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vaieri [72.5K]
3 years ago
15

Divide 98.66 grams by 10.3 mL

Chemistry
1 answer:
ivolga24 [154]3 years ago
8 0

9.58 g·mL⁻¹; 13 mL

In division problems, your answer can have no more significant figures than the number with the fewest significant figures.

98.66 g/10.3 mL = 9.578 640 777g·mL⁻¹ (by my calculator)

There are <em>four</em> significant figures in 98.66, but only <em>three</em> in 10.3.

You must round off your answer to <em>three</em> significant figures.

9.578 640 777 → 9.58 (to three significant figures)

∴ 98.66 g/10.3 mL = 9.58 g·mL⁻¹

______________________________________

When adding values, you must round your answer to the <em>same "place"</em> as the measurement with its <em>last significant figure furthest to the left</em>.  

   4.5   mL

+5.66 mL

+3      mL

13.16 mL

The “3” in “3 mL”has its last significant figure furthest to the left, so you round off the number to the “units" place. That's the last column that the measurements share.

13.16 → 13 (rounded to the “units” place)

∴ 4.5 mL + 5.66 mL + 3 mL = 13 mL

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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.
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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

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When temperatures are below freezing, the temperature at which air becomes saturated leading to the formation of frost is the
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Explanation:

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What is the pH of a 3.40 mM of acetic acid, 500 mL in which 50 mL of 1.00 M NaOAc has been added? Ka HOAc is 1.8 x 10-5? What pe
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Answer:

a) pH = 4.213

b) % dis = 2 %

Explanation:

Ch3COONa → CH3COO- + Na+

CH3COOH ↔ CH3COO- + H3O+

∴ Ka = 1.8 E-5 = ([ CH3COO- ] * [ H3O+ ]) / [ CH3COOH ]

mass balance:

⇒ <em>C</em> CH3COOH + <em>C</em> CH3COONa = [ CH3COOH ] + [ CH3COO- ]

<em>∴ C </em>CH3COOH = 3.40 mM = 3.4 mmol/mL * ( mol/1000mmol)*(1000mL/L)

∴ <em>C</em> CH3COONa = 1.00 M = 1.00 mol/L = 1.00 mmol/mL

⇒ [ CH3COOH ] = 4.4 - [ CH3COO- ]

charge balance:

⇒ [ H3O+ ] + [ Na+ ] = [ CH3COO- ] + [ OH- ]....is negligible [ OH-], comes from water

⇒ [ CH3COO- ] = [ H3O+ ] + 1.00

⇒ Ka = (( [ H3O+ ] + 1 )* [ H3O+ ]) / ( 3.4 - [ H3O+])) = 1.8 E-5

⇒ [ H3O+ ]² + [ H3O+ ] = 6.12 E-5 - 1.8 E-5 [ H3O+ ]

⇒ [ H3O+ ]² + [ H3O+ ] - 6.12 E-5 = 0

⇒  [ H3O+ ] = 6.12 E-5 M

⇒ pH = - Log [ H3O+ ] = 4.213

b) (% dis)* mol acid = <em>C</em> CH3COOH = 3.4

∴ mol CH3COOH = 500*3.4 = 1700 mmol = 1.7 mol

⇒ % dis = 3.4 / 1.7 = 2 %

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