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beks73 [17]
3 years ago
14

True or false: scientists make observations very carefully science

Chemistry
2 answers:
posledela3 years ago
8 0
True hope this helped u
Olenka [21]3 years ago
3 0
True, scientists do make observations very carefully.
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Gasoline has a density of 0.749 g/ml. how many pounds does 19.2 gallons of gasoline weigh?
Svetradugi [14.3K]
A good first step is writing the amount in terms of ml. 

19.2 gallons = 72.68 L = 72680 ml

that would mean it weighs 0.749*720680g = 54437.32ml = 54.437 L

hope that helps :)
6 0
2 years ago
Read 2 more answers
Give three examples of heterogeneous mixtures and three examples of solutions that you might use in everyday life.
beks73 [17]

Answer:

Explanation:

Heterogeneous mixtures: Milk, salt and pepper, smog, chocolate chip cookie, oil and water.

Solutions we might use everyday: Gatorade, Apple Juice, most wines and liquor, liquid detergent, coffee etc.

7 0
3 years ago
What is the balance equation for HgO(s)- Hg(l)+O2(g)
artcher [175]
The balanced equation is   2HgO --> 2Hg + O2
4 0
3 years ago
Does it make a difference if the jar is square or round? What about the size of the jar or glass?
morpeh [17]

no it doesn't make differance at all ...only the material used make difference

4 0
3 years ago
n unknown metal is either aluminum, iron or lead. If 150. g of this metal at 150.0 °C was placed in a calorimeter that contains
Nitella [24]

Answer : The metal used was iron (the specific heat capacity is 0.449J/g^oC).

Explanation :

In this problem we assumed that heat given by the hot body is equal to the heat taken by the cold body.

q_1=-q_2

m_1\times c_1\times (T_f-T_1)=-m_2\times c_2\times (T_f-T_2)

where,

c_1 = specific heat of unknown metal = ?

c_2 = specific heat of water = 4.184J/g^oC

m_1 = mass of unknown metal = 150 g

m_2 = mass of water = 200 g

T_f = final temperature of water = 34.3^oC

T_1 = initial temperature of unknown metal = 150.0^oC

T_2 = initial temperature of water = 25.0^oC

Now put all the given values in the above formula, we get

150g\times c_1\times (34.3-150.0)^oC=-200g\times 4.184J/g^oC\times (34.3-25.0)^oC

c_1=0.449J/g^oC

Form the value of specific heat of unknown metal, we conclude that the metal used in this was iron (Fe).

Therefore, the metal used was iron (the specific heat capacity is 0.449J/g^oC).

6 0
3 years ago
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