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AURORKA [14]
4 years ago
7

If the elastic modulus of cobalt (co) is 200 gpa, and the elastic modulus of tungsten carbide (wc) is 700 gpa, calculate the upp

er and lower bound estimates of the elastic modulus of a 10 vol% co – 90vol% wc composite.
Chemistry
1 answer:
Dmitry_Shevchenko [17]4 years ago
7 0
To determine the upper bond
Ec(u) = EmVm + EpVp
Em is the elastic modulus of cobalt.
E₁ is the elastic modulus of the particulate
Vm is the volume fraction of cobalt
Vp is the volume fraction of particulate
substitute
Ec(u) = 200 (Vm) + 700 (Vp)
To determine the lower bound
Ec (l) = EmEp/VmEp+ VpEm
Substitute
Ec (l) = 200(700)/Vm(700) + Vp (200)
Ec (l) = 1400/7Vm+2Vp
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Which one is the answer?
nadya68 [22]
I would go with C because that seems like the best answer choice
3 0
3 years ago
The combustion of ammonia in the presence of excess oxygen yields no2 and h2o: 4 nh3 (g) + 7 o2 (g) â 4 no2 (g) + 6 h2o (g) the
VLD [36.1K]
The answer would be 118.68 g.
Explanation for this is:4 moles of NH3 give 4 moles of NO2 
so 1mole of NH3 will give 1 mole of NO2 
43.9 grams of NH3 contains 2.58 moles 
so 2.58 moles will be produced of NO2 
which is 118.7 grams this the amount of oxygen that is used.
8 0
3 years ago
Fill in the coefficients that will balance the following reaction:
Burka [1]

Answer:

a0 = 2

a1= 9

a2= 6

a3= 8

Explanation:

The equation for the reaction is;

C3H7OH + O2 → CO2 + H2O

To balance the chemical equation we introduce coefficients;

Therefore the balanced chemical equation will be;

2C3H7OH + 9O2 → 6CO2 + 8H2O

Chemical equations are balanced to ensure the law of conservation of mass is obeyed, such that the mass of the reactants is equivalent to that of the products.

3 0
3 years ago
How many grams of NH3 can be dissolved in 50 grams of water at 50oC?
weqwewe [10]

15 grams of  NH3 can be dissolved

<h3>Further explanation</h3>

Given

50 grams of water at 50°C

Required

mass of NH3

Solution

Solubility is the maximum amount of a substance that can dissolve in some solvents. Factors that affect solubility  

  • 1. Temperature:
  • 2. Surface area:
  • 3. Solvent type:
  • 4. Stirring process:

We can use solubility chart (attached) to determine the solubility of NH3 at 50°C

From the graph, we can see that the solubility of NH3 in 100 g of water at 50 C is 30 g

So that the solubility in 50 grams of water is:

= 50/100 x 30

= 15 grams

5 0
3 years ago
Does adding 1 mol of NaCl to 1 kg of water lower the vapor pressure of water to the same extent, a lesser extent, or a greater e
igor_vitrenko [27]

Answer:

Adding 1 mol of NaCl to 1 kg of water lower the vapor pressure of water <em><u>to the same extent</u></em>  by adding 1 mol of C_06H_{12}O_6 to 1 kg of water.

Explanation:

1) Moles of NaCl ,n_1=1 mol

Mass of water = m= 1 kg = 1000 g

Moles of water = n_2=\frac{1000 g}{18 g/mol}=55.55 mol

Vapor pressure of the solution = p

Vapor pressure of the pure solvent that is water = p_o=17.5 Torr

Mole fraction of solute(NaCl)= \chi_1=\frac{n_1}{n_1+n_2}

\frac{p_o-p}{p_o}=\frac{n_1}{n_1+n_2}

\frac{17.5 Torr-p}{17.5 Torr}=\frac{1 mol}{1 mol+55.55 mol}

p=17.19 Torr

The vapor pressure for the NaCl solution at 17.19 Torr.

2) Moles of sucrose ,n_1=1mol

Mass of water = m  = 1 kg = 1000 g

Moles of water = n_2=\frac{1000 g}{18 g/mol}=55.55 mol

Vapor pressure of the solution = p'

Vapor pressure of the pure solvent that is water = p_o=17.5 Torr

Mole fraction of solute ( glucose)= \chi_1=\frac{n_1}{n_1+n_2}

\frac{p_o-p}{p_o}=\frac{n_1}{n_1+n_2}

\frac{17.5 Torr-p}{17.5 Torr}=\frac{1 mol}{1 mol+55.55 mol}

p'=17.19 Torr

The vapor pressure for the glucose solution at 17.19 Torr.

p = p' = 17.19 Torr

Adding 1 mol of NaCl to 1 kg of water lower the vapor pressure of water to the same extent  by adding 1 mol of C_06H_{12}O_6 to 1 kg of water.

3 0
3 years ago
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