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Elis [28]
3 years ago
9

What piece of lab equipment would you use to conduct a small chemical experiment?

Chemistry
2 answers:
Scrat [10]3 years ago
6 0

test tube

Explanation:

you state it is small and a test tube is the smallest container out of the answers you listed

Svetradugi [14.3K]3 years ago
3 0

Answer:

You would use a test tube to conduct the experiment.

Explanation:

All of the other options are to big for a smaller experiment.

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Please help I will give brainliest!
aleksklad [387]
I believe it’s the 3rd option
6 0
2 years ago
_______________ is when a compound containing carbon and hydrogen (and sometimes oxygen) combines with oxygen gas to produce car
kotykmax [81]

Answer:

C. Combustion

Explanation:

Combustion- an organic compound containing carbon, hydrogen and sometimes oxygen reacts  with oxygen gas to form carbon dioxide and water.

4 0
3 years ago
A reaction between an acid and a base is called a
Sergeeva-Olga [200]

Neutralization reaction

5 0
3 years ago
The enthalpy change for converting 1.00 mol of ice at -50.0 ∘c to water at 60.0∘c is ________ kj. the specific heats of ice, wat
guajiro [1.7K]
First, we have to get:

1- The heat required to increase T of ice from -50 to 0 °C:

according to q formula:

q1 = m*C*ΔT

when m is the mass of ice = mol * molar mass

                                             =  1 mol * 18 mol/g

                                            = 18 g

and C is the specific heat capacity of ice = 2.09 J/g-K

and ΔT change in temperature = 0- (-50) = 50°C

by substitution:

∴q1 = 18 g * 2.09 J/g-K *50°C

       = 1881 J = 1.881 KJ

2- the heat required to melt this mass of ice is :

q2 = n*ΔHfus 

when n is the number of moles of ice = 1 mol

and ΔHfus = 6.01 KJ/mol

by substitution:

q2 = 1 mol * 6.01 KJ/mol

     = 6.01 KJ

3- the heat required to increase the water temperature from 0°C to 60 °C is:

q3 = m*C*ΔT

when m is the mass of water = 18 g 

C is the specific heat capacity of water = 4.18 J/g-K

ΔT is the change of Temperature of water = 60°C - 0°C = 60°C

by substitution:

∴q3 = 18 g * 4.18 J/g-K * 60°C

      = 4514 J = 4.514 KJ

∴the total change of enthalpy = q1+q2+q3

                                                  = 1.881 KJ  +6.01 KJ + 4.514 KJ

                                                  = 12.405 KJ


5 0
3 years ago
Thorium-234 has a half-life of 24.1 days. How many grams of a 150 g sample would you have after 120.5 days?
chubhunter [2.5K]

Answer:

8.625 grams of a 150 g sample of Thorium-234  would be left after 120.5 days

Explanation:

The nuclear half life represents the time taken for the initial amount of sample  to reduce into half of its mass.

We have given that the half life of thorium-234 is 24.1 days. Then it takes 24.1 days for a Thorium-234 sample to reduced to half of its initial amount.

Initial amount of Thorium-234 available as per the question is 150 grams

So now  we start with 150 grams  of Thorium-234

150 \times \frac{1}{2}=24.1

75 \times \frac{1}{2} =48.2

34.5 \times \frac{1}{2} =72.3

17.25 \times \frac{1}{2} =96.4

8.625\times \frac{1}{2} =120.5

So after 120.5 days the amount of sample that remains is 8.625g

In simpler way , we can use the below formula to find the sample left

A=A_{0} \cdot \frac{1}{2^{n}}

Where

A_0  is the initial sample amount  

n = the number of half-lives that pass in a given period of time.

7 0
3 years ago
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