CHEMISTRY HIGH SCHOOL

A vineyard has 145 acres of Chardonnay grapes. A particular soil supplement requires 5.50 g for every square meter of vineyard. How many kilograms of the soil supplement are required for the entire vineyard? (Recall that 1 km2 = 247 acres.) Express your answer in kilograms to three significant figures.

Answers

Answer 1

Answer:

1.36.10 ^ 6 kg of supplements are required for the entire vineyard

Further explanation

7 main quantities have been determined based on international standards, namely:

1. Length, meters (m)

2. Time, second (s)

3. Mass, kilograms (kg)

4. Temperature, kelvin (K)

5. Light intensity, candela (cd)

6. Electric current, ampere (A)

7. Amount of substance, mol (m)

Derivative magnitude is a quantity derived from one or more principal quantities. So in addition to 7 principal quantities, other quantities are derived quantities

An area with the formula length x width is a unit derived from the length of the principal. The international standard unit is square meters (m²).

other area units: Km², hm², dam², m², dm², cm², and mm²

Hectare is an SI unit

1 Hectare is equal to 100 a (Are) or 10000 m² (ten thousand square meters) or 100 x 100 m 1 Hectares = 2.47 acres

In the question , there is a 145 acres vineyard, with the supplement is given 5.5 gram / m² vineyard.

So for 1 km², a supplement =

1 km² = 10⁶ m²

5.5 gram / m² = 5.5.10⁶ grams / km²

whereas 1 km² = 247 acres and vineyard = 145 acres so

$5.5.10^6*(145)/(247)$

= 3.2287.10⁶ grams

= 3.23.10³ kg (3 significant numbers: 3,2 and 3)

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Keywords: area, convert, acres

Answer 2

Answer:

The soil supplement required for the entire vineyard is $\boxed{3.23 * {{10}^3}{\text{ kg}}}$ .

Further Explanation:

There are two types of units. One is basic or fundamental while the other ones are derived units. Basic units cannot be further reduced and other quantities are expressed in these units. Derived units are those that can need basic units to express themselves. Area, density, volume and velocity are some examples of derived units.

Seven basic units are present in the SI system. These are as follows:

1. Meter (m)

2. Kilogram (kg)

3. Second (s)

4. Kelvin (K)

5. Ampere (A)

6. Mole (mol)

7. Candela (Cd)

Firstly, the area of vineyard has to be converted into ${\text{K}}{{\text{m}}^2}$ . The conversion factor for this is,

$1{\text{ acre}} = \left( {\frac{1}{{247}}} \right){\text{ K}}{{\text{m}}^2}$

Therefore the area of vineyard can be calculated as follows:

$\begin{aligned}{\text{Area of vineyard}} &= \left( {145{\text{ acres}}} \right)\left( {\frac{{1/247{\text{ K}}{{\text{m}}^2}}}{{1{\text{ acre}}}}} \right)\n&= 0.587045{\text{ K}}{{\text{m}}^2}\n\end{aligned}$

The area is again converted into ${{\text{m}}^{\text{2}}}$ . The conversion factor for this is,

$1{\text{ K}}{{\text{m}}^2} = {10^6}{\text{ }}{{\text{m}}^2}$

So the area of vineyard can be calculated as follows:

$\begin{aligned}{\text{Area of vineyard}}&= \left( {0.587045{\text{ K}}{{\text{m}}^2}}\right)\left( {\frac{{{{10}^6}{\text{ }}{{\text{m}}^2}}}{{1{\text{ K}}{{\text{m}}^2}}}}\right)\n&= 587045{\text{ }}{{\text{m}}^2}\n\end{aligned}$

The amount of supplement required for the entire vineyard can be calculated as follows:

$\begin{aligned}{\text{Amount of supplement required}}&= \left( {587045{\text{ }}{{\text{m}}^2}} \right)\left( {\frac{{5.50{\text{ g}}}}{{1{\text{ }}{{\text{m}}^2}}}} \right)\n&=3.2287475 * {10^6}{\text{ g}}\n\end{aligned}$

The amount of supplement is to be converted into kg. The conversion factor for this is,

$1{\text{ g}} = {\text{1}}{{\text{0}}^( - 3)}{\text{ kg}}$

Therefore the amount of supplement can be calculated as follows:

$\begin{aligned}{\text{Amount of supplement}}&= \left({3.2287475 * {{10}^6}{\text{ g}}} \right)\left( {\frac{{{{10}^( - 3)}{\text{ kg}}}}{{1{\text{ g}}}}} \right)\n&= 3.2287475 * {10^3}{\text{ kg}}\n&\approx 3.23 * {10^3}{\text{ kg}}\n\end{aligned}$

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Answer details:

Grade: Senior School

Subject: Chemistry

Chapter: Basic concepts of chemistry

Keywords: supplement, vineyard, 5.50 g, 3.23*10^3 kg, conversion factor, basic units, fundamental units, derived units, area, volume, density, kg, m, Km, acre.

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