One student uses a meterstick to measure the thickness of a textbook and obtains 4.3 cm 2 0.1 cm. Other studentsmeasure the thickness with vernier calipers and obtain four different measurements: (a) 4.32 cm i 0.01 cm. (b) 4.31 cm :0.01 cm. (c) 4.24 cm 1- 0.01 cm, and (d) 4.43 cm 2 0.01 cm. Which of these four measurements, if any, agree with thatobtained by the first student?

the first three measurement a, b and c are in good agreement with the measurement obtained by the first student.

Explanation:

One student uses a metre rule to measure the thickness of a textbook and finds it to be

4.3cm +- 0.1 cm. it implies that the thickness lies between 4.2cm  and 4.4cm

a) Another student measures the thickness of the vernier callipers and got

4.32cm +- 0.01cm. it implies that the thickness lies between 4.31cm and 4.33cm and as such the thickness range lies between what was measured by the first student

b) Another student measures the thickness of the vernier callipers and got

4.31cm +- 0.01cm. it implies that the thickness lies between 4.30cm and 4.32cm and as such the thickness range lies between ( 4.2cm  and 4.4cm) that was measured by the first student.

c) Another student measures the thickness of the vernier callipers and got

4.24cm +- 0.01cm. it implies that the thickness lies between 4.23cm and 4.25cm and as such the thickness range lies between ( 4.2cm  and 4.4cm) that was measured by the first student.

d) Another student measures the thickness of the vernier callipers and got

4.43cm +- 0.01cm. it implies that the thickness lies between 4.42cm and 4.44cm and as such the thickness range lies outside ( 4.2cm  and 4.4cm) that was measured by the first student.

Therefore, the first three measurement a, b and c are in good agreement with the measurement obtained by the first student.

u=13

v=30

s/t=u+1\2at

=13+431/2

=13+265.5

=278.5

  1.the balanced   chemical equation is as below

cu +   2 ag(no)₃ → cu(no₃)₂   +2 ag

2.the   much silver   produced   is   50.9 g

step 1: find the   moles of cu

moles =   mass÷   molar mass

  from periodic table the molar mass of cu = 63.5 g/mol

moles is therefore = 15.00 g÷ 63.5 g/mol = 0.236   moles

step 2: use the mole ratio to determine the   moles of ag

from equation above   cu: ag   is 1: 2

therefore the moles of ag   =0.236   moles   x 2/1 =0.472   moles

step 3: find the mass   of ag

mass = moles x molar mass

from periodic table   the molar mass of ag =107.87 g/mol

mass   = 0.472   moles x 107.87 g/mol = 50.9 g

3.   the   much   silver   collected worth is   1.797 oz

  convert   50.9 g   to oz

that is   1 gram = 0.0353 oz

          50.9 grams =?   oz

by cross multiplication

={(50.9 grams x 0.0353 oz) / 1   gram}   =1.797 oz