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olganol [36]
3 years ago
8

A sample of nitrogen gas has a mass of 48.6 grams. How many N2 molecules are there in the sample? molecules Submit Answer &

Next
Chemistry
1 answer:
marysya [2.9K]3 years ago
6 0

Answer:

There are 1.05  x 10²⁴ molecules in 48.6 g N₂

Explanation:

1 mol of N₂ has a mass of (14 g * 2) 28 g.

Then, 48.6 g of N₂ will be equal to (48.6 g *(1 mol/ 28 g)) 1.74 mol.

Since there are 6.022 x 10²³ molecules in 1 mol N₂, there will be

(1.74 mol *( 6.022 x 10²³ / 1 mol)) 1.05  x 10²⁴ molecules in 1.74 mol N₂ (or 48. 6 g N₂).

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Scrat [10]

Which two solutions, when mixed together, will undergo a double replacement reaction and form a white, solid substance?


1. NaCl(aq) and LiNO3(aq)

2. KCl(aq) and AgNO3(aq)  answer

3. KCl(aq) and LiCL(aq)

4. NaNO3(aq) and AgNO3(aq)


2 is the answer because AgCl is formed and that is a white ppt.

5 0
3 years ago
Read 2 more answers
Suppose that 0.48 g of water at 25∘C∘C condenses on the surface of a 55-gg block of aluminum that is initially at 25∘C∘C. If the
GarryVolchara [31]

Answer : The final temperature of the metal block is, 25^oC

Explanation :

heat_{absorbed}=heat_{released}

As we know that,  

Q=m\times c\times \Delta T=m\times c\times (T_{final}-T_{initial})

m_1\times c_1\times (T_{final}-T_1)=-[m_2\times c_2\times (T_{final}-T_2)]         .................(1)

where,

q = heat absorbed or released

m_1 = mass of aluminum = 55 g

m_2 = mass of water = 0.48 g

T_{final} = final temperature = ?

T_1 = temperature of aluminum = 25^oC

T_2 = temperature of water = 25^oC

c_1 = specific heat of aluminum = 0.900J/g^oC

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

Now put all the given values in equation (1), we get

55g\times 0.900J/g^oC\times (T_{final}-25)^oC=-[0.48g\times 4.184J/g^oC\times (T_{final}-25)^oC]

T_{final}=25^oC

Thus, the final temperature of the metal block is, 25^oC

6 0
3 years ago
Metallic elements can be recovered from ores that are oxides, carbonates, halides, or sulfides. Give an example of each type.
vredina [299]

Metallic elements which can be recovered from ores that are oxides, carbonates, halides, or sulfides are iron, zinc, silver and lead respectively.

Ores which is deposited in earth's crust which contain minerals and metals. Metals can be obtained economically and sold commercially.

<h3>How metals obtained from sulphide or carbonate ore? </h3>

As we get to know that it is easy to obtain metals from their oxides. So, firstly ores which is found in the form of carbonates and sulphide are converted into their oxides by using the process of calcination and roasting.

The metals which is in the middle of the activity series are moderately reactive. These metals are found in the crust of the earth is mainly found as oxides, sulphides, or carbonates.

A metal which can occur in the form of sulphide ore is lead.

A metal which can occur in the form of oxide ore is iron.

A metal which can occur in the form of carbonate ore is zinc.

A metal which can occur in the form of halides ore is silver.

Thus, we concluded that the metallic elements which can be recovered from ores that are oxides, carbonates, halides, or sulfides are iron, zinc, silver and lead respectively.

learn more about ore:

brainly.com/question/10306443

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6 0
1 year ago
The Element rhenium has two naturally occurring isotopes, 185 Re and 187 Re, with an average atomic mass of 186.207 amu. Rhenium
BigorU [14]
To calculate the average mass of the element, we take the summation of the product of the isotope and the percent abundance. In this case, the equation becomes 186.207=187*0.626+185*x where x is the percent abundance of 185. The answer is 0.374 or 37.4%. This can also be obtained by 100%-62.6%= 37.4%. 
8 0
3 years ago
Help?......please???
Flura [38]

Answer:im in middle school in know this but w x y = d

Explanation:so your answer is d :)

5 0
2 years ago
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