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irina [24]
3 years ago
8

Ann drove to the store 10 km north of her house and then drove to the library, which is 5 km south of the store. She drove a tot

al distance of 15 km. What was Ann's displacement? O A. 15 km south OB. 5 km north C. 5 km south D. 15 km north SUBMIT​
Chemistry
1 answer:
Diano4ka-milaya [45]3 years ago
6 0

Answer:

C = 5km North

Explanation:

d = 10 - 5 = 5km

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What is a wind turbine often used for?
kirill115 [55]

Answer:

A

Explanation:

Converting thermal energy into electrical energy

5 0
3 years ago
Read 2 more answers
How many molecules of co2,h2o,c3h8, and o2 will be present if the reaction goes to completion?
Nadya [2.5K]

Let initially there are 10 molecules of O2 and 3 molecules of C3H8 present

The reaction will be

C3H8(g)  + 5O2(g)  ----> 3CO2(g)   + 4H2O

so here oxygen molecules are limiting as for 3 molecules of C3H8 we need 15 molecules of O2

now the given 10 molecules of O2 will react with only 2 molecules of C3H8 and they will form six molecules of CO2 and 8 molecules of H2O

Hence answer is

molecules of CO2 formed = 6

Molecules of H2O formed = 8

molecules of C3H8 left = 1

molecules of O2 left = 0


7 0
3 years ago
Read 2 more answers
How many moles of potassium nitrate, KNO3 are present in a sample with a mass of 85.2 g?
Bingel [31]
0.843 moles would be the answer
5 0
3 years ago
H4X - pKa1 =1.5
Nataly [62]
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8 0
3 years ago
Ideal He gas expanded at constant pressure of 3 atm until its volume was increased from 9 liters to 15 liters. During this proce
kkurt [141]

Explanation:

The given data is as follows.

             P = 3 atm

                = 3 atm \times \frac{1.01325 \times 10^{5} Pa}{1 atm}  

                 = 3.03975 \times 10^{5} Pa

    V_{1} = 9 L = 9 \times 10^{-3} m^{3}    (as 1 L = 0.001 m^{3}),  

        V_{2} = 15 L = 15 \times 10^{-3} m^{3}

            Heat energy = 800 J

As relation between work, pressure and change in volume is as follows.

                  W = P \times \Delta V

or,                W = P \times (V_{2} - V_{1})

Therefore, putting the given values into the above formula as follows.

                  W = P \times (V_{2} - V_{1})

                      = 3.03975 \times 10^{5} Pa \times (15 \times 10^{-3} m^{3} - 9 \times 10^{-3} m^{3})

                      = 1823.85 Nm

or,                   = 1823.85 J

As internal energy of the gas \Delta E is as follows.

                     \Delta E = Q - W

                                  = 800 J - 1823.85 J

                                  = -1023.85 J

Thus, we can conclude that the internal energy change of the given gas is -1023.85 J.

8 0
3 years ago
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