krit.club logo

Where are we now and where might we be going - Conversion of different units

Grade 7IB

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

•

The International System of Units (SI) is the standard metric system used globally in science to ensure consistency. Base units include the meter (mm) for length, kilogram (kgkg) for mass, and second (ss) for time.

•

Metric prefixes are used to scale units by powers of 1010. Common prefixes include kilo- (10310^{3}), centi- (10−210^{-2}), milli- (10−310^{-3}), and micro- (10−610^{-6}).

•

Unit conversion involves multiplying a given value by a conversion factor (a fraction equal to 11) so that the unwanted units cancel out, leaving the desired unit.

•

Scientific notation is a way of expressing very large or very small numbers in the form a×10na \times 10^{n}, where 1≤a<101 \le a < 10. This is crucial when discussing distances in space or the size of atoms.

•

Temperature is commonly measured in degrees Celsius (∘C^{\circ}C) in the lab, but the SI base unit is Kelvin (KK). The two scales are related by a constant offset.

📐Formulae

1 km=1000 m1\text{ km} = 1000\text{ m}

1 m=100 cm=1000 mm1\text{ m} = 100\text{ cm} = 1000\text{ mm}

1 kg=1000 g1\text{ kg} = 1000\text{ g}

1 L=1000 mL=1000 cm31\text{ L} = 1000\text{ mL} = 1000\text{ cm}^{3}

T(K)=T(∘C)+273.15T(K) = T(^{\circ}C) + 273.15

Value in new unit=Value in old unit×Conversion Factor NewConversion Factor Old\text{Value in new unit} = \text{Value in old unit} \times \frac{\text{Conversion Factor New}}{\text{Conversion Factor Old}}

💡Examples

Problem 1:

An astronaut measures the length of a localized crater on the moon to be 4.5 km4.5\text{ km}. Convert this distance into meters (mm) and express it in scientific notation.

Solution:

4.5 km×1000 m1 km=4500 m4.5\text{ km} \times \frac{1000\text{ m}}{1\text{ km}} = 4500\text{ m} 4500 m=4.5×103 m4500\text{ m} = 4.5 \times 10^{3}\text{ m}

Explanation:

Since 1 km1\text{ km} equals 1000 m1000\text{ m}, we multiply the distance by 10001000. To convert to scientific notation, we move the decimal point three places to the left.

Problem 2:

A chemical reaction in a beaker produces 350 mL350\text{ mL} of gas. Convert this volume into liters (LL).

Solution:

350 mL×1 L1000 mL=0.35 L350\text{ mL} \times \frac{1\text{ L}}{1000\text{ mL}} = 0.35\text{ L}

Explanation:

To convert from a smaller unit (milliliters) to a larger unit (liters), we divide by the conversion factor of 10001000 because there are 1000 mL1000\text{ mL} in 1 L1\text{ L}.

Problem 3:

The surface temperature of a liquid is 25∘C25^{\circ}C. What is this temperature in Kelvin (KK)?

Solution:

T(K)=25+273.15=298.15 KT(K) = 25 + 273.15 = 298.15\text{ K}

Explanation:

The Kelvin scale starts at absolute zero. To convert from Celsius to Kelvin, we add 273.15273.15 to the Celsius value.