DIGESTION AND ENZYMES LAB Standard Curve (1 point) 3. You are building a go-cart

Question

DIGESTION AND ENZYMES LAB Standard Curve (1 point) 3. You are building a go-cart for the National Go-cart Championship Race. Knowing that heavier go-carts travel faster but consume more gas, you need to figure out the optimal go-cart weight This graph shows the relationship between the weights of last year's go-carts and the amount of gas they used. Use the graph to answer questions a-d. 250- 200- 50 50- 20 40 60 80 Car weight (kg 100 120 140 160 a. What overall conclusion can you draw from the data depicted in this graph? b. How much gas would a 40 kg go-cart use? Predict how much gas a 100 kg go-cart would use. Show how you determined this. c. You buy a tank that holds 200 ml of gas. What is the maximum weight your go-cart can have to consume exactly 200 ml of gas? d. BIO

Explanation / Answer

Ans1. A. From the graph it is very clearly visible that with increasing the cart weight there is increase in the requirement of the gas. Heavier the cart, larger the amount of gas required.

B. A 40 kg go cart will require 100ml of gas, as per graph.

C. A 100 kg Go cart will; use around 250 ml gase. We have determined by simply drawing a straight line joining the maximum points and then drawing a perpendicular to the 100kg. 250 ml is the point that joins at 100kg. So the required gas is approx 250ml.

D. Similarly as we have done in above case (C), by joining the points we and drawing perpendicular line from 200ml, we can say that a 70kg go cart can be used for this volume of gas.

2. Solute weight = 265gm

Volume of the solution = 725ml

Molecular weight of sugar = 180gm

So,

Concentration of solution = 265/180/725/1000

= 1.47/0.725

=2.027 M

3. Stock solution = 500g/l

Working solution required = ½ of 500g/l = 125g/l

Volume required = 100ml

Step 1: new solution has concentration = 125g/l

Step 2

M1V1 = M2 V2

500*X = 125 * 100

X = 25 ml

So, we will require to add 25 ml of stock solution followed by adding 75ml water to make final vol. 100ml to make 125g/l solution.

4. To make 1/10th of stock volume

500/10 = 50g/l will be the new concentration.

M1V1 = M2V2

500* X = 50 X 10

X = 1ml

So 1 ml of the stock will be required followed by adding 99 ml of water to make final volume 100ml and conc. of 50g/l.

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