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Sergio039 [100]
3 years ago
6

You add 4.7 gg iron to 27.10 mL of water and observe that the volume of iron and water together is 27.70 mLmL . Calculate the de

nsity of iron.
Chemistry
1 answer:
sattari [20]3 years ago
5 0

Answer : The density of iron is, 7.8 g/mL

Explanation : Given,

Mass of iron = 4.7 g

Volume of water = 27.10 mL

Volume of water and iron = 27.70 mL

First we have to calculate the volume of iron.

Volume of iron = Volume of water and iron - Volume of water

Volume of iron = 27.70 mL - 27.10 mL

Volume of iron = 0.6 mL

Now we have to calculate the density of iron.

\text{Density of iron}=\frac{\text{Mass of iron}}{\text{Volume of iron}}

Now put all the given values in this expression, we get:

\text{Density of iron}=\frac{4.7g}{0.6mL}

\text{Density of iron}=7.8g/mL

Thus, the density of iron is, 7.8 g/mL

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The correct answer is C
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4 years ago
P O R F A V O R
ale4655 [162]

Answer:

a. 45×10³ kg

b. 1.25 kg

c. 5443200 s

d. 2.69×10⁻⁴ m/s²

e. 2.57×10⁻⁶N

f. 7.48×10⁻³ m /s²

g. 2.45 Pa

h. 10 m/s

Explanation:

The SI units are: kg, m, s, N, K, A, Pa, J and cd

a. 1 g is the mass for 1 cm³. We convert the m³ to cm³

45 m³. 1×10⁶ cm³ / 1 m³ = 45×10⁶ cm³

By the way, 45×10⁶ cm³ = 45×10⁶ g

We convert the g to kg →  45×10⁶ g . 1 kg / 1000 g = 45×10³ kg

b. As 1 g = 1 cm³,  we convert the cm³ to g and then, the g to kg

1250 cm³ = 1250 g → 1250 g . 1kg / 1000 g = 1.25 kg

c. 1 day has 24 hours; 1 hour has 60 minutes; 1 minute has 60 seconds

1 hour has 3600 s. Then 24 h . 3600 s / 1 h = 86400 s

86400 s/d. 63 d = 5443200 s

d. 1 min² = 3600 s²

97 cm / 3600 s² = 0.0269 cm/s²  / 100 = 2.69×10⁻⁴ m/s²

e. 927 g.cm / min² / 3600 s² = 0.2575 g.cm/s² → dyn

We need to convert dyn to N

1 dyn = 10⁻⁵N → 0.2575  dyn . 10⁻⁵N / 1dyn = 2.57×10⁻⁶N

f. 1 m/s² = 12960 km/h²

12960 km/h² . 1 m/s² / 97 km/h² = 7.48×10⁻³ m /s²

g. 2500 g/cm² . 1kg / 1000 g = 2.5 kg/cm²

1 Pa = 1.02kg/cm²

2.5 kg/cm² . 1 Pa / 1.02 kg/cm² = 2.45 Pa          

h. 1 h = 3600 s

36 km / 3600 s = 0.01 km /s → 0.01 km . 1000 m / 1 km = 10

= 10 m/s

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3 years ago
If you observe a liquid turning into gas at 135 degrees Celsius what are you observing?
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4 years ago
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Calculate the pH of 1.00 L of the buffer 0.95 M CH3COONa/0.92 M CH3COOH before and after the addition of the following species.
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Answer: (a) pH = 4.774, (b) pH = 4.811 and (c) pH = 4.681

Explanation: (a) pH of the buffer solution is calculated using Handerson equation:

pH=pKa+log(\frac{base}{acid})

pKa for acetic acid is 4.76. concentration of base and acid are given as 0.95M and 0.92M. Let's plug in the values in the equation and calculate the pH of starting buffer.

pH=4.76+log(\frac{0.95}{0.92})

pH = 4.76 + 0.014

pH = 4.774

(b) When 0.040 mol of NaOH (strong base) are added to the buffer then it reacts with 0.040 mol of acetic acid and form 0.040 mol of sodium acetate.

Original buffer volume is 1.00 L. So, the original moles of sodium acetate will be 0.95 and acetic acid will be 0.92.

moles of acetic acid after addition of NaOH = 0.92 - 0.040 = 0.88

moles of sodium acetate after addition of NaOH = 0.95 + 0.040 = 0.99

Let's again plug in the values in the Handerson equation:

pH=4.76+log(\frac{0.99}{0.88})

pH = 4.76 + 0.051

pH = 4.811

(c) When 0.100 mol of HCl are added then it reacts with exactly 0.100 moles of sodium acetate(base) and form 0.100 moles of acetic acid(acid).

so, new moles of acetic acid = 0.92 + 0.100 = 1.02

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Let's plug in the values in the equation:

pH=4.76+log(\frac{0.85}{1.02})

pH = 4.76 - 0.079

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5 0
4 years ago
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V₂ = desired volume of final solution = 0.215g  = 0.215ml
Now putting the values in the formula;
0.098 x 20.01 = C₂ x 0.215
C₂ = 0.215 / 1.96098 = 0.109m = 0.11m
Thus, the answer is 0.11m.
4 0
3 years ago
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