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olga55 [171]
3 years ago
13

Identify the state(s) of matter that each property describes.

Chemistry
1 answer:
Zina [86]3 years ago
5 0

Answer:solid,liquid,gas,plasma

Explanation:

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If another student joins the game and stands behind the two students
Mandarinka [93]

Answer:

ha djd sidi dj sjejeisns

4 0
2 years ago
How many bromine atoms are present in 37.9 g of CH2Br2?
VARVARA [1.3K]

Answer:

The answer to your question is: 6.55 x 10 ²³ atoms of Br

Explanation:

CH2Br2 = 37.9 g

MW CH2Br2 = (12 x 1) + (2 x 1) + (80 x 2) = 174 g

                   174 g of CH2Br2 ------------------  160 g of Br2

                   37.9 g of CH2Br2   ---------------     x

                x = 37.9 x 160/174 = 34.85 g of Br

                      1 mol of Br -----------------   160 g Br2

                         x              ----------------    174 g Be2

               x = 174 x 1 /160 = 1.088 mol of Br2

                1 mol of Br -----------------  6.023 x 10 ²³ atoms

            1.088 mol of Br -------------    x

                    x = 1.088 x 6.023 x 10 ²³ / 1 = 6.55 x 10 ²³ atoms

4 0
3 years ago
The solubility table helps predict the products of what kind of reactions?
Ganezh [65]

Answer:

B. double replacement

Explanation:

Rate, or give brainliest if you can :)

3 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
4 years ago
In the
sweet-ann [11.9K]

Answer:

The acid must be a concentrated acid

Explanation:

Ethene is prepared in the laboratory by heating ethanol with excess concentrated tetraoxosulphate VI acid at 170°C . The reactionoccursc in two stages;

1) when the ethanol and sulphuric acid are mixed in a ratio of 1:2, ethyl hydrogentetraoxosulpate VI is formed

2) The compound formed in the first step is heated in the presence of excess concentrated sulphuric acid to give ethene and sulphuric acid.

The overall reaction can be perceived as the dehydration of ethanol. The gas produced (ethene) is usually passed through sodium hydroxide solution to remove any gaseous impurities present.

concentrated sulphuric acid is used in this process since it is a good dehydrating agent.

7 0
4 years ago
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