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lions [1.4K]
3 years ago
14

Round off each of the following numbers to two significant figures:

Chemistry
2 answers:
Kay [80]3 years ago
7 0
A) 5.2 x 10^2
B) 86.
C) 6.4 x 10^3
D) 5.0
E) 22.
F) 0.89
strojnjashka [21]3 years ago
7 0

Answer:

(a) 5.1 × 10²

(b) 86

(c) 6.4 × 10³

(d) 5.0

(e) 22

(f) 0.89

Explanation:

To round off a number, we consider the number that is to its right. If that number is 5 or higher, we increase the number of interest. If the number to the right is 4 or lower, the number of interest remains the same. When taking less significant figures, we might need to use scientific notation for large or small numbers.

<em>Round off each of the following numbers to </em><em>two</em><em> significant figures: </em>

<em>(a) 517</em>   → 5.1 × 10²

<em>(b) 86.3</em>  → 86

<em>(c) 6.382 × 10³ </em> → 6.4 × 10³

<em>(d) 5.0008</em>  → 5.0

<em>(e) 22.497</em>  → 22

<em>(f) 0.885</em> → 0.89

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Answer:

18.0 Ampere is the size of electric current that must flow.

Explanation:

Moles of electron , n = 550 mmol = 0.550 mol

1 mmol = 0.001 mol

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Charge on N electrons : Q

Q = N\times 1.602\times 10^{-19} C

Duration of time charge allowed to pass = T = 49.0 min = 49.0 × 60 seconds

1 min = 60 seconds

Size of current : I

I=\frac{Q}{T}=\frac{N\times 1.602\times 10^{-19} C}{49.0\times 60 seconds}

=\frac{n\times N_A\times 1.602\times 10^{-19} C}{49.0\times 60 seconds}

I=\frac{0.550 mol\times 6.022\times 10^{23} mol^{-1}\times 1.602\times 10^{-19} C}{49.0\times 60 seconds}=18.047 A\approx 18.0 A

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determine the freezing point depression of a solution that contains 30.7 g glycerin (c3h8o3, molar mass
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The freezing point depression of a solution containing 30.7 g of glycerin  is  calculated as -1.65°C

Equating :

It is given that,

Given mass of glycerin is = 30.7 grams (Solute)

Volume of water = 376 mL

K_{f}or molar -freezing-depression point is = 1.86°C/m

Molar mass of glycerin = 92.09 g/mole

Now, to work out the value, the mass of water should be known. Thus, to calculate, the formula used will be:

Mass = Density X Volume

Mass = 1.0 g/mL X 376 mL

Mass = 376 g or 0.376 Kg

Using the formula of melting point depression, the equation becomes:

             ΔT_{f} = i ×K_{f} ×m

T⁰-T_{s}  = i *K_{f} *\frac{mass of glycerin}{molar mass of glycerin * mass of water     in     kg}

in which,

ΔT_{f} = change in freezing point

ΔT_{s} = freezing point of solution that has to be find

ΔT° = freezing point of water ()

Since, glycerin is a non-electrolyte, the Van't Hoff factor will be 1.

Substituting the values in the above equation:

0⁰C₋T_{s} = 1 ×1.86°C/m ×\frac{30.7}{92.09g/mol * 0.376kg}

T_{s} = -1.65°C

Thus, the freezing point depression of a solution is  -1.65°C

<h2 />

Freezing point depression

Freezing point depression is a colligative property observed in solutions that results from the introduction of solute molecules to a solvent. The freezing points of solutions are all less than that of the pure solvent and is directly proportional to the molality of the solute

Is melting point elevation or depression?

Boiling point elevation is that the raising of a solvent's boiling point due to the addition of a solute. Similarly, melting point depression is the lowering of a solvent's freezing point due to the addition of a solute. In fact, because the boiling point of a solvent increases, its melting point decreases

Learn more about freezing point depression :

brainly.com/question/26525184

#SPJ4

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