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FromTheMoon [43]
3 years ago
6

Please help. im freaking out rn. i have like 40 missing assignments please

Chemistry
1 answer:
Katarina [22]3 years ago
5 0

Answer:

I'm pretty sure its the one that says very little at the beginning but if I get it wrong I'm sorry

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How many kj of heat are needed to completely vaporize 1.30 moles of h2o? the heat of vaporization for water at the boiling point
Cerrena [4.2K]
The Kj of heat that  are needed to completely vaporize 1.30  moles of H2O  if  the heat of vaporization  for water is 40.6 Kj/mole  is calculated as  below

Q(heat) = moles x heat  of vaporization)

=1.30 mol  x40.6 kj/mol= 52.78 Kj  is needed
6 0
3 years ago
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A student is trying to identify a clear liquid that he finds in a beaker in the classroom. He thinks the liquid might be either
kumpel [21]

Simply mulitply the volume by the density. As we shall see, this is dimensionally consistent.

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3 0
3 years ago
2. Graph the following information in a LINE graph. Label and number the x and y-axis approHDILI.
Harman [31]
A) the independent variable is the variable that doesn’t rely on another variable. For this question the independent would be the days.
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4 0
3 years ago
The _____ heats the earth. this creates convection currents in the _____ and ________. the heating of these areas create ______
andrew-mc [135]

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sun, air, water, wind and ocean current

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As sun is the major source of energy on earth, the earth gets heated due to solar energy, this creates convection currents in the Air and Water, the heating of these areas create wind and ocean currents due to difference in temperature and pressure.

5 0
3 years ago
How many grams of Co2 are produced from 12 g of CH4 and 133 g of O2
Andru [333]

Answer:

m_{CO_2}=33gCO_2

Explanation:

Hello,

In this case the undergoing chemical reaction is:

CH_4+2O_2\rightarrow CO_2+2H_2O

Next we identify the limiting reactant with the given amounts of methane and oxygen by computing the available moles of methane and the moles of methane consumed by the 133 g of oxygen:

n_{CH_4}^{available}=12g*\frac{1mol}{16g} =0.75molCH_4\\n_{CH_4}^{consumed\ by\ O_2}=133gO_2*\frac{1molO_2}{32gO_2}*\frac{1molCH_4}{2molO_2}=2.08molCH_4

In such a way, the limiting reactant is methane with 0.75 moles which produce the following mass of carbon dioxide:

m_{CO_2}=0.75molCH_4*\frac{1molCO_2}{1molCH_4} *\frac{44gCO_2}{1molCO_2} \\\\m_{CO_2}=33gCO_2

Best regards.

7 0
3 years ago
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