Answer:
8 GRAMS OF OXYGEN WILL BE REQUIRED TO PRODUCE 9 GRAMS OF WATER MOLECULES.
Explanation:
The equation for the reaction is;
2 H + O2 ----> 2 H2O
(H = 1, O = 16)
From the reaction, it can be observed that 1 mole of O2 reacts to form 2 moles of H2O
At STP, using the molar masses of the elememts and compound in the reaction, 32 g of O2 will reacts to form 36 g (18 * 2 g) of H2O
32 g of O2 = 36 g OF H2O
if 9 grams of H2O is produced, the mass of oxygen required to produce it will be:
( 32 * 9 / 36 ) g of O2
= (288 / 36) g
= 8 g of O2
So, 8 grams of Oxygen will be required to produce 9 grams of water.
The % composition when 10g of magnesium combine with 4g of nitrogen is 71.43% magnesium and 28.57 % nitrogen
calculation
% composition = mass of an element / total mass x100
mass of magnesium = 10 g
mass of nitrogen = 4g
calculate the total mass used
= 10g of Magnesium + 4 g of nitrogen = 14 grams
% composition for magnesium is therefore = 10/14 x100 = 71.43 %
% composition for nitrogen is therefore = 4 /14 x100 = 28.57 %
A drought is a period of drier-than-normal conditions that results in water-related problems.<span> When rainfall is less than normal for several weeks, months, or years, the flow of streams and rivers declines, water levels in lakes and reservoirs fall, and the depth to water in wells increases.</span>
pretty sure its B thank me later
Answer:
The answer to your question is: Initial temperature of copper = 67.1°C
Explanation:
Data
mass Copper = 248 g
volume Water = 390 ml
T1 water = 22.6°C
T2 = 39.9°C
T1 copper = ?
Specific heat water = 1 cal/g°C
Specific heat copper = 0.092 cal/g°C
Formula copper water
Heat is negative for copper because it releases heat
- mCp(T2 - T1) = mCp(T2 - T1)
- (248)(39.9 - T1) = 390 (1)((39.9 - 22.6) Substitution
-9895.2 + 248T1 = 390(17.3) Simplification
-9895.2 + 248T1 = 6747
248 T1 = 6747 + 9895.2
248 T1 = 16642.2
T1 = 16642.2 / 248
T1 = 67.1 °C Result
