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Serga [27]
3 years ago
13

Which scientist was first to extensively develop the theory of continental drift?

Chemistry
2 answers:
fiasKO [112]3 years ago
7 0

Answer:

Alfred Wegener

Explanation:

K12

Simora [160]3 years ago
4 0

A. Alfred Wegener

hope that helps you  :) ✨

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Iron(III) oxide is formed when iron combines with oxygen in the air. How many grams of Fe2O3 are formed when 25.9 g of Fe reacts
zhuklara [117]
The balanced chemical reaction would be

<span>4Fe(s) + 3O2(g) ---> 2Fe2O3(s) 

We are given the amount of Fe to be used in the reaction. We use this as the starting point for the calculations. We do as follows:

25.9 g Fe (1 mol / 55.85 g) ( 2 mol Fe2O3 / 1 mol Fe ) (159.69 g / mol ) = 262.48 g Fe2O3 produced</span>
6 0
3 years ago
Read 2 more answers
How do you do empirical formula
mihalych1998 [28]

Answer:

ez

Explanation:

Step 1: Obtain the mass of each element present in grams. Element % = mass in g = m.

Step 2: Determine the number of moles of each type of atom present. ...

Step 3: Divide the number of moles of each element by the smallest number of moles. ...

Step 4: Convert numbers to whole numbers.

7 0
3 years ago
Suppose you are helping mccarthy choose her sample sites. you have the resources to conduct the study at only 4 sites. pick the
VARVARA [1.3K]

I dont know the answer

4 0
3 years ago
What is one pro and one con of mining?
blondinia [14]

Answer:

Pro exercise con suffication

Explanation:

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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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