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mylen [45]
2 years ago
12

What is the quire drought of hotdog water and why does it taste wired

Chemistry
1 answer:
Liula [17]2 years ago
3 0

Answer:

Oil trout

Explanation:

Hotdog water with toes

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What type of forces allow for an object to stay at rest or continue at constant velocity.
galina1969 [7]

Answer:Inertia is the property of a body to remain at rest or to remain in motion with constant velocity. Some objects have more inertia than others because the inertia of an object is equivalent to its mass.

Explanation:

5 0
2 years ago
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If the temperature of a gas increases, describe what should happen to the pressure of the gas. Assume the volume and amount of g
Sliva [168]
The pressure will increase with decreasing volume. if they remain constant, that is. 
4 0
3 years ago
What is the molarity of 62grams of ammonium in 5liters of water?
Viefleur [7K]

Answer:

Explanation:

First we need to find how many moles of ammonium weigh 62 grams.

Molar mass of NH4 = (14.0)+(4*1.0) grams

or 18.0 grams/mole

62 (g)/18(g/mole) = 3.444... moles of NH4

If it is dissolved in 5 litres of water, the concentration will be 3.444moles/5L

or 0.6888 M.

7 0
3 years ago
the density of aluminum is 2.70g/cm^3. a piece of aluminum foil has a volume of 54.0 cm^3. what is the mass of this piece of alu
Neporo4naja [7]
The  mass  of  aluminium  foil   is   calculated   as  follows
mass  =  density  x  volume
density =  2.70  g/cm^3
volume  54  cm^3
mass  of   aluminium  foil  is  therefore  =  2.70  g/cm^3  x  54  cm^3  =145.8  grams
cm^3  cancel   out  each   other
6 0
3 years ago
Seawater is typically 3.5% salt and has a density of 1.03 g/mL. How many grams of salt would be needed to prepare enough seawate
Bas_tet [7]

Answer:

Amount of salt needed is around 2.3*10³ g

Explanation:

The salt content in sea water = 3.5 %

This implies that there is 3.5 g salt in 100 g sea water

Density of seawater = 1.03 g/ml

Volume of seawater = volume of tank = 62.5 L = 62500 ml

Therefore, the amount of seawater required is:

=Density*Volume = 1.03g/ml*62500ml = 6.44*10^{4} g

The amount of salt needed for the calculated amount of seawater is:

=\frac{6.44*10^{4}g\ water*3.5g\ salt }{100g\ water} =2254 g =2.3*10^{3} g

8 0
3 years ago
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