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RideAnS [48]
2 years ago
10

How many grams of carbon dioxide are produced when 16 g of methane and 48 g of oxygen gas combust

Chemistry
1 answer:
larisa [96]2 years ago
6 0
I believe the correct answer is 11 g
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Mwehehe poide bang pakiss?
Contact [7]

Answer:

what confused

Explanation:

6 0
3 years ago
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A 24.00 mL sample of a solution of Pb(ClO3)2 was diluted with water to 52.00 mL. A 17.00 mL sample of the dilute solution was fo
lara [203]

Answer:

0.238 M

Explanation:

A 17.00 mL sample of the dilute solution was found to contain 0.220 M ClO₃⁻(aq). The concentration is an intensive property, so the concentration in the 52.00 mL is also 0.220 M ClO₃⁻(aq). We can find the initial concentration of ClO₃⁻ using the dilution rule.

C₁.V₁ = C₂.V₂

C₁ × 24.00 mL = 0.220 M × 52.00 mL

C₁ = 0.477 M

The concentration of Pb(ClO₃)₂ is:

\frac{0.477molClO_{3}^{-} }{L} \times \frac{1molPb(ClO_{3})_{2}}{2molClO_{3}^{-}} =0.238M

6 0
3 years ago
How old is the earth in millions of years
Paul [167]
A million years old
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6 0
3 years ago
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A 248-g piece of copper is dropped into 390 mL of water at 22.6 °C. The final temperature of the water was measured as 39.9 °C.
max2010maxim [7]

Answer:

The answer to your question is: Initial temperature of copper = 67.1°C

Explanation:

Data

mass Copper = 248 g

volume Water = 390 ml

T1 water = 22.6°C

T2           = 39.9°C

T1 copper = ?

Specific heat water = 1 cal/g°C

Specific heat copper = 0.092 cal/g°C

Formula       copper             water

Heat is negative for copper because it releases heat

                  - mCp(T2 - T1) = mCp(T2 - T1)                  

                  - (248)(39.9 - T1) = 390 (1)((39.9 - 22.6)           Substitution

                 -9895.2 + 248T1 = 390(17.3)                             Simplification

                 -9895.2 + 248T1 = 6747

                 248 T1 = 6747 + 9895.2

                 248 T1 = 16642.2

                 T1 = 16642.2 / 248

                 T1 = 67.1 °C                                                         Result

6 0
3 years ago
If 12.5 grams of strontium hydroxide is reacted with 150 mL of 3.5 M carbonic acid, identify the limiting reactant.
vesna_86 [32]

Answer:

Sr(OH)2

Explanation:

We'll begin by calculating the number of mole of carbonic acid in 150mL of 3.5 M carbonic acid solution. This is illustrated below:

Molarity = 3.5M

Volume = 150mL = 150/1000 = 0.15L

Mole of carbonic acid, H2CO3 =..?

Mole = Molarity x Volume

Mole of carbonic acid, H2CO3 = 3.5 x 0.15 = 0.525 mole.

Next, we shall convert 0.525 mole of carbonic acid, H2CO3 to grams.

Mole of H2CO3 = 0.525 mole

Molar mass of H2CO3 = (2x1) + 12 + (16x3) = 62g/mol.

Mass of H2CO3 =..?

Mass = mole x molar mass

Mass of H2CO3 = 0.525 x 62 = 32.55g

Next, we shall write the balanced equation for the reaction. This is given below:

Sr(OH)2 + H2CO3 → SrCO3 + 2H2O

Next, we shall determine the mass of Sr(OH)2 and H2CO3 that reacted from the balanced equation. This is illustrated below:

Molar mass of Sr(OH)2 = 88 + 2(16 + 1) = 88 + 2(17) = 122g/mol

Mass of Sr(OH)2 from the balanced equation = 1 x 122 = 122g

Molar mass of H2CO3 = (2x1) + 12 + (16x3) = 62g/mol.

Mass of H2CO3 from the balanced equation = 1 x 62 = 62g.

From the balanced equation above, 122g of Sr(OH)2 reacted with 62g of H2CO3.

Finally, we shall determine the limiting reactant as follow:

From the balanced equation above, 122g of Sr(OH)2 reacted with 62g of H2CO3.

Therefore, 12.5g of Sr(OH)2 will react with = (12.5 x 62)/122 = 6.35g.

We can see evidently from the calculations made above that it will take 6.35g out 32.55g of H2CO3 to react with 12.5g of Sr(OH)2. Therefore, Sr(OH)2 is the limiting reactant and H2CO3 is the excess reactant

5 0
3 years ago
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