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Diano4ka-milaya [45]
4 years ago
9

Butane is also a molecular compound. what is its formula? how many atoms does it contain?

Chemistry
1 answer:
pishuonlain [190]4 years ago
5 0
The formula of butane is C4H10 but I don't how many atoms it contains though
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Explanation:

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Various pieces of safety equipment are used in the lab to provide protection against injury. Which of the following pieces of la
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The lab safety equipment that is least likely to be used if proper eye protection is used is d. fire extinguisher<span>. A fire extinguisher serves to put fires out by cutting off the oxygen to a fire and operates by a spraying action. Even if an injury occurs, a fire extinguisher is extremely unlikely to be used.</span>
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A compound is found to contain 7.523% phosphorus and 92.48% iodine by weight. what is the empirical formula for this compound?
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The answer is PI3
Ratio of 1:3
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3 years ago
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How many liters of fluorine gas, at 298 K and 0.98 atm, will react with 23.5 grams of potassium metal? Show all of the work used
Sliva [168]
For the reaction 2 K + F2 --> 2 KF,
consider K atomic wt. = 39
23.5 g of K = 0.603 moles, hence following the molar ratio of the balanced equation, 0.603 moles of potassium will use 0.3015 moles of F2. (number of moles, n = 0.3015)

Now, following the ideal gas equation, PV = nRT
P = 0.98 atm
V = unknown
n = 0.3015 moles
R = 82.057 cm^3 atm K^-1mole^-1 (unit of R chosen to match the units of other parameters; see the reference below)
T = 298 K
Solving for V,
V = (nRT)/P = (0.3015 mol * 82.057 cm^3 atm K^-1 mol^-1 * 298 K)/(0.98 atm)
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3 0
4 years ago
Calculate the final temperature of the system: A 50.0 gram sample of water initially at 100 °C and a 100 gram sample initially a
katrin2010 [14]

Answer:

The final temperature of the system is 42.46°C.

Explanation:

In this problem we assumed that heat given by the hot body is equal to the heat taken by the cold body.

q_1=-q_2

m_1\times c\times (T_f-T_1)=-(m_2\times c\times (T_f-T_2))

where,

c = specific heat of water= 4.18J/g^oC

m_1 = mass of water sample with 100 °C= 50.0 g

m_2 = mass of water sample with 13.7 °C= 100.0 g

T_f = final temperature of system

T_1 = initial temperature of 50 g of water sample= 100^oC

T_2 = initial temperature of 100 g of water =13.7^oC

Now put all the given values in the given formula, we get

50.0 g\times 4.184 J/g^oC\times (T_f-100^oC)=-(100 g\times 4.184 J/g^oC\times (T_f-13.7^oC))

T_f=42.46^oC

The final temperature of the system is 42.46°C.

5 0
4 years ago
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