A student is determining the density of an unknown metal with a mass of 32.56 g. The student partially fills a graduated cylinder with water and measures the volume of the water by itself as 14.78 mL. The student then adds the metal to the water and measures the new volume as 20.44 mL. What is the identity of the metal

The metal is Tin (Sn)

Explanation:

The following data were obtained from the question:

Mass of metal = 32.56 g

Volume of water = 14.78 mL

Volume of water + metal = 20.44 mL

Next, we shall determine the volume of the metal. This is illustrated below:

Volume of water = 14.78 mL

Volume of water + metal = 20.44 mL

Volume of metal =..?

Volume of metal = (Volume of water + metal) – (Volume of water )

Volume of metal = 20.44 – 14.78

Volume of metal = 5.66mL

Next, we shall determine the density of the metal.

Density of a substance is defined as the mass of the substance per unit volume of the substance. Mathematically, the density of a substance is expressed as:

Density = Mass / volume

With the above formula, we can obtain the density of the metal as follow:

Mass of metal = 32.56 g

Volume of metal = 5.66mL

Density =.?

Density = Mass /volume

Density = 32.56g/5.66mL

Density of the metal = 5.75g/mL

Comparing the density of the metal with standard density of elements, the metal is Tin (Sn)

1. the question asks the most appropriate, which seems to rule out "all of the above." your choice appears to be the correct one.

2. the protocol described is good for identifying correlation, but not cause and effect. for cause/effect, you need to devise a protocol that will assign the treatment (or not) to groups that have essentially the same "demographics" with respect to all the other variables. your choice regarding control group seems appropriate. random assignment of treatment (choice d) may be appropriate also. (i think you can argue that b and d are just different descriptions of the same method.)

3. d will randomize the birds, but may not guarantee there are 10 in each group. e will do both, so seems the better choice.

comment on question 2

based on the problem statement, it seems clear that confident, attentive drivers are more likely to use daytime running lights. so, the protocol might need to be devised in such a way that the lights would be turned on or off without the driver's knowledge. (an attentive driver will figure it out.) assigning a confident, attentive driver to a group required to drive without running lights might mess with his/her confidence and attentiveness. you don't want your experiment to do that. (nor do you want the experiment to subject a driver to greater risk of accident as a consequence of driving without running lights.) experiments involving human safety are always problematic.

answer:

the answer is 6 valence electrons.

explanation:

valence is the # of electrons on the outer most shell of an atom.

oxygen has 6.

although, the valency of oxygen is 2, because it only needs to make 2 covalent bonds, or gain 2 electrons, to achieve the stable octet configuration of a noble gas.

that's it!

if you need further assistance, let me know: )

~ trueboss72