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Soloha48 [4]
3 years ago
13

An economical substitute for vitreous silica is a high-silica glass made by leaching the B2O3-rich phase from a two-phase borosi

licate glass. (The resulting porous microstructure is densified by heating.) A typical starting composition is 81 wt % SiO2, 4 wt % Na2O, 2 wt % Al2O3, and 13 wt % B2O3. A typical final composition is 96 wt % SiO2, 1 wt % Al2O3, and 3 wt % B2O3. How much product (in kilograms) would be produced from 50 kg of starting material, assuming no SiO2 is lost by leaching
Chemistry
1 answer:
stich3 [128]3 years ago
7 0

Answer:48kg of SiO2, 0.5kg of Al2O3, and 1.5kg of B2O3

Will be the final product

Explanation:

I) 96wt% of SiO2 will amount to 96/100*50 = 0.96*50=48kg of SiO2

ii) 1wt% of Al2O3 will amount to 1/100*50 = 0.01*50=0.5kg of Al2O3

III) 3wt% of B2O3 will amount to 3/100*50 = 0.03*50=1.5kg of B2O3..

The overall product form 48+ 0.5+1.5= 50kg

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Calculate the molecular mass or formula mass (in amu) of each of the following substances: (a) BrN3 amu (b) C2H6 amu (c) NF2 amu
irakobra [83]

Answer:

Shown below

Explanation:

a) for BrN3

80+3(14)=122amu

b) forC2H6

2(12) + 6(1) = 30amu

C) for NF2

14+2(19) = 52amu

D) Al2S3

2(27) + 3(32)= 150amu

E) for Fe(NO3)3

56 + 3 [14+3(16)] =242amu

F) Mg3N2

3(24) + 2(14)= 100amu

G) for (NH4)2CO3

2[14 +4(1)] +12 +3(16)=96amu

5 0
3 years ago
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What is the measurement for momentum?
sertanlavr [38]
The unit of momentum is the product of the units of mass and velocity.
8 0
3 years ago
_P4 + ____O2 → __P2O3<br><br> How do I balance this equation?
Murljashka [212]

Explanation:

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5 0
3 years ago
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Consider the equilibrium: HCOOH(aq) + F-(aq) &lt;----&gt; HCOO-(aq) + HF (aq) Given that the Ka of HCOOH = 1.8 x 10-4 and the Ka
Aleks [24]

Hey there!:

K = Ka * Kb / Kw

Ka = 1.8*10⁻⁴

Kb = 10⁻¹⁴ / 6.8*10⁻⁴

K =  1.8*10⁻⁴ * ( 10⁻¹⁴/ 6.8*10⁻⁴ ) * ( 1 / 10⁻¹⁴ )

K =  = 1.8 / 6.8

K = 0.265

Answer A

Therefore:

K is less than on the forward reaction is not favorable .


Hope That helps!

8 0
3 years ago
The rate of effusion of an unknown gas was measured and found to be 11.9 mL/min. Under identical conditions, the rate of effusio
iren2701 [21]

Answer : The correct option is, (B) CO_2

Solution :

According to the Graham's law, the rate of effusion of gas is inversely proportional to the square root of the molar mass of gas.

R\propto \sqrt{\frac{1}{M}}

or,

(\frac{R_1}{R_2})=\sqrt{\frac{M_2}{M_1}}       ..........(1)

where,

R_1 = rate of effusion of unknown gas = 11.9\text{ mL }min^{-1}

R_2 = rate of effusion of oxygen gas = 14.0\text{ mL }min^{-1}

M_1 = molar mass of unknown gas  = ?

M_2 = molar mass of oxygen gas = 32 g/mole

Now put all the given values in the above formula 1, we get:

(\frac{11.9\text{ mL }min^{-1}}{14.0\text{ mL }min^{-1}})=\sqrt{\frac{32g/mole}{M_1}}

M_1=44.2g/mole

The unknown gas could be carbon dioxide (CO_2) that has approximately 44 g/mole of molar mass.

Thus, the unknown gas could be carbon dioxide (CO_2)

5 0
4 years ago
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