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3 years ago

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The density of red wine is 1.01g/cm3 and the density of olive oil is 0.92g/cm3.A salad dressing that consisted of these two ingr

edients was left sitting too long.The ingredients separated in the bottle ,which ingredient is which layer?Explain.

1 answer:

Sauron [17]3 years ago

3 0

the red wine is the densest so, it will be on bottom olive oil will be on the top. because its less than the red wine

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Ghella [55]

Answer:

I would say C

Explanation: Hope this helped

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3 years ago

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Calculate the initial rate for the formation of c at 25 ∘c, if [a]=0.50m and [b]=0.075m

Nataly_w [17]

The rate of formation of a product depends on the the concentrations of the reactants in a variable way.

If two products, call them A and B react together to form product C, a general equation for the formation of C has the form:

rate = k*[A]^m * [B]^n

Where the symbol [ ] is the concentration of each compound.

Then, plus the concentrations of compounds A and B you need k, m and n.

Normally you run controled trials in lab which permit to calculate k, m and n .

Here the data obtained in the lab are:

<span>Trial      [A]     [B]     Rate </span><span>
<span>            (M)    (M)      (M/s) </span>
<span>1     0.50 0.010      3.0×10−3 </span>
<span>2        0.50 0.020       6.0×10−3 </span>
<span>3       1.00 0 .010       1.2×10−2</span></span>

Given that for trials 1 and 2 [A] is the same you can use those values to find n, in this way

rate 1 = 3.0 * 10^ -3 = k [A1]^m * [B1]^n

rate 2 = 6.0*10^-3 = k [A2]^m * [B2]^n

divide rate / rate 1 => 2 = [B1]^n / [B2]^n

[B1] = 0.010 and [B2] = 0.020 =>

6.0 / 3.0 =( 0.020 / 0.010)^n =>

2 = 2^n => n = 1

Given that for data 1 and 3 [B] is the same, you use those data to find m

rate 3 / rate 1 = 12 / 3.0 = (1.0)^m / (0.5)^m =>

4 = 2^m => m = 2

Now use any of the data to find k

With the first trial: rate = 3*10^-3 m/s = k (0.5)^2 * (0.1) =>

k = 3.0*10^-3 m/s / 0.025 m^3 = 0.12 m^-2 s^-1

Now that you have k, m and n you can use the formula of the rate with the concentrations given

rate = k[A]^2*[B] = 0.12 m^-2 s^-1 * (0.50m)^2 * (0.075m) = 0.0045 m/s = 4.5*10^=3 m/s

Answer: 4.5 * 10^-3 m/s

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3 years ago

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Calculate the energy in joules and calories required to heat 25.0 g of water from 12.5C to 25.7C

Alex Ar [27]

The heat (Q) required to raise the temp of a substance is:<span>Q=m∗Cp∗ΔT</span><span> where m is the mass of the object (25.0g in this case), Cp is the specific heat capacity of the substance (for water Cp = 1.00cal/gC, or 4.18J/gC,
and Dt is the change in temp.
You'll have to solve this twice, once with the Cp in calories, and once with the Cp in joules.
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3 years ago

What should you do if the high power objective lens touches or breaks the cover slip?

saul85 [17]

Tell your teacher. They'll know what to do and it's best to report it to them.

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3 years ago

Based on the three formulas shown, use one of them to solve for the purple yellow and red box and explain how you did it.

zysi [14]

P = 11.133 atm (purple)

T = -236.733 °C(yellow)

n = 0.174 mol(red)

<h3>Further explanation </h3>

Some of the laws regarding gas, can apply to ideal gas (volume expansion does not occur when the gas is heated),:

Boyle's law at constant T, P = 1 / V
Charles's law, at constant P, V = T
Avogadro's law, at constant P and T, V = n

So that the three laws can be combined into a single gas equation, the ideal gas equation

In general, the gas equation can be written

$\large {\boxed {\bold {PV = nRT}}}$

where

P = pressure, atm

V = volume, liter

n = number of moles

R = gas constant = 0.08206 L.atm / mol K

T = temperature, Kelvin

To choose the formula used, we refer to the data provided

Because the data provided are temperature, pressure, volume and moles, than we use the formula PV = nRT

Purple box

T= 10 +273.15 = 373.15 K

V=5.5 L

n=2 mol

$\tt P=\dfrac{nRT}{V}\\\\P=\dfrac{2\times 0.08205\times 373.15}{5.5}\\\\P=11.133~atm$

Yellow box

V=8.3 L

P=1.8 atm

n=5 mol

$\tt T=\dfrac{PV}{nR}\\\\T=\dfrac{1.8\times 8.3}{5\times 0.08205}\\\\T=36.42~K=-236.733^oC$

Red box

T = 12 + 273.15 = 285.15 K

V=3.4 L

P=1.2 atm

$\tt n=\dfrac{PV}{RT}\\\\n=\dfrac{1.2\times 3.4}{0.08205\times 285.15}\\\\n=0.174~mol$

3 0

3 years ago