48]
<span> As a liquid is heated, its vapor pressure increases until the vapor pressure equals the pressure of the gas above it. Bubbles of vaporized liquid (i.e., gas) form within the bulk liquid and then rise to the surface where they burst and release the gas. (At the boiling temperature the vapor inside a bubble has enough pressure to keep the bubble from collapsing.) In order to form vapor, the molecules of the liquid must overcome the forces of attraction between them.<span> The temperature of a boiling liquid remains constant, even when more heat is add.</span></span>
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3 years ago
What is the Si base unit of mass?
miv72 [106K]

The SI unit of temperature is the kelvin (K), which spans the same temperature change as the degree Celsius. The Kelvin scale is a thermodynamic scale, meaning that its zero point is at absolute zero rather than the freezing point of water. The second reference point for this scale as it is currently defined is the triple point of water, which is a unique point on the phase diagram of water (a specific combination of pressure and temperature) where ice, liquid water and water vapor are all in equilibrium. The triple point is assigned the temperature of 273.16 K.

The old centigrade scale used the freezing and boiling temperatures of water as its reference points, with one degree centigrade equal to 1/100 of the temperature span between the freezing and boiling points of water. The definition of the Kelvin scale was chosen to make the kelvin the same size as the centigrade degree.

The Celsius scale is defined in terms of the Kelvin scale but is equivalent to the old centigrade scale, which it replaces. It is convenient for reporting weather and cooking temperatures and so on, but is not particularly useful for scientific purposes. For instance, the behavior of gases which approximate ideal gases is such that at zero degrees C they experience a volume change of 1/273 for a one degree change in temperature. This observation provided one of the first indications for the value of absolute zero.

When using the ideal gas law:

PV = nRT

where P is pressure

V is volume

n is the quantity of gas in moles

R is a constant

T is the temperature

it is necessary to use a thermodynamic scale, usually Kelvin.

Another thermodynamic scale, the Rankine scale, has a relationship to the Fahrenheit temperature scale analogous to that between the Kelvin and Celsius scales.

4 0
3 years ago
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