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3 years ago

How many liters of oxygen are required to burn 3.86 l of carbon monoxide?

Chemistry

2 answers:

Sergio039 [100]3 years ago

7 0

Answer:

Explanation:

3.86 L CO x 1 mol O2 x 1mole O2 x 22.4 L O2 = 1.93 L O2

22.4 L CO 2 mole CO 1 mole O2

3241004551 [841]3 years ago

5 0

1.93 liters. Hope this helps. :)

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LAB: predicting products

kondor19780726 [428]

Answer:

1) synthesis MgI2

2) double replacement CuS + (HCl)2

3) double replacement, not sure ab the formula sorry

4 0

3 years ago

1. An electric power plant uses energy from burning coal to generate steam at 450 °C. The plant is cooled by 20 °C water

kap26 [50]

Answer:

Explanation:

Efficiency of the electric power plant is $e=1-\frac{T_{2}}{T_{1}}$

Here Temperature of hot source $T_{1} = 450^{o}C=450+273=723 K$

and Temperature of sink $T_{1} = 20^{o}C=20+273=293 K$

Hence the efficiency is $e=1-\frac{293}{723}=0.5947=59.47%$

Now another formula for thermal efficiency Is

$e_{therm}=\frac{Q_{1}-Q_{2}}{Q_{1}}=\frac{W}{Q_{1}}$

Here QI is the of heat taken from source 100 MJ ; Q2 of heat transferred to the sink (river) to be found

W is the of work done and W = QI -Q2

Hence From $e_{therm}=\frac{Q_{1}-Q_{2}}{Q_{1}}=\frac{W}{Q_{1}}$

$W=e(Q_{1})=(0.5947)(100)=59.47MJ$

Hence the of heat transferred to the river Is $Q_{2} -W = (100 -59.47=40.53$

6 0

3 years ago

Which is an example of making a quantitative observation?

Elanso [62]

Answer:

A) measuring mass of metal used in a reaction

Explanation:

Quantitive observations is data involving statistics and numerical values.

4 0

3 years ago

By what factor would the average kinetic energy of the particles by increased if the temperature of a substance was increased fr

natita [175]

Average kinetic energy of a particle :
0.5 mv^2 = kT/2
so the kinetic energy = kT/2
assuming the same value of K
T1 = -49 + 273 = 224
T2 = 287 + 273 = 560

E2 / E1 = kT2 / 2 / kT1 / 2
E2 / E1 = T2 / T1
E2 / E1 = 560 / 224 = 2.5
so the average kinetic energy of the particle increases by 2.5

3 0

3 years ago

What is the molar solubility of marble (i.e., [ca2 ] in a saturated solution in normal rainwater, for which ph=5.60? express you

Brut [27]

Missing in your question :

Ksp of(CaCO3)= 4.5 x 10 -9

Ka1 for (H2CO3) = 4.7 x 10^-7

Ka2 for (H2CO3) = 5.6 x 10 ^-11

1) equation 1 for Ksp = 4.5 x 10^-9

CaCO3(s)→ Ca +2(aq) + CO3-2(aq)

2) equation 2 for Ka1 = 4.7 x 10^-7

H2CO3 + H2O → HCO3- + H3O+

3) equation 3 for Ka2 = 5.6 x 10^-11

HCO3-(aq) + H2O(l) → CO3-2 (aq) + H3O+(aq)

so, form equation 1& 2&3 we can get the overall equation:
CaCO3(s) + H+(aq) → Ca2+(aq) + HCO3-(aq)

note: you could get the overall equation by adding equation 1 to the inverse of equation 3 as the following:
when the inverse of equation 3 is :

CO3-2 (aq) + H3O+ (aq) ↔ HCO3- (aq) + H2O(l) Ka2^-1 = 1.79 x 10^10
when we add it to equation 1
CaCO3(s) ↔ Ca2+(aq) + CO3-2(aq) Ksp = 4.5 x 10^-9

∴ the overall equation will be as we have mentioned before:
when H3O+ = H+

CaCO3(s) + H+(aq) ↔ Ca2+ (aq) + HCO3-(aq) K= 80.55

from the overall equation:

∴K = [Ca2+][HCO3-] / [H+]

when we have [Ca2+] = [HCO3-] so we can assume both = X

∴K = X^2 / [H+]

when we have the PH = 5.6 so we can get [H+]

PH = - ㏒[H+]
5.6 = -㏒[H]
∴[H] = 2.5 x 10^-6

so, by substitution on K expression:

∴ 80.55 = X^2 / (2.5 x10^-6)

∴X = 0.0142

∴[Ca2+] = X = 0.0142

6 0

3 years ago