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Setler [38]
3 years ago
8

whenever a chemical is dispensed from its original container into a secondary container the secondary container must be labeled

to show the product name and information true or false
Chemistry
1 answer:
spayn [35]3 years ago
7 0
The answer to this question is true.

If you remove a liquid from its container to a new container, you should put the product name so you will know what is the liquid inside the container.
If you don't do this, you might find a container with unknown liquid. It will be hard to determine what substance inside the container just by observing it. If it dangerous it might harm the others too.
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What is the volume of an object with the mass of 7.9 grams in the density of 2.28g/ml.
Schach [20]

Answer:

3.7mL is the volume of the object

Explanation:

To convert the mass of any object to volume we must use density that is defined as the ratio between mass of the object and the space that is occupying. For an object that weighs 7.9g and the density is 2.28g/mL, the volume is:

7.9g * (1mL / 2.28mL) =

<h3>3.7mL is the volume of the object</h3>
7 0
3 years ago
What is a type of mechanical weathering in which rocks are broken by the expansion of water as it freezes in joints, pores, or b
Stolb23 [73]

Answer:

irozion

Explanation:

7 0
2 years ago
Read 2 more answers
A gas held at constant volume is heated from -5 degrees C to 50 degrees C, if the initial pressure is 1.0 atm, what is the new p
Hunter-Best [27]
1.21 atm, assuming the gas behaves ideally
8 0
3 years ago
How to revise in just 1 whole day for chemistry combined science is it possible to get a good grade
Pavlova-9 [17]

it will be hard, but you can do it. Just study given the materials for the course. Understand enthalpy and entropy, and various types of bonding and you'll be fine.

4 0
3 years ago
When 7.085 grams of a hydrocarbon, CxHy, were burned in a combustion analysis apparatus, 21.71 grams of CO2 and 10.37 grams of H
Taya2010 [7]

Answer:

- Empirical:

C_3H_7

- Molecular:

C_6H_{14}

Explanation:

Hello,

In this case, based on the information regarding the combustion, the moles of carbon turn out:

n_C=21.71gCO_2*\frac{1molCO_2}{44gCO_2}*\frac{1molC}{1molCO_2}=0.493molC

Moreover, the moles of hydrogen:

n_H=10.37gH_2O*\frac{1molH_2O}{18gH_2O}*\frac{2molH}{1molH_2O}=1.152molH

Thus, the subscripts of carbon and hydrogen in the hydrocarbon turn out:

C=\frac{0.4934}{0.4934}=1\\H=\frac{1.15222}{0.4934}=2.335\\CH_{2.335}

Now, looking for a suitable whole number we obtain the following empirical formula as 2.335 times 3 is 7 for hydrogen:

C_3H_7

In such a way, that compound has a molar mass of 43 g/mol, thus, the whole compound's molar mass is 86.18 g/mol for which the molecular formula is twice the empirical one, therefore:

C_6H_{14}

Which is hexane.

Best regards.

6 0
3 years ago
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