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Nuetrik [128]
3 years ago
12

When heating a chemical solution over an open flame in a test tube you should - A. keep a cork in the test tube so the solution

cannot spill out. B. put your safety goggles on once the solution gets really hot. C. use several paper towels to pick up the hot test tube when it is ready. D. make sure the test tube is not pointing towards you or your classmates.
Chemistry
1 answer:
Alla [95]3 years ago
3 0

Hello.

The answer is:  A. keep a cork in the test tube so the solution cannot spill out

B and C both could be very deadly and it shouldnt be D because if u have an open flame you shouldnt open it even if its not towards anyone.

have a nice day

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What mass of butane in grams is necessary to produce 1.5×103 kj of heat what mass of co2 is produced?
kari74 [83]
The heat of reaction (i.e. combustion) of butane (C_{4} H_{10}) when reacted with oxygen (O_{2})  is -2658 kJ/mol butane, and the chemical reaction is given by: 

C_{4} H_{10} + \frac{13}{2} O_{2} ---> 4 CO_{2}  + 5 H_{2}O

The mass of butane required in the reaction is based on the heat produced by the reaction, which is given to be -1,500 kJ. The minus sign is added because the reaction releases heat (exothermic), which means that the products are in a "lower energy state" than the reactants. 

Dividing this with the heat of reaction per mole of butane reacted would give the number of moles butane required. Then, multiplying the answer with the molar mass of butane which is 58 grams/mole, will give the mass of butane required. 

Moles of butane = [(-1,500 kJ)/(-2658 kJ/mol butane)]
Moles of butane = 0.5643 moles butane

Mass of butane  = 0.5643 moles butane * 58 grams/mol butane
Mass of butane  = 32.73 grams butane

The mass of carbon dioxide (CO_{2}) can be determined by multiplying the moles of butane (C_{4} H_{10}) with the mole ratio of (CO_{2}) produced to the (C_{4} H_{10}) reacted, and then with the molar mass of (CO_{2}), which is 44 grams/mole. 

Mass of carbon dioxide produced 
    = 0.5643 moles butane * [4 moles CO_{2}/ 1 mole C_{4} H_{10}] * 44 grams/mole CO_{2}

Mass of carbon dioxide produced  
    = 99.32 grams CO_{2}

Thus, the mass of butane required is 32.73 grams, and the mass of carbon dioxide produced from the reaction of this amount of butane is 99.32 grams. 
                
4 0
3 years ago
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125 ml of nitrogen gas is collected at 70.0 degrees Celsius. The pressure
dybincka [34]

Answer: Volume of the gas at STP is 22.53 L.

Explanation:

Given : Volume = 125 mL   (as 1 mL = 0.001 L) = 0.125 L

Temperature = 70^{o}C = (70 + 273) K = 343 K

Pressure = 125 kPa = 125 kPa \times \frac{0.01 atm}{1 kPa} = 1.25 atm

According to the ideal gas equation, the volume of given nitrogen gas is calculated as follows.

PV = nRT

where,

P = pressure

V = volume

n = number of moles

R = gas constant = 0.0821 L atm/mol K

T = temperature

Substitute the values into above formula as follows.

1.25 atm \times V = 1 mol \times 0.0821 L atm/mol K \times 343 K\\V = \frac{1 mol \times 0.0821 L atm/mol K \times 343 K}{1.25 atm}\\= \frac{28.1603}{1.25} L\\= 22.53 L

Hence, volume of the gas at STP is 22.53 L.

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