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Korolek [52]
3 years ago
12

What’s a synonym for control group

Chemistry
1 answer:
viktelen [127]3 years ago
3 0

Answer:

comparison group

reference group

comparator group

pilot group

monitoring group

Explanation: you could use one or another

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4 Al + 3O2 → 2Al2O3 If 14.6 grams Al are reacted, how many liters of O2 at STP would be required?
melomori [17]

Answer: 9.08 L

Explanation:

To calculate the moles :

\text{Moles of solute}=\frac{\text{given mass}}\times{\text{Molar Mass}}    

\text{Moles of} Al=\frac{14.6g}{27g/mol}=0.54moles

4Al+3O_2\rightarrow 2Al_2O_3

According to stoichiometry :

4 moles of Al require  = 3 moles of O_2

Thus 0.54 moles of Al will require=\frac{3}{4}\times 0.54=0.405moles  of O_2

Standard condition of temperature (STP)  is 273 K and atmospheric pressure is 1 atm respectively.

According to the ideal gas equation:

PV=nRT

P = Pressure of the gas = 1 atm

V= Volume of the gas = ?

T= Temperature of the gas = 273 K      

R= Gas constant = 0.0821 atmL/K mol

n=  moles of gas= 0.405

V=\frac{nRT}{P}=\frac{0.405\times 0.0821\times 273}{1}=9.08L

Thus 9.08 L of O_2 at STP would be required

6 0
3 years ago
Conservation of Mass - Combustion
Rashid [163]

Answer:

i think its b

Explanation:

3 0
3 years ago
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What type of matter is salt water?
algol13
Salt water is a Homogeneous Mixture......
8 0
4 years ago
Add the polynomials (7x^3-2x^2-12)+(3x^3-8x^2+10x)
laiz [17]

Answer:

10 x ^3 − 10 x ^2 + 10 x − 12

Explanation:

4 0
3 years ago
For the reaction, a → b, the rate constant is 0.0208 m-1 sec-1. How long would it take for [a] to decrease from 0.100 to 0.0450
goldfiish [28.3K]

First, assume the order of the given reaction is n, then the rate of reaction i.e. \frac{dx}{dt}=k\times[A]^{n}

where, dx is change in concentration of A in small time interval dt and k is rate constant.

According to units of rate constant, the reaction is of second order.

\frac{1}{[A]_{t}}-\frac{1}{[A]_{o}} = kt   (second order formula)

Put the values,

\frac{1}{0.04590 m}-\frac{1}{0.100 m} =0.0208 m^{-1}s^{-1} \times t  

 22.23 m -10 m =0.0208 m^{-1}s^{-1} \times t

\frac{12.23 m}{0.0208 m^{-1}s^{-1}} = t

t= 587.9 s

Hence, time taken is 587.9 s






5 0
3 years ago
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