25
Explanation:
The coefficient 5 applis to each element in the molecule, so there are 5 times 1 H atom, 5 times 1 N atom, and ,since O has a 3 subscript, there are 5 times 3 O atoms. Therefore there are 25 atoms.
Answer:
The reaction can produce 287 grams of iron(II) carbonate
Explanation:
To solve this question we must find the moles of iron(II) chloride that react. Using the chemical equation we can find the moles of iron(II) carbonate and its mass -Molar mass FeCO3: 115.854g/mol-
<em>Moles FeCl2:</em>
1.24L * (2.00mol / L) = 2.48 moles FeCl2
As 1 mol FeCl2 produce 1 mol FeCO3, the moles of FeCO3 = 2.48 moles
<em>Mass FeCO3:</em>
2.48mol * (115.854g / mol) =
<h3>The reaction can produce 287 grams of iron(II) carbonate</h3>
Answer:
Methane
Explanation:
The gas that you could keep in an outdoor storage tank in winter in Alaska is Methane.
The reason is the extreme low temperature during the winter. The boiling point of butane is 44 ºF ( -1ºC) and that of propane is a higher -43.6 º F but still within the range of average minimum winter temperature in Alaska (-50 ªF). Therefore we will have condensation in the tanks and not enough gas pressure.
Methane having a boling point of -259 ºF will not condense at the low wintertime temperatures in Alaska.
Answer:
A scientific problem is something you don't understand but you can do an experiment to help you understand. Scientific problems are usually based on observation of scientific phenomena. Here is some advice to help you identify a scientific problem you can address by designing your own experiment.
Explanation:
Example of a scientific problem are Clean Drinking Water.
Answer:
The mass of ice required to melt to lower the temperature of 353 mL of water from 26 ⁰C to 6 ⁰C is 85.4197 kg
Explanation:
Heat gain by ice = Heat lost by water
Thus,
Heat of fusion + 
Where, negative sign signifies heat loss
Or,
Heat of fusion + 
Heat of fusion = 334 J/g
Heat of fusion of ice with mass x = 334x J/g
For ice:
Mass = x g
Initial temperature = 0 °C
Final temperature = 6 °C
Specific heat of ice = 1.996 J/g°C
For water:
Volume = 353 mL
Density of water = 1.0 g/mL
So, mass of water = 353 g
Initial temperature = 26 °C
Final temperature = 6 °C
Specific heat of water = 4.186 J/g°C
So,
345.976x = 29553.16
x = 85.4197 kg
Thus,
<u>The mass of ice required to melt to lower the temperature of 353 mL of water from 26 ⁰C to 6 ⁰C is 85.4197 kg</u>
