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RSB [31]
3 years ago
10

Can someone label these?

Chemistry
1 answer:
STatiana [176]3 years ago
8 0
1. Igneous roc
2. Weathering & erosion
3. Sediments
4. Sedimentary rock
5. Melting
6. Heat & pressure

Hope this is helpful
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lana [24]

Answer: You Would have to .

Explanation:

4 0
3 years ago
How many of the following molecules are polar? Pcl5 cos xeo3 sebr2
klasskru [66]

Answer:

<em>Three (3) of the molecules are polar: </em>CoS,<em> </em>XeO_{3},<em> </em>SeBr_{2}<em>.</em>

Explanation:

Polar substances have their elements held together by a covalent bond that contain partially positive and negative charges, which results in a difference in the charges' electronegativity difference (usually ranging between 0.4 and 0.7).

  • PCl5 is <u>non-polar</u> with a symmetric geometry
  • CoS is <u>polar</u>
  • XeO3 is <u>polar</u>, with a trigonal pyramidal molecular geometric
  • SeBr2 is <u>polar</u> as the difference their electronegativity is about ).5
5 0
3 years ago
Ammonia, NH3NH3 , can react with oxygen to form nitrogen gas and water. 4NH3(aq)+3O2(g)⟶2N2(g)+6H2O(l) 4NH3(aq)+3O2(g)⟶2N2(g)+6H
solmaris [256]

Answer:

36.37% is the percent yield of the reaction.

Explanation:

4NH_3(aq)+3O_2(g)\rightarrow 2N_2(g)+6H_2O(l)

1)0.650 L nitrogen gas  , at 295 K and 1.01 bar.

Let the moles of nitrogen gas be n.

Pressure of the gas ,P=  1.01 bar = 0.9967 atm (1 bar = 0.9869 atm)

Temperature of the gas = T =  295 K

Volume of the gas = V = 0.650 L

Using an ideal gas equation:

PV=nRT

n=\frac{PV}{RT}=\frac{0.9967 atm\times 0.650 L}{0.0821 atm L/mol K\times 295 K}=0.0267 mol

2) Moles of ammonia gas=\frac{2.53 g}{17 g/mol}=0.1488 mol

Moles of oxygen gas =\frac{3.53 g}{32 g/mol}=0.1101 mol

According to reaction ,3 mol of oxygen reacts with 4 mol of ammonia.

Then,0.1101 mol of oxygen will react with:

\frac{4}{3}\times 0.1101 mol=0.1468 mol of ammonia.

Hence, oxygen gas is in limiting amount and act as limiting reagent.

3) Theoretical yield of nitrogen gas :

According to reaction, 3 mol of oxygen gas gives 2 moles of nitrogen gas.

Then 0.1101 mol of oxygen will give:

\frac{2}{3}\times 0.1101 mol=0.0734 mol of nitrogen.

Theoretical yield of nitrogen gas = 0.0734 mol

Experimental yield of nitrogen as calculated in part (1) = 0.0267 mol

Percentage yield:

\frac{\text{Experiential yield}}{\text{Theoretical yield}}\times 100

Percentage yield of the reaction:

\frac{ 0.0267 mol}{0.0734 mol}\times 100=36.37\%

36.37% is the percent yield of the reaction.

3 0
3 years ago
Ik i already asked this but i need diff point of views
salantis [7]

Answer:

Is it prescribe to you?If so than yes if not then no need to

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