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Norma-Jean [14]
3 years ago
10

You would need these tools to set up a filtration device​

Chemistry
1 answer:
marysya [2.9K]3 years ago
3 0

1/2 L of swamp water (or tap water with mud or dirt added)

2 L soda pop bottle with its lid

2 L plastic soda pop bottle—cut in half (have an adult help with cutting)

1000 ml beaker

2 20 oz cups

1 tbsp alum (aluminum potassium sulfate)

2 c fine sand

1 c coarse sand

1 c small pebbles

1 filter paper or coffee filter

1 rubber band

1 large spoon for stirring

spoon for scooping the alum

stopwatch or timer

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A _________________ is the path of energy transfer from producer to consumers.
Verizon [17]

Answer: Food chain

Explanation: none

3 0
3 years ago
The price of tomatoes is 1.8 lb/$. How many $ would you need to buy 8.1 lb?​
sesenic [268]
If you divide 8.1 by 1.8 that leads to 4.5
7 0
3 years ago
What is the specific heat capacity of an unknown metal if 75.00 g of the metal absorbs 418.6J of heat and the temperature rises
EleoNora [17]

Answer:

The specific heat capacity of the unknown metal is 0.223 \frac{J}{g*C}

Explanation:

Calorimetry is the measurement and calculation of the amounts of heat exchanged by a body or a system.

There is a direct proportional relationship between heat and temperature. The constant of proportionality depends on the substance that constitutes the body as on its mass, and is the product of the specific heat by the mass of the body. So, the equation that allows calculating heat exchanges is:

Q = c * m * ΔT

where Q is the heat exchanged by a body of mass m, made up of a specific heat substance c and where ΔT is the temperature variation.

In this case, you know:

  • Q= 418.6 J
  • c= ?  
  • m= 75 g
  • ΔT= 25 C

Replacing:

418.6 J= c* 75 g* 25 C

Solving:

c=\frac{418.6 J}{75 g*25 C}

c= 0.223 \frac{J}{g*C}

<u><em>The specific heat capacity of the unknown metal is 0.223 </em></u>\frac{J}{g*C}<u><em></em></u>

<u><em> </em></u>

<u><em></em></u>

3 0
2 years ago
29.5 g of mercury is heated from 32°C to 161°C, and absorbs 499.2 joules of heat in the process. Calculate the specific heat cap
Finger [1]

Answer:

c = 0.13 j/ g.°C

Explanation:

Given data:

Mass of mercury = 29.5 g

Initial temperature = 32°C

Final temperature = 161°C

Heat absorbed = 499.2 j

Solution:

Formula:

Q = m.c. ΔT

Q = amount of heat absorbed or released

m = mass of given substance

c = specific heat capacity of substance

ΔT = change in temperature

Q = m.c. ΔT

ΔT  = T2 - T1

ΔT  = 161°C - 32°C

ΔT  = 129 °C

Q = m.c. ΔT

c = Q / m. ΔT

c = 499.2 j / 29.5 g. 129 °C

c =  499.2 j / 3805.5 g. °C

c = 0.13 j/ g.°C

5 0
3 years ago
3.80 moles of oxygen are used up in the reaction. How many moles of water are produced?5.How many grams of oxygen does it take t
r-ruslan [8.4K]

Answer:

2 KClO3 (s) = 2 KCl (s) + 3 O2 (g)

2.5 g x g

Explanation:

x g O2 = 2.5 g KClO3 x (1 mol KClO3) x (3 mol O2) x (32 g O2) = 0.98 g O2

(122.5 g KClO3) (2 mol KClO3) (1 mol O2)

2 KClO3 (s) 2 KCl (s) + 3 O2 (g)

2.5 g x g

x g KCl = 2.5 g KClO3 x (1 mol KClO3) x (2 mol KClO3) x (74.5 g KCl) = 1.52 g KCl

(122.5 g KClO3) (2 mol KClO3) (1 mol KCl)

2 KClO3 (s) 2 KCl (s) + 3 O2 (g)

x mol 10 mol

x mol KClO3 = 10 mol O2 x (2 mol KClO3) = 6.7 mol KClO3

(3 mol O2)

7 0
3 years ago
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