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3 years ago

Significant digits are the number of digits that reflect the precision of a measurement or number.

Chemistry

1 answer:

Delvig [45]3 years ago

5 0

Answer:

True

Explanation:

Significant digits are numbers that helps to present the precision of measurements calculations.

Numbers that do not contribute to the precision of a reading should not be counted as significant.

There are rules of assigning significant numbers:

Leading or trailing zeros are insignificant and should only be counted as a place holder.
All non-zero digits are significant
Zeroes between non-zero digits are significant.
Leading zeros in a decimal are significant before the number.
All the numbers in a scientific notation are significant.

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A 0.15 M NaOH solution has a volume of 0.125 L but is then diluted to 0.15 L. What is the concentration of the new solution

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Answer:

.125 M

Explanation:

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now dilute to 150 cc (by adding 25 cc)

.01875M / (150/1000) = .125M

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2 years ago

What is the difference between traffic lights with red arrows and those with solid red lights?

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3 years ago

500 mL of a solution contains 1000 mg of CaCl2. Molecular weight of CaCl2 is 110 g/mol. Specific gravity of the solution is 0. C

dangina [55]

Answer:

a) 0,2% w/v

b) r=500

c) 0,0182 M

d) 0,0145 m

e) 0,0137 equivalents

Explanation:

a) % w/v means mass of solute in grams per 100 mililiter of solution. Thus:

%w/v= $\frac{1,000 g CaCl2}{500mL}$ ×100 = 0,2%w/v

b) Ratio strength is a way to express concentration. For w/v is in 1g of solute r mililiters of solution have. Thus, r = 500 because we have in the first 1 g of CaCl₂ in 500 mL of solution.

c) Molarity is moles of solute per liter of solution, thus:

1,000 g of CaCl₂ × $\frac{1mol}{110g}$ = 9,09×10⁻³ moles of CaCl₂

500 mL of solution × $\frac{1L}{1000mL}$ = 0,500 L of solution

M = $\frac{9,09x10^{-3} moles }{0,500 L}$ = 0,0182 M

d) Molality is moles of solute per kg of solution.

Specific gravity is the ratio between density of the solution and density of a reference substance (Usually water). With a specific gravity of 0,8:

kg of solution = 0,500 L of solution × $\frac{0,8 kg}{1L}$ = 0,625 kg of solution

m = $\frac{9,09x10^{-3}moles }{0,625 kg}$ = 0,0145 m

e) In a salt, equivalents are the number of moles ables to replace one mole of charge. In CaCl₂ is ¹/₂ because with ¹/₂ moles of CaCl₂ it is possible to replace 1 mole of charges. Thus, in 1,5 L there are:

1,5 L × $\frac{0,0182 CaCl2 moles}{1L}$ × $\frac{1equivalent}{2 moles}$ = 0,0137 equivalents

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