All categories

2 years ago

The concentration of Si in an Fe-Si alloy is 0.25 wt%. What is the concentration in kilograms of Si per cubic meter of alloy?

Chemistry

1 answer:

kodGreya [7K]2 years ago

5 0

Answer : The concentration of Si in kilograms is, $19.55kg/m^3$

Explanation :

As we are given that, the concentration of Si in an Fe-Si alloy is 0.25 wt% that means:

Weight of Si = 0.25 g = 0.00025 kg

Weight of Fe = 100 - 0.25 = 99.75 g = 0.09975 kg

Density of Si = $2.32g/cm^3=2.32\times 10^6g/m^3$

Density of Fe = $7.87g/cm^3=7.87\times 10^6g/m^3$

Now we have to calculate the concentration in kilograms of Si per cubic meter of alloy.

Concentration of Si in kilograms = $\frac{\text{Weight of Si in 100 g of alloy}}{\text{Volume of 100 g of alloy}}$

Concentration of Si in kilograms = $\frac{\text{Weight of Si in 100 g of alloy}}{\frac{\text{Wight of Fe}}{\text{Density of Fe}}+\frac{\text{Wight of Si}}{\text{Density of Si}}}$

Now put all the given values in this expression, we get:

Concentration of Si in kilograms = $\frac{0.00025kg}{\frac{99.75g}{7.87\times 10^6g/m^3}+\frac{0.25g}{2.23\times 10^6g/m^3}}$

Concentration of Si in kilograms = $19.55kg/m^3$

Thus, the concentration of Si in kilograms is, $19.55kg/m^3$

You might be interested in

This tea kettle shows a change in the state of matter of water. Which part of the water cycle represents the same change in stat

jekas [21]

I think D, because water evaporates. Once it gets hot. Then condensation. I think

3 0

2 years ago

Read 2 more answers

Describe the hybrid orbitals used by the central atom and the types of bonds formed in o3

Tems11 [23]

Hybridization in ozone, O3......

...O = O ........ 1 lone pair on central O, 2 lone pairs on terminal O 
../ 
O .................. 3 lone pairs on terminal O 

I didn't show the second of two resonance structures in which the single and double bonds are reversed. In reality, both bonds are identical have a bond order of 1.5 due to delocalized pi-bonding. 

The central atom exhibits sp2 hybridization since there is trigonal planar electron pair geometry. The notion of hybrid orbitals was "invented" by Linus Pauling in the 1930's as a way of explaining the geometry of molecules, primarily the geometry of carbon compounds. 

If the electron pair geometry is linear, the hybridization is sp. 
If the electron pair geometry is trigonal planar, the hybridization is sp2. 
If the electron pair geometry is tetrahedral, the hybridization is sp3. 

The notion that there is sp3d and sp3d2 because of d-orbital participation has been debunked. Chemists know today that there is no d-orbital involvement in hypervalent molecules regardless of what some out-of-date textbooks and some teachers' dusty old notes may say. Instead, the best explanation involves 3-center, 4-electron bonding.

8 0

2 years ago

4.82 g of an unknown metal is heated to 115.0∘C and then placed in 35 mL of water at 28.7∘C, which then heats up to 34.5∘C. What

nikitadnepr [17]

Answer: The specific heat of metal is 2.34 J/g°C

Explanation:

To calculate the mass of water, we use the equation:

$\text{Density of substance}=\frac{\text{Mass of substance}}{\text{Volume of substance}}$

Density of water = 1 g/mL

Volume of water = 35 mL

Putting values in above equation, we get:

$1g/mL=\frac{\text{Mass of water}}{35mL}\\\\\text{Mass of water}=(1g/mL\times 35mL)=35g$

When metal is dipped in water, the amount of heat released by metal will be equal to the amount of heat absorbed by water.

$Heat_{\text{absorbed}}=Heat_{\text{released}}$

The equation used to calculate heat released or absorbed follows:

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

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

where,

q = heat absorbed or released

$m_1$ = mass of metal = 4.82 g

$m_2$ = mass of water = 35 g

$T_{final}$ = final temperature = 34.5°C

$T_1$ = initial temperature of metal = 115°C

$T_2$ = initial temperature of water = 28.7°C

$c_1$ = specific heat of metal = ?

$c_2$ = specific heat of water = 4.186 J/g°C

Putting values in equation 1, we get:

$4.82\times c_1\times (34.5-110)=-[35\times 4.186\times (34.5-28.7)]$

$c_1=2.34J/g^oC$

Hence, the specific heat of metal is 2.34 J/g°C

4 0

2 years ago

The ΔG°f of atomic oxygen is 230.1 kJ/mol. Find ΔG° for the following dissociation reactionO2 (g) <--> 2O (g)then calculat

marusya05 [52]

Answer:

Kc = 2.145 × 10⁻⁸¹

Explanation:

Let's consider the following reaction:

O₂(g) ⇄ 2O(g)

The standard Gibbs free energy for the reaction (ΔG°) can be calculated using the following expression:

ΔG° = Σnp. ΔG°f(p) - Σnp. ΔG°f(p)

where,

ni are the moles of products and reactants

ΔG°f(p) are the standard Gibbs free energy of formation of products and reactants

In this case,

ΔG° = 2 × ΔG°f(O) - 1 × ΔG°f(O₂)

ΔG° = 2 × 230.1 kJ/mol - 1 × 0 kJ/mol

ΔG° = 460.2 kJ/mol

With this information, we can calculate the equilibrium constant (Kc) using the following expression:

$Kc=e^{-\Delta G \°/R.T } = e^{-460.2 kJ/mol/(8.314 \times 10^{-3}kJ/mol.K) \times 298K }=2.145 \times 10^{-81}$

3 0

2 years ago

Assuming equal concentrations and complete dissociation, rank these aqueous solutions by their freezing points. NH4l, CoBr3, Na2

seraphim [82]

Answer:

NaI > Na2SO4 > Co Br3

meaning that NaI has the highest freezing point, and Co Br3 has the lowest freezing point.

Explanation:

The freezing point depression is a colligative property.

That means that it depends on the number of solute particles dissolved.

The formula to calculate the freezing point depression of a solution of a non volatile solute is:

ΔTf = i * Kf * m

Where kf is a constant, m is the molality and i is the van't Hoff factor.

Molality, which is number of moles per kg of solvent, counts for the number of moles dissolved and the van't Hoff factor multipllies according for molecules that dissociate.

The higher the number of molecules that dissociate, the higher the van't Hoff, the greater the freezing point depression and the lower the freezing point.

As the question states that you assume equal concentrations (molality) and complete dissociation you just must find the number of ions generated by each solute, in this way:

NH4 I → NH4(+) + I(-) => 2 ions

Co Br3 → Co(+) + 3 Br(-) => 4 ions

Na2SO4 → 2Na(+) + SO4(2-) => 3 ions.

So, Co Br3 is the solute that generate more particles and that solution will exhibit the lowest freezing point among the options given, Na2SO4 is next and the NaI is the third. Ordering the freezing point from higher to lower the rank is NaI > Na2SO4 > CoBr3, which is the answer given.

4 0

2 years ago