Classify each of the following values by the number of significant figures present in each. 8.314 J/mol-K 6.022x10 23 mol-1 12 mL 1.2 years 7.0 m2 50, cm 0,005 L 40.7 g 670.90 in 0.6258 mg 2.205 lb 3270 ft 90 m 5 significant 4 figuressignificant 3 figuressignificant 2 figuressignificant 1 figures

5 significant figures: 670.90 in.

4 significant figures: 8.314 J/mol-K, 6.022x10 23 mol-1, 0.6258 mg, 2.205 lb and 3270 ft.

3 significant figures: 40.7 g.

2 significant figures: 12 mL, 1.2 years, 7.0 m2, 50, cm and 90 m.

1 significant figure: 0,005 L.

Explanation:

Hello,

In this case, we proceed as follows:

* 8.314 J/mol-K has 4 significant figures since 8, 3, 1 and 4 are significant (nonzero).

* 6.022x10 23 mol-1 has 4 significant figures since 6, 0, 2 and 2 are significant including the zero as it is at the right of the first nonzero digit (6).

* 12 mL has 2 significant figures since 1 and 2 are significant (nonzero).

* 1.2 years has 2 significant figures since 1 and 2 are significant (nonzero).

* 7.0 m2 has 2 significant figures since 1 and 2 are significant (nonzero).

* 50, cm has 2 significant figures since 1 and 2 are significant (nonzero).

* 0,005 L has 1 significant figure since 5 is the only nonzero digit.

* 40.7 g has 3 significant figures since 4, 0 and 7 are significant including the zero as it is at the right of the first nonzero digit (4).

* 670.90 in has 5 significant figures since 6, 7, 0, 9 and 0 are significant including the zeros as they are at the right of the first nonzero digit (6).

* 0.6258 mg has 4 significant figures since the first zero is at the left of the first nonzero digit, therefore, it is not included.

* 2.205 lb has 4 significant figures since 2, 2, 0 and 5 are significant including the zero as it is at the right of the first nonzero digit (2).

* 3270 ft has 4 significant figures since 3, 2, 7 and 0 are significant including the zero as it is at the right of the first nonzero digit (3).

* 90 m has 2 significant figures since 1 and 2 are significant (nonzero).

Regards.

the answer is 1260

you spend 7 hours in school for 1 school day with 5 school days in 1 school week with 36 school weeks in a school year.

7*5*36= 1260 it's all about multiplying

as given the equation is

2al + 6 hcl > 2 alcl3 + 3 h2(g)

as per equation two moles of aluminum reacts with six moles of hcl to give two moles of alcl3 and three moles of hydrogen gas

given: three moles of al are present and 7 moles of hcl

for three moles of aluminum nine moles of hcl will be required

however as 7 moles of hcl are present so the limiting reagent is hcl

hcl is going to decide the number of moles of product formed

when six moles of hcl are present = 3 moles of h2 are formed

when one mole of hcl is present == 3/6 mole sof h2 will be formed

when 7 moles of hcl are present = 3 x 7/ 6 moles of h2 will be formed = 3.5 moles of h2