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borishaifa [10]
2 years ago
15

Classify the following as a heterogeneous mixture, homogeneous mixture (solution), or a pure substance: lemonade with no pulp A.

heterogeneous mixture B. homogeneous mixture/solution C. pure substance​
Chemistry
1 answer:
IceJOKER [234]2 years ago
7 0

Answer:

b

Explanation:

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I could really use some help with this question, thank you in advance!​
Taya2010 [7]
C because
Explanation Plato
6 0
2 years ago
How many grams of CO2 are in 3.6 mol of CO2?
hammer [34]

There are 158.4 grams of CO2 in 3.6 mol of CO2.

<h3>HOW TO CALCULATE MASS?</h3>

The mass of a substance can be calculated by multiplying the number of moles of the substance by its molar mass. That is;

mass of CO2 = no. of moles × molar mass

According to this question, there are 3.6 moles of CO2.

mass of CO2 = 3.6 moles × 44g/mol

mass of CO2 = 158.4g.

Therefore, there are 158.4 grams of CO2 in 3.6 mol of CO2

Learn more about mass at: brainly.com/question/15959704

6 0
2 years ago
Read 2 more answers
What island chain is South America was the source of many of Darwin’s insights
photoshop1234 [79]
The Galapagos Islands
8 0
3 years ago
What is the path of electron around nucleus.?
mezya [45]
There is no path of electrons around the nucleus. There are however things called orbitals where you are likely to find electrons.
4 0
3 years ago
A 1.52 g sample of KCIO3 is reacted according to the balanced equation below. How many liters of O2 is produced at a pressure of
ipn [44]

Answer:

0.486 L

Explanation:

Step 1: Write the balanced reaction

2 KCIO₃(s) ⇒ 2 KCI (s) + 3 O₂(g)

Step 2: Calculate the moles corresponding to 1.52 g of KCIO₃

The molar mass of KCIO₃ is 122.55 g/mol.

1.52 g × 1 mol/122.55 g = 0.0124 mol

Step 3: Calculate the moles of O₂ produced from 0.0124 moles of KCIO₃

The molar ratio of KCIO₃ to O₂ is 2:3. The moles of O₂ produced are 3/2 × 0.0124 mol = 0.0186 mol

Step 4: Calculate the volume corresponding to 0.0186 moles of O₂

0.0186 moles of O₂ are at 37 °C (310 K) and 0.974 atm. We can calculate the volume of oxygen using the ideal gas equation.

P × V = n × R × T

V = n × R × T/P

V = 0.0186 mol × (0.0821 atm.L/mol.K) × 310 K/0.974 atm = 0.486 L

7 0
3 years ago
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