All categories

3 years ago

34 grams of carbon react with an unlimited amount of H2O. The reaction is: C + H2O → CO + H2 What is the ending substance?

Chemistry

2 answers:

katovenus [111]3 years ago

7 0

H2c+o
hope this helps :0

Tamiku [17]3 years ago

6 0

The answer is hydrogen.

You might be interested in

3. Write a hypothesis that explains your inference about the acidity of the paper. How might you test your hypothesis by extendi

vredina [299]

Answer:

google

Explanation:

go to the plus thing on your screen and search it. (please dont attack me im doing a dare >:( )

8 0

3 years ago

Balance the following equation;C<img src="https://tex.z-dn.net/?f=C5H12%28g%29%2BO2%28g%29%3DC02%28g%29%2BH2O%28l%29" id="TexFor

Bumek [7]

put 8 in front of the oxygen in the reactants side to make it 16 molecules then put a 5 in front of the co2 in the product side to balance the carbon atoms then put a 6 in front of the H20 on the product side this balances both the hydrogen and oxygen atoms here is a representation

C5H12(g)+8O2(g)=5CO2(g)+6H20

4 0

2 years ago

How can we fix the affects the affects of acid rain

dybincka [34]

Adaptation actually and also following control measures on how to avoid it from happening

6 0

3 years ago

Chemical digestion would not be possible without _____, which speed up the rate of chemical reactions in your body?

Mashutka [201]

Enzymes in our body increases the speed of chemical digestion.

7 0

3 years ago

Read 2 more answers

Molybdenum (Mo) has a body centered cubic unit cell. The density of Mo is 10.28 g/cm3. Determine (a) the edge length of the unit

ser-zykov [4K]

Answer:

For a: The edge length of the unit cell is 314 pm

For b: The radius of the molybdenum atom is 135.9 pm

Explanation:

For a:

To calculate the edge length for given density of metal, we use the equation:

$\rho=\frac{Z\times M}{N_{A}\times a^{3}}$

where,

$\rho$ = density = $10.28g/cm^3$

Z = number of atom in unit cell = 2 (BCC)

M = atomic mass of metal (molybdenum) = 95.94 g/mol

$N_{A}$ = Avogadro's number = $6.022\times 10^{23}$

a = edge length of unit cell =?

Putting values in above equation, we get:

$10.28=\frac{2\times 95.94}{6.022\times 10^{23}\times (a)^3}\\\\a^3=\frac{2\times 95.94}{6.022\times 10^{23}\times 10.28}=3.099\times 10^{-23}\\\\a=\sqrt[3]{3.099\times 10^{-23}}=3.14\times 10^{-8}cm=314pm$