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Nitella [24]
3 years ago
11

What is the density of a 4400. g/cm^3gram brick that is 5.00 inches x 3.00 inches x 2.00 inches?

Chemistry
1 answer:
erica [24]3 years ago
7 0
Density 727838382737737383
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Consider the reaction 2 al + Fe2O3 to 2Fe + Al2O3. If 60.0g of Al is reacted with excess Fe2O3, determine the amount (in moles)
olganol [36]

 The  amount  of  Al2O3  in moles=  1.11 moles    while in  grams   = 113.22 grams


    <em><u>calculation</u></em>

     2 Al  + Fe2O3 → 2Fe  + Al2O3

    step  1: find the moles of Al  by  use of <u><em>moles= mass/molar  mass  </em></u>formula

    =  60.0/27= 2.22  moles


    Step 2: use the mole ratio to determine the  moles of Al2O3.

 The  mole ratio  of Al : Al2O3 is  2: 1 therefore the moles of Al2O3= 2.22/2=1.11  moles


Step 3:    finds the mass  of  Al2O3  by us of  <u><em>mass= moles x molar mass</em></u><em> </em>formula.

The molar  mass of Al2O3  =  (2x27)  +( 16 x3) = 102  g/mol

mass is therefore=  102  g/mol  x 1.11= 113.22 grams


             

7 0
2 years ago
Which mixture can be separated based on differences in particle size of the components?
Aleksandr [31]
Heterogenous mixtures can separates due to differences in size of components
7 0
3 years ago
Balanced symbol equation for reaction between sodium carbonate solution and dilute sulphuric acid?
Dmitry [639]

Answer:

The molecular equation for the reaction betweensodium carbonate and sulfuric acid is: 1. Na2CO3(aq)+H2SO4(aq)→Na2SO4(aq)+CO2(g)+H2O(l) N a 2 C O 3 ( a q ) + H 2 S O 4 ( a q ) → N a 2 S O 4 ( a q ) + C O 2 ( g ) + H 2 O ( l ) .

Explanation:

3 0
2 years ago
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Answer: time

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7 0
2 years ago
Consider the reaction: P(s) + 5/2 Cl2(g)PCl5(g) Write the equilibrium constant for this reaction in terms of the equilibrium con
Pani-rosa [81]

Answer: The equilibrium constant for the overall reaction is K_a\times K_b

Explanation:

Equilibrium constant is defined as the ratio of concentration of products to the concentration of reactants each raised to the power their stoichiometric ratios.

a) P(s)+\frac{3}{2}Cl_2(g)\rightarrow PCl_3(g)

K_a=\frac{[PCl_3]}{[Cl_2]^{\frac{3}{2}}}

b) PCl_3(g)+Cl_2(g)\rightarrow PCl_5(g)

K_b=\frac{[PCl_5]}{[Cl_2]\times [PCl_3]}

For overall reaction on adding a and b we get c

c) P(s)+\frac{5}{2}Cl_2(g)\rightarrow PCl_5(g)

K_c=\frac{[PCl_5]}{[Cl_2]^\frac{5}{2}}

K_c=K_a\times K_b=\frac{[PCl_3]}{[Cl_2]^{\frac{3}{2}}}\times \frac{[PCl_5]}{[Cl_2]\times [PCl_3]}

The equilibrium constant for the overall reaction is K_a\times K_b

4 0
3 years ago
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