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tia_tia [17]
3 years ago
6

Rachel loves gardening. This summer, her 10 ft x 10 ft garden space is divided equally among tomatoes, lettuce, and strawberries

. a) The instructions for tomato fertilizer suggest applying 1500 lb/ac. How many kilograms of fertilizer should she apply to her tomatoes? How DoWe Ce Convert
Chemistry
1 answer:
Ray Of Light [21]3 years ago
7 0

Answer:

Rachel needs 0.52 kg of fertilizer.

Explanation:

Since the surface is a <em>square</em> of 10 ft of size its area is A = 10 ft x 10 ft = 1.0 x 10² ft². <em>One third</em> of that space is devoted to tomatoes (<em>"space is divided equally among tomatoes, lettuce, and strawberries"</em>), that is, 1/3 x 100 ft² = 33 ft².

We calculate the kilograms of fertilizer required we need to know 2 equivalences:

1 ac = 43,560 ft²

1 lb = 0.4535 kg

Then, we can use proportions to find out the amount of fertilizer used:

33ft^{2} \frac{1ac}{43,560ft^{2} } .\frac{1,500lb}{1ac} .\frac{0.4535kg}{1lb} =0.52kg

You might be interested in
Convert 5,500 mg to kg. Show your work to receive full credit.
Ivanshal [37]
5500 mg = 0.0055 kg. is the answer
7 0
3 years ago
What is a measure of the amount of matter in an object? (2 points) Select one: a. density b. mass c. gravity d. weight
agasfer [191]

mass is the answer 110% sure hope this helps

please mark brainliest

3 0
3 years ago
Read 2 more answers
Karen and nbsp; measures the initial volume of water as 12.5ml and nbsp; she then adds a 2.3 gram rubber stopper and the volume
Ivenika [448]

Answer:

8.3mL

Explanation:

According to the question, the initial volume of a measured amount of water is 12.5mL. Karen then adds a rubber stopper that weighs 2.3grams. This increases the volume of the water.

The increase in the volume of the water is attributed to the rubber stopper. Hence, to calculate the volume of the rubber stopper, we subtract the initial volume of the water from the final volume. i.e.

Volume of added rubber stopper = final volume of water - initial volume of water

= 20.8 - 12.5

= 8.3mL

Hence, the volume of the rubber stopper is 8.3mL

8 0
4 years ago
The anticancer drug Platinol (cisplatin), Pt(NH3)2Cl2, reacts with the cancer cell's DNA and interferes with its growth. (a) Wha
USPshnik [31]

Answer:

(a) = 65.018% (b) = 1.807,83 grams

Explanation:

(a)

The mass % of platinum (pt) in platinol can be obtained applying the general percentage formula (portion of interest / total), in this case the platinol its composed of a single platinum (Pt) mole with molecular weight of 195.088 g/mol, two nitrogen (N)  mole of 28.012 g/mole each. six hydrogen (H) mole of 1.007 g/m and two chlorine (Cl) of 35.453 g/mole

N= 14.006 g/mole x 2 = 28.012

H= 1.0078 g/mole x 6 = 6.046

Cl= 35.453 g/mole x 2 = 70.906

Pt= 195. 088 g/mole x 1 = 195.088

the molecular weight of platinol its = 300.052g

% of Pt in platinol its = ((195.088 g/mole) /(300.052 g/mole))* 100= 65.018%

(b)

Assumption: 851/9 its in fact 851 $/gram

to know how many grams of Platinum can be bought it's necessary to divided the total money between the price of platinum, knowing how many grams can be bought and knowing the percentage of platinum in platinol, it's possible to know how many grams of platinol can be made from $1.00 million

grams of platinum $1'000.000/(851$/g) = 1.175,09 grams of Pt

grams of platinol =

1.175,09 grams of pt/ 0.65018 (grams of pt/grams of platinol)

grams of platinol = 1.807,83 grams of platinol

7 0
4 years ago
How do you make 20 mL of 25 M HCl given 1.00 M HCI?<br><br> Add_mL of acid
Norma-Jean [14]

Answer:

To prepare 20 mL of 0.25 M HCl from 1.00 M HCl, 5 mL of the 1.00 M HCl is measured and transferred in a volumetric flask or beaker containing about 10 mL of water. Then, water is added to make it up to the 20mL mark.

Explanation:

The question is not correct, because the molarity of HCl cannot be greater than 12.2 M.

Also, a higher concentration of an acid can not be prepared by any dilution method,  but can only be prepared by distillation procedures requiring great care and expertise.

Therefore,  the following assumptions are made about the question :

The initial acid concentration, M1 = 1.00 M

Required acid concentration, M2 = 0.25 M

Volume of required acid solution, V2 = 20 mL

Using the dilution formula: M1V1 = M2V2

Volume of initial acid solution, V1 must be found then.

Making V1 subject of the formula; V1 = M2V2/M1

V1 = 0.25 × 20 / 1.00

V1 = 5 mL

To prepare 20 mL of 0.25 M HCl from 1.00 M HCl, 5 mL of the 1.00 M HCl is measured and transferred in a volumetric flask or beaker containing about 10 mL of water. Then, water is added to make it up to the 20mL mark.

5 0
3 years ago
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