This is a lab assignment, you need to do something physically to find your answers
Answer:
1. Delete lion, bison, reindeer, and giraffe
2. Delete cactus, oak, evergreen,
Explanation:
1. Lions live in the African savanna, bison live in open fields with lots of grass, reindeer live in Antarctica and other very cold places, and giraffes also live in the areas of the African savanna where there are trees.
2. Cacti live in the desert, oaks and evergreens live in open grassy areas.
Hope this helps
Enzymes affect the rate of the reaction in both the forward and reverse directions; the reaction proceeds faster because less energy is required for molecules to react when they collide. Thus, the rate constant (k) increases. Figure 3: Lowering the Activation Energy of a Reaction by a Catalyst.
<em>Best of luck,</em>
<em>-Squeak</em>
Answer:
When chlorine (as a gas or dissolved in water) is added to sodium bromide solution, the chlorine takes the place of the bromine. Because chlorine is more reactive than bromine, it displaces bromine from sodium bromide. The solution turns brown. ... The chlorine has gone to form sodium chloride.
Answer:
14.533 grams of solid precipitate of mercury(II) dichromate will form.
Explanation:
Moles of mercury(II) acetate = 
Moles of sodium dichromate = 
According to reaction , 1 mole of sodium dichromate reacts with 1 mole of mercury(II) acetate , then 0.045906 moles of sodium dichromate will recat with :
of mercury(II) acetate
This means that mercury(II) acetate is present in an excess amount and sodium dichromate is present in limiting amount.So, amount of precipitate will depend upon moles of sodium dichromate.
According to reaction , 1 mole of sodium dichromate gives 1 mole of mercury(II) dichromate , then 0.045906 moles of sodium dichromate will give :
of mercury(II) dichromate
Mass of 0.045906 moles of mercury(II) dichromate:
0.045906 mol × 316.59 g/mol = 14.533 g
14.533 grams of solid precipitate of mercury(II) dichromate will form.
