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3 years ago

In which of the following situations is the greatest amount of work accomplished?

Chemistry

2 answers:

Dahasolnce [82]3 years ago

7 0

work = force x distances

A - moving 2 newton's up 0.6 meter = 1.2 joules
B - moving 4 newton's up 0.6 meter = 2.4 joules
C - moving 6 newton's up 0.3 meter = 1.8 joules
D - moving 9 newton's up 0.3 meter = 2.7 joules
The greatest amount of work is in example D.

D is your answer.

UNO [17]3 years ago

6 0

The answer is D “A boy lifts a 9-newton box 3.0 meters”

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Why is the addition of cold distilled water to the container holding the aspirin you are trying to isolate a necessary step and

Rudik [331]

Answer:

The aspirin is more soluble in ethanol than water - the water helps the crystals separate from solution.

Explanation:

Hope this helps!

3 0

2 years ago

I know how to do electron configuration, but I think I’m doing the rest wrong. Answers and explanations would be much appreciate

Natasha_Volkova [10]

Your answers seem great so far, except for a tiny issue: With the ionic symbols, try to get into the habit of using "+", with metals, like sodium, and try to use the integer first. So, for example, a potassium ion would be K^+, while an oxide ion would be O^2-

Let's take aluminium as an example I'll work through:

Aluminium, with it's atomic number of 13, will have an electronic configuration of 1s2 2s2 2p6 1s2 2p1. So it would have 2, 8, 3 electrons in the first three energy levels, respectively.

Usually, if an elemental atom has a valence electron (highest energy level electron) count less than 4, it almost always will lose electrons. Since aluminium has 3, it will also lose the electrons.

It loses the 3 valence electrons, and so will end up with 10 electrons.

Since the atomic number also tells how many protons it has, we know that an aluminium atom has 13 protons, which doesn't change.

Since the size of the charges of a proton and an electron are the same, with protons being positive and the electrons being negative, an aluminium ion would have a charge of +3, and the Ionic symbol would be Al^3+

Hope I helped! xx

4 0

3 years ago

A thermometer having first-order dynamics with a time constant of 1 min is placed in a temperature bath at 100oF. After the ther

sveticcg [70]

Answer:

(a) See below

(b) 103.935 °F; 102.235 °F

Explanation:

The equation relating the temperature to time is

$T = T_{0} + \Delta T\left (1 - e^{-t/\tau} \right )$

1. Calculate the thermometer readings after 0.5 min and 1 min

(a) After 0.5 min

$\begin{array}{rcl}T & = & T_{0} + \Delta T\left (1 - e^{-t/\tau} \right )\\ & = & 100 + 10\left (1 - e^{-0.5/1} \right )\\ & = & 100 + 10\left (1 - e^{-0.5} \right )\\ & = & 100 + 10 (1 - 0.6065)\\ & = & 100 + 10(0.3935)\\ & = & 100 + 3.935\\ & = & 103.935\,^{\circ}F\\\end{array}$

(b) After 1 min

$\begin{array}{rcl}T & = & T_{0} + \Delta T\left (1 - e^{-t/\tau} \right )\\ & = & 100 + 10\left (1 - e^{-1/1} \right )\\ & = & 100 + 10\left (1 - e^{-1} \right )\\ & = & 100 + 10 (1 - 0.3679)\\ & = & 100 + 10(0.6321)\\ & = & 100 + 6.321\\ & = & 106.321\,^{\circ}F\\\end{array}$

2. Calculate the thermometer reading after 2.0 min

T₀ =106.321 °F

ΔT = 100 - 106.321 °F = -6.321 °F

t = t - 1, because the cooling starts 1 min late

$\begin{array}{rcl}T & = & T_{0} + \Delta T\left (1 - e^{-(t - 1)/\tau} \right )\\ & = & 106.321 - 6.321\left (1 - e^{-(2 - 1)/1} \right )\\ & = & 106.321 - 6.321\left (1 - e^{-1} \right )\\ & = & 106.321 - 6.321 (1 - 0.3679)\\ & = & 106.321 - 6.321 (0.6321)\\ & = & 106.321 - 3.996\\ & = & 102.325\,^{\circ}F\\\end{array}$

3. Plot the temperature readings as a function of time.

The graphs are shown below.

6 0

3 years ago

Bf3 can be octet or not

makvit [3.9K]

Answer:

(BH3 follows the octet rule by dimerizing, as Hadi Kurniawan AR pointed out.) For H and He, an "octet" = 2 electrons. Boron does prefer to follow the octet rule, in that it likes to form borate compounds such as NaBH4. It also is happy to form compounds with elements with lone pairs.

7 0

3 years ago

Imagine a 15kg block moving with a speed of 20m/s. Calculate the kinetic energy of this block.

SSSSS [86.1K]

The kinetic energy formula is;

$KE=\frac{m.v^2}{2}$

The variable $m$ represents the mass. Its unit is kilogram. We are informed that the mass of the object is $15kg$ . The variable $v$ represents the linear velocity. Its unit is meter per second. We are informed that the linear velocity of the object is $20m/s$ . Let's find the kinetic energy of the object by substituting these values in the formula.

$KE=\frac{15kg.(20m/s)^2}{2}$
$2.KE=15kg.400m^2/s^2$
$2.KE=(6000)kgm^2/s^2$
$KE=(3000)kgm^2/s^2=3000J$

5 0

1 year ago