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3 years ago

5

The lower the percent of energy required for a machine to perform work, the more energy efficient it is. Please select the best

answer from the choices provided T F

2 answers:

natita [175]3 years ago

5 0

Answer:

True

Explanation: just took the test

Anvisha [2.4K]3 years ago

3 0

Hey You!

The Correct Answer Is: True.

I Really Hope This Helped You, Good Luck With Your Studies! =)

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To locate objects in their environments, bats in flight and porpoises under water both use ultrasound waves with frequencies tha

Katarina [22]

Answer: The porpoise would hear its echo first because sound travels faster in the water than in air.

Explanation:

3 0

3 years ago

During science lab, Carl notices that when he adds water to his solid sample of anhydrous copper

Bond [772]

Answer:

The dissociation of copper sulfate into ions is an exothermic chemical reaction that releases heat into the surroundings.

Explanation:

Some of the potential energy stored in the solid sample of anhydrous copper sulfate is released as heat as the sample dissolves and dissociates into ions in the water. This is due to the large lattice energy of the crystalline copper sulfate.

hope this helps

4 0

3 years ago

A student mixes 33.0 mL of 2.70 M Pb ( NO 3 ) 2 ( aq ) with 20.0 mL of 0.00157 M NaI ( aq ) . How many moles of PbI 2 ( s ) prec

GalinKa [24]

<u>Answer:</u> The moles of precipitate (lead (II) iodide) produced is $1.57\times 10^{-5}$ moles

<u>Explanation:</u>

To calculate the number of moles for given molarity, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}$ .....(1)

<u>For lead (II) nitrate:</u>

Molarity of lead (II) nitrate solution = 2.70 M

Volume of solution = 33.0 mL = 0.033 L (Conversion factor: 1 L = 1000 mL)

Putting values in equation 1, we get:

$2.70M=\frac{\text{Moles of lead (II) nitrate}}{0.033L}\\\\\text{Moles of lead (II) nitrate}=(2.70mol/L\times 0.0330L)=0.0891mol$

<u>For NaI:</u>

Molarity of NaI solution = 0.00157 M

Volume of solution = 20.0 mL = 0.020 L

Putting values in equation 1, we get:

$0.00157M=\frac{\text{Moles of NaI}}{0.020L}\\\\\text{Moles of NaI}=(0.00157mol/L\times 0.0200L)=3.14\times 10^{-5}mol$

For the given chemical reaction:

$Pb(NO_3)_2(aq.)+2NaI(aq.)\rightarrow PbI_2(s)+2NaNO_3(aq.)$

By Stoichiometry of the reaction:

2 moles of NaI reacts with 1 mole of lead (II) nitrate

So, $3.14\times 10^{-5}$ moles of NaI will react with = $\frac{1}{2}\times 3.14\times 10^{-5}=1.57\times 10^{-5}mol$ of lead (II) nitrate

As, given amount of lead (II) nitrate is more than the required amount. So, it is considered as an excess reagent.

Thus, NaI is considered as a limiting reagent because it limits the formation of product.

By Stoichiometry of the reaction:

2 moles of NaI produces 1 mole of lead (II) iodide

So, $3.14\times 10^{-5}$ moles of NaI will produce = $\frac{1}{2}\times 3.14\times 10^{-5}=1.57\times 10^{-5}moles$ of lead (II) iodide

Hence, the moles of precipitate (lead (II) iodide) produced is $1.57\times 10^{-5}$ moles

4 0

3 years ago

A. 207 kJ<br> B. 4730 kJ<br> O C. 9460 kJ<br> O D. 414 kJ

slavikrds [6]

Answer:

C = 9460 Kj

Explanation:

Given data:

Mass of copper = 2kg

Latent heat of vaporization = 4730 Kj/Kg

Energy required to vaporize 2kg copper = ?

Solution:

Equation

Q= mLvap

by putting values,

Q= 2kg × 4730 Kj/Kg

Q = 9460 Kj

3 0

3 years ago

Look at the above table.

MariettaO [177]

Answer:

<em> 1</em>. A. 0

<em>2</em>. B. 7

<em>3. </em>C<em>.</em><em> </em>4

Explanation:

1. charge is equal to the number of protons minus the number of electrons!

2. neutrons is equal to mass number minus atomic number!

3. valence electrons equal 4!

Hope this helped you! :)

7 0

3 years ago