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anyanavicka [17]

3 years ago

11

What energy form is a bow and arrow??

1 answer:

lakkis [162]3 years ago

3 0

Hello Naphya,

Question - What Energy Form Is A Bow And Arrow

Answer - Potential Energy

Why - "The energy is stored in the limbs in the form of potential energy; when the string is released the energy stored into the limbs is released, most of which is absorbed by the arrow."

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A solution is prepared by mixing 50.0 mL toluene (C6H5CH3 d=0.867 g/mL) with 125 mL Benzene (C6H6 d=0.874 g/mL). Assuming that t

Tatiana [17]

Answer:

mass % = 28.4%

mole fraction = 0.252

molality = 3.08 molal

molarity = 2.69 M

Explanation:

Step 1: Data given

Volume of toluene = 50.0 mL = 0.05 L

Density toluene = 0.867 g/mL

Molar mass toluene = 92.14 g/mol

Volume of benzene = 125 mL = 0.125 L

Density benzene = 0.874 g/mL

Molar mass benzene = 78.11 g/mol

Step 2: Calculate masses

Mass = density * volume

Mass toluene = 50.0 mL * 0.867 g/mL = 43.35 g

Mass benzene = 125 mL * 0.874 g/mL = 109.25 g

Step 3: Calculate number of moles

Moles = mass / molar mass

Moles toluene = 43.35 grams /92.14 g/mol = 0.470 5 moles

Moles benzene = 109.25 grams / 78.11 g/mol = 1.399 moles

Step 4: Calculate molarity of toluene

Molarity = moles / volume

Molarity toluene = 0.4705 moles / 0.175 L = 2.69 M

Step 5: Calculate mass % of toluene

Mass % = (43.35 grams / (43.35 + 109.25) )*100 % = 28.4 %

Step 6: Calculate mole fraction of toluene

Mole fraction toluene = Moles toluene / total number of moles

Mole fraction toluene = 0.4705 / (0.4705 + 1.399) = 0.252

Step 7: Molality of toluene

Molality = number of moles / mass

Molality of toluene = 0.4705 moles / (0.04335 + 0.10925)

Molality of toluene = 3.08 molal

4 0

3 years ago

He isotope 62(over28ni has the largest binding energy per nucleon of any isotope. calculate this value from the atomic mass of n

irina [24]

The atomic mass of the isotope Ni ( 62 over 28 ) = 61.928345 amu.
Mass of the electrons: 28 · 5.4584 · 10^(-4 ) amu = 0.0152838 amu ( g/mol )
Mass of the nuclei:
61.928345 amu - 0.0152838 amu = 61.913062 amu (g/mol)
The mass difference between a nucleus and its constituent nucleons is called the mass defect.
For Ni ( 62 over 28 ): Mass of the protons: 28 · 1.00728 amu = 28.20384 amu
Mass of the neutrons: 34 · 1.00866 amu = 34.299444 amu
In total : 62.49828 amu
The mass defect = 62.49828 - 61.913062 = 0.585218 amu
Nucleus binding energy:
E = Δm · c² ( the Einstein relationship )
E = 0.585218 · ( 2.9979 · 10^8 m/s )² · 1 / (6.022 · 10^23) · 1 kg / 1000 g =
= 0.585218 · 8.9874044 · 10 ^16 : (6.022 · 10^23) · 0.001 =
= ( 5.2595908 : 6.022 ) · 0.001 · 10^(-7 ) =
= 0.0008733 · 10^(-7) J = 8.733 · 10^(-11) J
The nucleus binding energy per nucleon:
8.733 · 10^(-11) J : 62 = 0.14085 · 10 ^(-11) =
= 1.4085 · 10^(-12) J per nucleon.

4 0

3 years ago

A chemical formula represents:

Ivenika [448]

Answer:

no

Explanation:

it is a molecular formula that represents the number atoms and type of atoms

7 0

3 years ago

2. Sitting on a bench top are several samples: lithium metal (d = 0.53 g/mL), gold (d = 19.3 g/mL), aluminum (d = 2.70 g/mL), an

Snezhnost [94]

Answer:

The sample of lithium occupies the largest volume.

Explanation:

Given the densities for the four elements, we have the expression $d=\frac{m}{V}$ that shows the relationship between mass and Volume to express the density of an element.

For each element we have:

$d_{lithium}=\frac{m_{lithium}}{V_{lithium}}=0.53g/mL$

$d_{gold}=\frac{m_{gold}}{V_{gold}}=19.3g/mL$

$d_{aluminum}=\frac{m_{aluminum}}{V_{aluminum}}=2.70g/mL$

$d_{lead}=\frac{m_{lead}}{V_{lead}}=11.3g/mL$

The problem says that all the samples have the same mass, so:

$m_{lithium}=m_{gold}=m_{aluminum}=m_{lead}=m$

it means that m is a constant

Now, solving for the Volume in each element and with m as a constant, we have:

$V_{lithium}=\frac{m}{d_{lithium}}$

$V_{lithium}=\frac{1}{0.53\frac{g}{mL}} *m$

$V_{lithium}=1.88\frac{mL}{g}*m$

$V_{gold}=\frac{m}{d_{gold}}$

$V_{gold}=\frac{1}{19.3\frac{g}{mL}} *m$

$V_{gold}=5.18*10^{-2}\frac{mL}{g}*m$

$V_{aluminum}=\frac{m}{d_{aluminum}}$

$V_{aluminum}=\frac{1}{2.70\frac{g}{mL}} *m$

$V_{aluminum}=3.70*10^{-1}\frac{mL}{g}*m$

$V_{lead}=\frac{m}{d_{lead}}$

$V_{lead}=\frac{1}{11.3\frac{g}{mL}} *m$

$V_{lead}=8.85*10^{-2}\frac{mL}{g}*m$

If we assume m = 1g, we find that:

$V_{lithium}=1.88mL$

$V_{gold}=5.18*10^{-2}mL$

$V_{aluminum}=3.70*10^{-1}mL$

$V_{lead}=8.85*10^{-2}mL$

So we can see that the sample of lithium occupies the largest volume with 1.88mL

Note that m only can take positive values, so if you change the value of m, always will be the lithium which occupies the largest volume.

4 0

2 years ago

determine the element of lowest atomic number that contains a complete d subshell in the ground state.

viktelen [127]

Answer:

The element with the lowest atomic number that contains a complete d-subshell in the ground state is copper.

Source:

Determine the element of lowest atomic number that contains a complete d subshell in the ground... - Study.com

4 0

2 years ago