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WINSTONCH [101]
3 years ago
12

Change the following word equation into balanced formula equation

Chemistry
1 answer:
Natali [406]3 years ago
6 0

Answer:

Fe + CuO ----> FeO+ Cu

Explanation:

A word equation is a chemical equation in which the reactants and products are represented by words rather than the respective symbols of the species involved in the reaction.

Generally, when writing a balanced reaction equation, it is normal to ensure that the number of atoms of each element on the reactant side is the same as the number of atoms of the same element on the product side. This has been ensured in the reaction above.

The reaction above is possible because iron is higher than copper in the electrochemical series hence iron can displace copper from its compounds.

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What is the decay mode of radium-226?
inn [45]

Answer:

Radium-226 is a radioactive decay product in the uranium-238 decay series and is the precursor of radon-222. Radium-228 is a radioactive decay product in the thorium-232 decay series. Both isotopes give rise to many additional short-lived radionuclides, resulting in a wide spectrum of alpha, beta and gamma radiations.

6 0
3 years ago
How many moles of nitrogen are present at STP if the volume is 846L
Artyom0805 [142]
A mole of any gas occupied 22.4 L at STP. So, the number of moles of nitrogen gas at STP in 846 L would be 846/22.4 = 37.8 moles of nitrogen gas.

Alternatively, you can go the long route and use the ideal gas law to solve for the number of moles of nitrogen given STP conditions (273 K and 1.00 atm). From PV = nRT, we can get n = PV/RT. Plugging in our values, and using 0.08206 L•atm/K•mol as our gas constant, R, we get n = (1.00)(846)/(0.08206)(273) = 37.8 moles, which confirms our answer.
3 0
3 years ago
Please help me with these?
Julli [10]

Question 5 is the second one.


8 0
3 years ago
How many grams of solid Ca(OH)2 (74.1 g/mol) are required to make 500 ml of a 3 M solution?
mars1129 [50]

Answer:

111.15 g are required to prepare 500 ml of a 3 M solution

Explanation:

In a 3 M solution of Ca(OH)₂ there are 3 moles of Ca(OH)₂ per liter solution. In 500 ml of this solution, there will be (3 mol/2) 1.5 mol Ca(OH)₂.

Since 1 mol of Ca(OH)₂ has a mass of 74.1 g, 1.5 mol will have a mass of

(1.5 mol Ca(OH)₂ *(74.1 g / 1 mol)) 111.15 g. This mass of Ca(OH)₂ is required to prepare the 500 ml 3 M solution.

3 0
3 years ago
What is 10% of 70 so I kmow
vaieri [72.5K]

10 x 70 = 100 x Part, or

700 = 100 x Part 

Now, divide by 100 and get the answer:

Part = 700 / 100 = <span>7</span>

7 0
4 years ago
Read 2 more answers
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