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Taya2010 [7]
3 years ago
7

Answer nb 1 and 2 to be marked as Brainliest ​

Chemistry
1 answer:
tatuchka [14]3 years ago
8 0

Answer:

Nb 1= mmmmmmmmmmmmmmmmmmmmmmmmmmmmmmm

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Read the temperatures shown to the nearest 0.5°C.<br> I’ll mark you as brainlister
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The temperature to the nearest 0.5°C is 98.5°C

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Determine where to dispose of each type of waste.
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Syringe tips and titrant solution in a container,  unused solid reagent in the manufacturer's container and  broken flask in the cardboard box.

Syringe tips are dispose into the disposal containers present in the lab or hospitals.  Unused solid reagent is dispose in the manufacturer's container and then label the container with a hazardous waste tag so that people can't use it and stay away from it.  

Broken flask is dispose in the cardboard box or a container.  Titrant solution can also be dispose by placing in a container which is leak proof and closed the opening tightly so that it can't pollute and damaged the outside world.

Learn more: brainly.com/question/24610949

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It is endothermic and entropy increases.

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3 years ago
How many grams of hydrogen are necessary to react with 2.85 mol CO
Lynna [10]

CO+2 H2=CH3OH

2.85 mol Co x (2mol H2/1 mol Co)=5.70 mol just concert to grams

5.70 mol H2 x (2 g H2/1 mol H2) =11.40 grams of H2

6 0
3 years ago
The radiator in a car is filled with a solution of 60% antifreeze and 40% water. The manufacturer of the antifreeze suggests tha
lakkis [162]

Answer:

0.6 Liters of coolant should be drained and 0.6 Liter of water should be added.

Explanation:

Volume of the radiator = 3.6 L

Percentage of antifreeze = 60%

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\frac{60}{100}\times 3.6 = 2.16 L

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Given,optimal cooling of the engine is obtained with only 50% antifreeze.

So, now we want to reduce the percentage of antifreeze from 60% to 50 %

Volume of coolant removed = x

Volume of water added = x

Volume of anti freeze removed = 60% of(x) = 0.6x

Volume of antifreeze left in radiator  =50% of 3.6 L = \frac{50}{100}\times 3.6=1.8 L

1.8 Liter is the desired volume of the antifreeze.

Total volume - Removed volume = desired  volume

2.16 L - 60% of( x) = 1.8 L

2.16 L-0.6x=1.8 L

x=\frac{2.16 l- 1.8 L}{0.6}=0.6 L

0.6 Liters of coolant should be drained and 0.6 Liter of water should be added.

4 0
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