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Paha777 [63]
3 years ago
15

If .75 moles of ammonia is needed, how many grams of nitrogen will be consumed?

Chemistry
1 answer:
MrMuchimi3 years ago
5 0

We have to get the amount of nitrogen to be consumed to get 0.75 moles of ammonia.

The amount of nitrogen (in grams) required to prepare 0.75 moles of ammonia is: 10.5 grams.

Ammonia (NH₃) can be prepared from nitrogen (N₂) as per following balanced chemical reaction-

N₂ (g) + 3H₂ (g) ⇄ 2NH₃ (g)

According to the above reaction, to prepare 2 moles of ammonia, one mole of nitrogen is required. Hence, to prepare 0.75 moles of ammonia, \frac{1 X 0.75}{2} moles = 0.375 moles of nitrogen is required.

Molar mass of nitrogen is 28 grams, i.e, mass of one mole of nitrogen is 28 grams, so mass of 0.375 moles of nitrogen is 0.375 X 28 grams=10.5 grams of nitrogen.

Therefore, the amount of nitrogen (in grams) required to prepare 0.75 moles of ammonia is 10.5 grams.


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First, we have to get how many grams of C & H & O in the compound:
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- the mass of H atom on H2O = mass of H2O*molar mass of H / molar mass of H2O
                                                 =0.2835 * 2 / 18 = 0.0315 g
- the mass of O = the total mass - the mass of C atom - the mass of H atom
                          =  0.3 - 0.142 - 0.0315 = 0.1265 g
Convert the mass to mole by divided by molar mass
C(0.142/12) H(0.0315/2) O(0.1265/16) 
C(0.0118) H(0.01575) O(0.0079) by dividing by the smallest value 0.0079
C1.504 H3.99 O1 by rounding to the nearst fraction
C3/2 H4/1 )1/1 multiply by 2 
∴ the emprical formula C3H8O2
           



6 0
3 years ago
Silver metal can be prepared by reducing its nitrate, AgNO3 with copper according to the following equation:
Lerok [7]

%yield = 88.5%

<h3>Further explanation</h3>

Given

Reaction

Cu(s) + 2 AgNO₃(aq) → Cu(NO₃)₂(aq) + 2Ag(s)

Required

The percent yield

Solution

mol AgNO₃(MW=169,87 g/mol) :

= mass : MW

= 127 : 169.87

= 0.748

mol Ag from equation :

= 2/2 x mol AgNO₃

= 2/2 x 0.748

= 0.748

Mass Ag (theoretical) :

= mol x Ar Ag

= 0.748 x 108

= 80.784

% yield = (actual/theoretical) x 100%

%yield = 71.5/80.784 x 100%

<em>%yield = 88.5%</em>

7 0
2 years ago
What is the nucleus of an atom?
Leno4ka [110]

Answer:

The atomic nucleus is the small, dense region consisting of protons and neutrons at the center of an atom, discovered in 1911 by Ernest Rutherford based on the 1909 Geiger–Marsden gold foil experiment. 

5 0
2 years ago
43 milliliters of water weighs 43 g. what is the density of the water?
Anastaziya [24]

Answer:

\rho =1g/mL

Explanation:

Hello,

In this case, since the density is defined as the ratio between the mass and the volume as shown below:

\rho =\frac{m}{V}

We can compute the density of water for the given 43 g that occupy the volume of 43 mL:

\rho =\frac{43g}{43mL}=1g/mL

Regards.

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 = ?

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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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