Answer:
Igneous rock, or magma-tic rock, is one of the three main rock types, the others being sedimentary and metamorphic. Igneous rock is formed through the cooling and solidification of magma or lava. The magma can be derived from partial melts of existing rocks in either a planet's mantle or crust
Explanation:
b. False
Explanation:
Burning of alcohol produces different products as it depends on the type and nature of alcohol which is being burnt.
Some alcohols upon burning produce toxic gases or products whereas some alcohols on burning does not produce any toxic substances.
For example, when ethanol is burnt then it reacts with oxygen and results in the formation of water and carbon dioxide.
Thus, we can conclude that the statement burning alcohol for fuel is a very efficient process is a false statement.
To dissolve a 3.94g copper penny, we would need approximately 331 milliliters of 1.50 M HNO3, taking into consideration the stoichiometry of the copper-nitric acid reaction.
Stoichiometry is required to solve this chemistry problem. The balanced equation for the reaction between copper and nitric acid is: 3Cu(s) + 8HNO3(aq) → 2NO(g) + 3Cu(NO3)2(aq) + 4H2O(l)
From this, we can conclude that every mole of copper reacts with eight moles of nitric acid. The molecular weight of copper is approximately 63.546 g/mole, so the moles of copper in an old penny can be calculated by mass (3.94 g) divided by the molar mass (63.546 g/mol).
This calculation shows that there are approximately 0.062 moles of copper in a penny. Since we determined that eight moles of nitric acid are required to react with one mole of copper, eight times the moles of copper gives the moles of nitric acid required, which is 0.496 moles.
Since molarity (M) = moles/liters, we can find out the volume of 1.50 M HNO3 required by rearranging the formula to be Liters= moles/Molarity. So, 0.496 moles divided by 1.50 M gives roughly 0.331 liters, which can be converted to approximately 331 milliliters.
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Control Group:
In the experiment, the Experimental Group represents ducks in cold water, while the Control Group represents ducks in room temperature water. The scientist is testing whether water temperature affects duck swimming speed.
In the experiment described, the scientist is exploring whether the temperature of the water affects the speed of swimming of ducks. The Experimental Group in this scenario represents the ducks that are swimming in cold water. The scientist expects that these ducks will swim faster, based on the original hypothesis. On the other hand, the Control Group represents the ducks that are swimming in room temperature water. This group serves a basis for comparison to help the scientist determine what changes (if any) occur when the variable (water temperature) is altered.
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