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zavuch27 [327]
1 year ago
12

A garbage collector collects food wrappers and aluminum cans from a park. what type of waste is the garbage collector collecting

?
Chemistry
1 answer:
BartSMP [9]1 year ago
4 0

Answer:

Explanation:

Food wrappers and aluminium wrappers are widely use for packing various eatable items.

For example, in our homes our mother wraps our food for lunch in aluminium foil.

As it helps in keeping the food warm for hours.

Thus, we can conclude that out of the given options, a packaging waste is the garbage collector collecting.

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Calculate the theoretical value for the number of moles of CO2 that should have been produced in each balloon assuming that 1.45
musickatia [10]

Answer:

For 1 antacid tablet (in ballon1) we get .0173 moles of CO2

for 2 tablets (in balloon 2) we get: 2*0,0173=  0.0346 moles of CO2

For 3 tablets (in balloon 3) we get 3* 0.0173 = 0.0519 moles of CO2

Explanation:

The complete question:

Calculate the theoretical value for the number of moles of CO2 that should have been produced in each balloon assuming that 1.45 g of NaHCO3 is present in an antacid tablet. Use stoichiometry (a mole ratio conversion must be present) to find your answers (there should be three: one answer for each balloon).

Balloon 1 had 1 antacid tab

Baloon 2 had 2

Balloon 3 had 3

Step 1: Data given

1.45 g of NaHCO3 is present in an antacid tablet

Molar mass of NaHCO3 = 84.00 g/mol

Step 2: The balanced equation

NaHCO3 + H2O → NaOH + H2O + CO2

Step 3: Calculate moles of NaHCO3

Moles NaHCO3 = mass NaHCO3 / molar mass NaHCO3

1.45g / 84.0 g/mol = .0173 moles

Step 4: Calculate moles CO2

For 1 mol NaHCO3 we need 1 mol H2O to produce 1 mol NaOH 1 mol H2O and 1 mol CO2

For 0.0173 moles NaHCO3 we'll get 0.0173 moles CO2

so for 1 antacid tablet we get .0173 moles of CO2

for 2 tablets we get: 2*0,0173 =  0.0346 moles of CO2

For 3 tablets we get 3* 0.0173 = 0.0519 moles of CO2

 

5 0
3 years ago
Determine the number of atoms in 41.0 grams of calcium, Ca. (The mass of one mole of calcium is 40.08 g.)
Dovator [93]

Hi there!

Let me know if you have questions about my answer:

6.15 × 10²⁴

Explanation:

To find the number of atoms, divide the mass of calcium (m) by its molar mass (MM). This will give you the number of moles of calcium (n).

n = m/MM            Start with a formula

=\frac{41.0g}{40.08g/mol}             Substitute values from the question

= \frac{41.0g*1mol}{40.08g}             Simplify

= \frac{41.0mol}{40.08}                 Cancel out the units "g"

= 1.022...mol          Number of moles of calcium, with 4 significant figures

Then, multiply the number of moles of calcium (n) by Avogadro's number (N_{A}) to find the number of calcium atoms.

Number.of.Ca=n*N_{A}      Start with a formula

= 1.022mol*\frac{6.022*10^{23}Ca}{1mol}      Substitute values from the question

= \frac{1.022mol*6.022*10^{23}Ca}{1mol}           Simplify

= 1.022*(6.022*10^{23}Ca)     Cancel out the units "mol"

= 6.154*10^{24} Ca                   Multiply, leave one extra digit to help round. "4" tells you to round down.

= 6.15*10^{24} Ca                     Final answer to 3 significant figures*

*When multiplying and dividing numbers, your final answer will have the same number of significant figures as the number with the least number of significant figures in the question.

40.08 has 4 significant figures. 41.0 has 3 significant figures. So, you want 3 significant figures in your final answer.

I hope this helps!

Learn more about Avogadro's number here:

brainly.com/question/1445383

Check out another answer that uses significant figures:

brainly.com/question/3721164

7 0
2 years ago
HELP! URGENT Which of the following best states the relationship between erosion and deposition?
MariettaO [177]
The answer is A: When the energy transporting sediments diminishes, the sediments settle in a low-lying area; therefore, deposition always follows erosion
7 0
3 years ago
Calculate the number of joules released when 72.5 grams of water at 95.0 degrees Celsius cools to a final temperature of 28.0 de
Airida [17]
Water's specific heat capacity is 4200 J/Kg°C
95-28=67
72.5grams in kg is 0.0725kg
Energy = 67×0.0725×4200
Energy = 20,401.5 J or 20.4015 kJ
7 0
3 years ago
What was Anton van Lee
solniwko [45]

I would say the answer is A.

5 0
2 years ago
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