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Ostrovityanka [42]
3 years ago
13

From the mass spectrum of the compound, the molecule was determined to have a molar mass of 907 g/mol. what is its molecular for

mula?
Physics
1 answer:
Molodets [167]3 years ago
3 0
Please elaborate more on your question so I can help you
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If someone said to describe weather what would you tell them?<br> Please help.
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The state of the atmosphere at a given place and time as regards to heat, dryness sunshine, wind, rain, etc.
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How does friction act as an unbalanced force for objects in motion?
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Friction is a force that acts in a direction opposite to motion. If there were not force of friction, a baseball player would never stop and would be continuously moving in the direction of motion. Balanced forces do not cause a nonmoving object to start moving.

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3 years ago
Geronimo wants to move an object 12 meters. Calculate the net work done by the object with an applied force of 150 N and a frict
valina [46]

Answer:

1476 J

Explanation:

From the question,

Net Work done = Net force× distance moved by net force.

W' = (F-F')×d................... Equation 1

Where W' = Net work done, F = force applied, F' = Frictional force, d = distance moved.

Given: F = 150 N, F' = 37 N, d = 12 m

Substitute these values into equation 1

W' = (150-37)×12

W' = 123×12

W' = 1476 J.

hence the Net Work done by the object is 1476 J

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3 years ago
Which have scientists learned from studying seismic waves that travel through earth’s interior?
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3 years ago
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Three metal fishing weights, each with a mass of 1.00x102 g and at a temperature of 100.0°C, are placed in 1.00x102 g of water a
worty [1.4K]

Answer:

Approximately 0.253\; {\rm J \cdot g^{-1} \cdot K^{-1}} assuming no heat exchange between the mixture and the surroundings.

Explanation:

Consider an object of specific heat capacity c and mass m. Increasing the temperature of this object by \Delta T would require Q = c\, m \, \Delta T.

Look up the specific heat of water: c(\text{water}) = 4.182\; {\rm J \cdot g^{-1} \cdot K^{-1}}.

It is given that the mass of the water in this mixture is m(\text{water}) = 1.00 \times 10^{2}\; {\rm g}.

Temperature change of the water: \Delta T(\text{water}) = (45 - 35)\; {\rm K} = 10\; {\rm K}.

Thus, the water in this mixture would have absorbed :

\begin{aligned}Q &= c\, m\, \Delta T \\ &= 4.182\; {\rm J \cdot g^{-1}\cdot K^{-1}} \\ &\quad \times 1.00 \times 10^{2}\; {\rm g} \times 10\; {\rm K} \\ &= 4.182 \times 10^{3}\; {\rm J}\end{aligned}.

Thus, the energy that water absorbed was: Q(\text{water}) = 4.182 \times 10^{3}\; {\rm J}.

Assuming that there was no heat exchange between the mixture and its surroundings. The energy that the water in this mixture absorbed, Q(\text{water}), would be the opposite of the energy that the metal in this mixture released.

Thus: Q(\text{metal}) = -Q(\text{water}) = -4.182 \times 10^{3}\; {\rm J} (negative because the metal in this mixture released energy rather than absorbing energy.)

Mass of the metal in this mixture: m(\text{metal}) = 3 \times 1.00 \times 10^{2}\; {\rm g} = 3.00 \times 10^{2}\; {\rm g}.

Temperature change of the metal in this mixture: \Delta T(\text{metal}) = (100 - 45)\; {\rm K} = 55\; {\rm K}.

Rearrange the equation Q = c\, m \, \Delta T to obtain an expression for the specific heat capacity: c = Q / (m\, \Delta T). The (average) specific heat capacity of the metal pieces in this mixture would be:

\begin{aligned}c &= \frac{Q}{m\, \Delta T} \\ &= \frac{-4.182 \times 10^{3}\; {\rm J}}{3.00 \times 10^{2}\; {\rm g} \times (-55\; {\rm K})} \\ &\approx 0.253\; {\rm J \cdot g^{-1} \cdot K^{-1}}\end{aligned}.

6 0
2 years ago
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