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oee [108]
3 years ago
10

5. (10 Points) Write a synthesis reaction where one of the reactants is aluminum.

Chemistry
1 answer:
fgiga [73]3 years ago
7 0

Answer:

A reaction in which two or more reactants combine to form only one product is called a synthesis reaction.

Explanation:

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Which of these correctly defines the pOH of a solution? a.) pOH=log[H3O+] b.) pOH=-log[H30+] c.) pOH=log[OH-] d.) pOH=-log[OH-]
denis23 [38]
The answer is d) pOH = - lg[OH-]. Like pH expression: pH = -lg[H+] or pH = - lg[H3O+]. The pOH is a definition similar to the pH. So the expression is also similar. We only need to change H+ to OH-.
3 0
3 years ago
What is the empirical formula of a carbon-oxygen compound, given that a 95.2 g sample of the compound contains 40.8 g of carbon
algol [13]

Answer:

95.2 - 40.8 = 54.4 g of oxygen

number of moles = mass (g)/ Mr

no. of moles of carbon = 40.8/12 = 3.4

no. of moles of oxygen = 3.4

divide both by smallest value which is 3.4 and you’ll get 1 mole of carbon and 1 mole of oxygen therefore the empirical formula is CO

Explanation:

hope this helps :)

4 0
2 years ago
Calculate the mass of water produced when 1.92 g of butane reacts with excess oxygen.
never [62]
C4h10+6.5o2=4co2+5h2o
moles of butane=1.92/58=0.0331 moles
moles of water=0.1655 moles\
as the butane and water has 1 is to 5 molar ratio
0.1655=mass/18
mass=2.98 g
mass of water produced = 2.98 g
4 0
3 years ago
How many moles of a solute is contained in 500mL of a 2.5 M solution?
Pavlova-9 [17]

Answer:

100M

Explanation:

8 0
3 years ago
Calculate the change in pH when 71.0 mL of a 0.760 M solution of NaOH is added to 1.00 L of a solution that is 1.0O M in sodium
Eddi Din [679]

Explanation:

It is known that pK_{a} value of acetic acid is 4.74. And, relation between pH and pK_{a} is as follows.

                    pH = pK_{a} + log \frac{[CH_{3}COOH]}{[CH_{3}COONa]}

                          = 4.74 + log \frac{1.00}{1.00}

So, number of moles of NaOH = Volume × Molarity

                                                   = 71.0 ml × 0.760 M

                                                    = 0.05396 mol

Also, moles of  CH_{3}COOH = moles of CH_{3}COONa

                                          = Molarity × Volume

                                          = 1.00 M × 1.00 L

                                          = 1.00 mol

Hence, addition of sodium acetate in NaOH will lead to the formation of acetic acid as follows.

            CH_{3}COONa + NaOH \rightarrow CH_{3}COOH

Initial :    1.00 mol                                  1.00 mol

NaoH addition:               0.05396 mol

Equilibrium : (1 - 0.05396 mol)    0           (1.00 + 0.05396 mol)

                    = 0.94604 mol                       = 1.05396 mol

As, pH = pK_{a} + log \frac{[CH_{3}COONa]}{[CH_{3}COOH]}

               = 4.74 +  log \frac{0.94604}{1.05396}

               = 4.69

Therefore, change in pH will be calculated as follows.

                         pH = 4.74 - 4.69

                               = 0.05

Thus, we can conclude that change in pH of the given solution is 0.05.

8 0
3 years ago
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