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professor190 [17]
3 years ago
13

How many grams of ammonia (NH3) can be produced when 25 g of nitrogen gas reacts with 28 grams of hydrogen gas

Chemistry
1 answer:
Elis [28]3 years ago
7 0

Answer:

34 g of NH₃

Explanation:

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When placed in a freezer, liquid water turns into solid ice. Why is this considered a physical change?
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This is considered a physical change because the water does not change its chemical properties - it is still water. It is simply just moving its shape such as gas form, liquid form, or solid form. 
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Answer:

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Explanation:

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6 0
3 years ago
How many moles does 205g of helium, He contains ?
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3 years ago
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Find the empirical formula of the following compounds:
Aneli [31]

The empirical formula of the following compounds 0.903 g of phosphorus combined with 6.99 g of bromine.

<h3>What is empirical formula?</h3>

The simplest whole number ratio of atoms in a compound is the empirical formula of a chemical compound in chemistry. Sulfur monoxide's empirical formula, SO, and disulfur dioxide's empirical formula, S2O2, are two straightforward examples of this idea. As a result, both the sulfur and oxygen compounds sulfur monoxide and disulfur dioxide have the same empirical formula.

<h3>How to find the empirical formula?</h3>

Convert the given masses of phosphorus and bromine into moles by multiplying the reciprocal of their molar masses. The molar masses of phosphorus and bromine are 30.97 and 79.90 g/mol, respectively.

Moles phosphorus = 0.903 g phosphorus \frac{mol phosphorus}{ 30.97 g phosphorus}= 0.0293 mol

Moles bromine 6.99 g bromine\frac{mol bromine}{79.90 g bromine}=0.0875 mol

The preliminary formula for compound is P0.0293Bro.0875. Divide all the subscripts by the subscript with the smallest value which is 0.0293. The empirical formula is P1.00Br2.99 ≈ P₁Br3 or PBr3

To learn more about empirical formula visit:

brainly.com/question/14044066

#SPJ4

8 0
1 year ago
A compound contains 64.27% carbon, 7.19% hydrogen, and 28.54% oxygen. the molar mass is 168.19 g/mol. what is the molecular form
coldgirl [10]
Empirical formula is the simplest ratio of whole numbers of components in a compound 
calculating for 100 g of compound 
                                             C                             H                             O
mass                                  64.27 g                   7.19 g                     28.54 g
number of moles        64.27 g / 12 g/mol      7.19 g/1 g/mol     28.54 g / 16 g/mol 
                                        = 5.356 mol           = 7.19 mol           = 1.784 mol 
divide by least number of moles  
                                     5.356 / 1.784            7.19 / 1.784         1.784 / 1.784
                                      = 3.002                     4.03                     = 1.000
rounded off to nearest whole number 
 C - 3
 H - 4
 O - 1
empirical formula - C₃H₄O
 
 mass of empirical formula = 12 g/mol  x 3 + 1 g/mol x 4 + 16 g/mol x 1 = 56 g
molecular mass = 168.19 g/mol 
molecular formula is the actual ratio of elements making up the compound 
number of empirical units = molar mass of molecule / empirical mass
      empirical units = 168.19 g/mol  / 56 g = 3.00
there are 3 empirical units making up the molecular formula 
molecular formula = 3 x C₃H₄O

molecular formula = C₉H₁₂O₃

                 
5 0
4 years ago
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