How many atoms of aluminum (Al) are in 2.34 mol of aluminum?

 " " atoms of aluminum (Al) .

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Explanation:

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Using a technique called "dimensional analysis" ; as follows:

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→  "  

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The units "mol Al" ["moles of aluminum"] cancel out; and we have:

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→  " atoms of Al [aluminum].

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Note that by definition: 1 (one) "mole" (abbreviated "mol") of anything consists of " " units " of that particular thing.

 This number:  6.022*10^{23} —is known as:  "Avogadro's number".

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→  [ 2.34 * 6.022 * 10^23] ;

     = 1.409148 * 10^24 ;

                → Round to 3 (Three) significant figures;

                →  since: "2.34" has 3 (Three) significant figures;

                  →  to get: "  atoms of aluminum (Al).

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Hope this is helpful to you!  Wishing you the best!

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answer: c. the experiment shows that the red substance experienced a chemical change.

explanation: chemical change is the change in which chemical composition of the substance changes. physical change is the change in which there is no change in chemical composition.

chemical changes are accompanied by following changes:

a) precipitate formation

b) energy is released or absorbed

c) formation of bubbles

d) change in color

as it is given that the red color has disappeared and there is production of gas, it means the red substance has undergone a chemical change.

e

when 25.0 ml of 0.700 mol/l naoh was mixed in a calorimeter with 25.0 ml of 0.700 mol/l hcl, both initially at 20.0 °c, the temperature increased to 22.1 °c. the heat capacity of the calorimeter is 279 j/°c.

the equation for the reaction is:

naoh + hcl → nacl + h₂o

moles of hcl = 0.0250 l hcl ×

0.700

mol hcl

1 l hcl = 0.0175 mol hcl

volume of solution = (25.0 + 25.0) ml = 50.0 ml

mass of solution = 50.0 ml soln ×

1.00

ml soln

= 50.0 g soln

δt=t2-t1

= (22.1 – 20.0) °c = 2.1 °c

the heats involved are

heat from neutralization + heat to warm solution + heat to warm calorimeter = 0

q1+q2+q3=0

nδh+mcδt+cδt=0

0.0175 mol ×

δh

+ 50.0 g × 4.184 j·g⁻¹°c⁻¹ × 2.1 °c + 279 j°c⁻¹ × 2.1 °c = 0

0.0175 mol ×

δh

+ 439.32 j + 585.9 j = 0

0.0175 mol ×

δh

= -1025.22 j

δh=1025.22j0.0175 mol= -58 600 j/mol = -58.6 kj/mol