CH4+(x)O2=CO2 +(Y)H2O
C=1 +H=4 +O=? = C=1 +O=2+? +H=?
H=4>>Y=2
C=1 +H=4 +O=? = C=1 +O=(2+2) +H=4
C=1 +H=4 +O=4 = C=1 +O=4 +H=4
O=4>>X=2
CH4+(2)O2 =CO2 +(2)H2O
Answer:
The particles of the medium just vibrate in place.
Explanation:
As they vibrate, they pass the energy of the disturbance to the particles next to them, which pass the energy to the particles next to them, and so on. Particles of the medium don't actually travel along with the wave.
Answer:
I would use calorimetric to determine the specific heat and I would measure the mass of a sample
Explanation:
I would use calorimetry to determine the specific heat.
I would measure the mass of a sample of the substance.
I would heat the substance to a known temperature.
I would place the heated substance into a coffee-cup calorimeter containing a known mass of water with a known initial temperature.
I would wait for the temperature to equilibrate, then calculate temperature change.
I would use the temperature change of water to determine the amount of energy absorbed.
I would use the amount of energy lost by substance, mass, and temperature change to calculate specific heat.
Scientific Notation Is Basically The Shorthand For Writing Numbers. Scientists Use It When They Want To Write Numbers That Are Very Big Or Very Small.
Alright sorry you're getting the answer hours later, but i can help with this.
so you're looking for specific heat, the equation for it is <span>macaΔTa = - mbcbΔTb with object a and object b. that's mass of a times specific heat of a times final minus initial temperature of a equals -(mass of b times specific heat of b times final minus initial temperature of b)
</span>so putting in your values is, 755g * ca * (75 celsius - 84.5 celsius) = -(50g * cb * (75 celsius - 5 celsius))
well we know the specific heat of water is always 4180J/kg celsius, so put that in for cb
with a bit of simplification to the equation by doing everything on each side first you have, -7172.5 * ca = -14630000
divide both sides by -7172.5 so you can single out ca and you get, ca= 2039.74
add units for specific heat which are J/kg celsius and the specific heat of the material is 2039.74 J/kg celsius