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Harrizon [31]
3 years ago
6

Please help 8th grade science i need #2 please!

Chemistry
2 answers:
aivan3 [116]3 years ago
6 0
#2 onshahebsisbunwubsydtebwownsddhiiw
elena-14-01-66 [18.8K]3 years ago
3 0
Im not sure what this is give a better understanding pls
You might be interested in
Why are silver, lead, and mercury so special? (why do they always precipitate out and never dissolve) PLEASE BE DETAILED BUT SIM
lara [203]
Silver and lead are special elements, where silver is insoluble in all halogen anions (like AgCl, AgI, AgBr)
and lead is insoluble in sulphates and halogen anions ( PbSO4, PbCl2, etc.)
Mercury is special because it is the only metal that is a liquid at room temperature
hope this helps!!
8 0
3 years ago
The combustion of gasoline produces carbon dioxide and water. Assume gasoline to be pure octane (C8H18) and calculate the mass (
Mariulka [41]

Answer:

3.09kg

Explanation:

First, let us write a balanced equation for the reaction. This is illustrated below:

2C8H18 + 25O2 —> 16CO2 + 18H2O

Molar Mass of C8H18 = (12x8) + (18x1) = 96 + 18 = 114g/mol

Mass of C8H18 from the balanced equation = 2 x 114 = 228g

Converting 228g of C8H18 to kg, we obtained:

228/1000 = 0.228kg

Molar Mass of CO2 = 12 + (2x16) = 12 + 32 = 44g/mol

Mass of CO2 from the balanced equation = 16 x 44 = 704g

Converting 704g of CO2 to kg, we obtained:

704/1000 = 0.704kg

From the equation,

0.228kg of C8H18 produced 0.704kg of CO2.

Therefore, 1kg of C8H18 will produce = 0.704/0.228 = 3.09kg of CO2

6 0
3 years ago
URGENT!!!What volume of a 0.88 M solution can be made using 130. Grams of FeCI^2?
Likurg_2 [28]

Answer:

1.16L can be made

Explanation:

Molarity = Mol / Volume

Volume = Mol / Molarity

Let's determine the moles of salt, with that mass:

130 g FeCl₂ . 1mol / 126.75 g = 1.02 moles of FeCl₂

Volume = 1.02 mol / 0.88 mol/L → 1.16L

5 0
2 years ago
How many daltons is in an ion with 15 protons, 16 neutrons and 17 electrons?
Vera_Pavlovna [14]
31  
A dalton is the same as an atomic mass unit. And an atomic mass unit is approximately the mass of a nucleon (proton or neutron) such that the mass is 1 g/mol. So in this problem you have 15 protons and 16 neutrons, so the number of daltons is 15 + 16 = 31.
8 0
3 years ago
A buffer contains 0.19 mol of propionic acid (C2H5COOH) and 0.26 mol of sodium propionate (C2H5COONa) in 1.20 L. You may want to
Snezhnost [94]

Explanation:

It is known that pK_{a} of propionic acid = 4.87

And, initial concentration of  propionic acid = \frac{0.19}{1.20}

                                                                       = 0.158 M

Concentration of sodium propionate = \frac{0.26}{1.20}[/tex]

                                                             = 0.216 M

Now, in the given situation only propionic acid and sodium propionate are present .

Hence,      pH = pK_{a} + log(\frac{[salt]}{[acid]})

                       = 4.87 + log \frac{0.216}{0.158}

                        = 4.87 + log (1.36)

                        = 5.00

  • Therefore, when 0.02 mol NaOH is added  then,

     Moles of propionic acid = 0.19 - 0.02

                                              = 0.17 mol

Hence, concentration of propionic acid = \frac{0.17}{1.20 L}

                                                                 = 0.14 M

and,      moles of sodium propionic acid = (0.26 + 0.02) mol

                                                                  = 0.28 mol

Hence, concentration of sodium propionic acid will be calculated as follows.

                        \frac{0.28 mol}{1.20 L}

                           = 0.23 M

Therefore, calculate the pH upon addition of 0.02 mol of NaOH as follows.

             pH = pK_{a} + log(\frac{[salt]}{[acid]})

                       = 4.87 + log \frac{0.23}{0.14}

                        = 4.87 + log (1.64)

                        = 5.08

Hence, the pH of the buffer after the addition of 0.02 mol of NaOH is 5.08.

  • Therefore, when 0.02 mol HI is added  then,

     Moles of propionic acid = 0.19 + 0.02

                                              = 0.21 mol

Hence, concentration of propionic acid = \frac{0.21}{1.20 L}

                                                                 = 0.175 M

and,      moles of sodium propionic acid = (0.26 - 0.02) mol

                                                                  = 0.24 mol

Hence, concentration of sodium propionic acid will be calculated as follows.

                        \frac{0.24 mol}{1.20 L}

                           = 0.2 M

Therefore, calculate pH upon addition of 0.02 mol of HI as follows.

             pH = pK_{a} + log(\frac{[salt]}{[acid]})

                       = 4.87 + log \frac{0.2}{0.175}

                        = 4.87 + log (0.114)

                        = 4.98

Hence, the pH of the buffer after the addition of 0.02 mol of HI is 4.98.

7 0
3 years ago
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