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monitta
3 years ago
6

During the winter months, many locations experience snow and ice storms. It is a common practice to treat roadways and sidewalks

with salt, such as CaCO3 . If a 11.3 kg bag of CaCO3 is used to treat the sidewalk, how many moles of CaCO3 does this bag contain?
Chemistry
1 answer:
Luden [163]3 years ago
5 0

Answer:

The number of moles of CaCO3 on the bag is 112.90 moles

Explanation:

number mole (n) = mass (m) divided by molecular mass (Mm)

Mm of CaCO3 = 100.0869 g/mole

mass in grams = 11.3 Kg x (10^3 g/1 Kg) = 11300 grams

number of moles (n) = 11300 grams divided by 100.0869 grams per mole = 112.90 moles of CaCO3 in the bag.

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kompoz [17]

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10-10 above the 0 is positive and the ones below is - (negitive)

Explanation:

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3 years ago
What volume of a 2.25 M sodium chloride solution will contain 4.58 moles of sodium chloride
Rzqust [24]

Answer:

Option E. 2.04 L

Explanation:

Data obtained from the question include:

Molarity of NaCl = 2.25 M

Mole of NaCl = 4.58 moles

Volume =..?

Molarity is simply defined as the mole of solute per unit litre of the solution. It is represented mathematically as:

Molarity = mole /Volume

With the above formula, we can obtain the volume of the solution as follow:

Molarity = mole /Volume

2.25 = 4.58/volume

Cross multiply

2.25 x volume = 4.58

Divide both side by 2.25

Volume = 4.58/2.25

Volume = 2.04 L

Therefore, the volume of the solution is 2.04 L

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3 years ago
Which level is made up of organisms that break down dead organisms?
mario62 [17]

Answer:

Explanation:

decomposers

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3 years ago
Identify each of these substances as acidic, basic, or neutral. pure water, pH = 7.0 lake water, pH = 6.5 baking soda solution,
Alex_Xolod [135]

Answer:

pure water, pH = 7.0 (Neutral)

lake water, pH = 6.5 (Acidic)

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The degree of acidicity or alkalinity of a solution can be determined on a pH meter. A pH below 7 is acidic; a pH of 7 is neutral; a pH value of above 7 is alkaline.

5 0
3 years ago
Read 2 more answers
The reactant concentration in a second-order reaction was 0.560 m after 115 s and 3.30×10−2 m after 735 s . what is the rate con
Aloiza [94]
Use the formula for second order reaction:

\frac{1}{C} = \frac{1}{C_0} + kt

C = concentration at time t 
C0 =  initial conc.
k = rate constant
t = time

1st equation :   \frac{1}{0.56} = \frac{1}{C_0} + 115k

2nd Equation: \frac{1}{0.033} = \frac{1}{C_0} + 735k

Find \frac{1}{C_0} from 1st equation and put it in 2nd equation:

\frac{1}{0.033} = \frac{1}{0.56} - 115k + 735k

\frac{1}{0.033} - \frac{1}{0.56} = 620k

k = 0.046
3 0
3 years ago
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