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3 years ago

Balance the equation in the box. Click in the answer box to activate the palette. N2(g) + H2(g) → NH3(g)

Chemistry

1 answer:

lesantik [10]3 years ago

7 0

Answer:

N2(g) + 3H2(g) → 2 NH3(g)

Explanation:

N2(g) + H2(g) → NH3(g)

We start equaling the number of N atoms in both sides multiplying by 2 the NH3.

N2(g) + H2(g) → 2 NH3(g)

So we equals the H atoms (there are six in products sites)

N2(g) + 3 H2(g) → 2 NH3(g)

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What is the reason for the trend in lonization Energy? Select all that apply.

Nataliya [291]

Answer:

Explanation:

A, D,E and B i think is the correct answers

7 0

3 years ago

A 50.0 mL sample containing Cd2+ and Mn2+ was treated with 56.3 mL of 0.0500 M EDTA. Titration of the excess unreacted EDTA requ

egoroff_w [7]

Answer:

The concentration of Cd2+ is 0.0175 M

The concentration of Mn2+ is 0.0305 M

Explanation:

Step 1: Data given

A 50.0 mL sample contains Cd2+ and Mn2+

volume of 0.05 M EDTA = 56.3 mL

Titration of the excess unreacted EDTA required 13.4 mL of 0.0310 M Ca2+.

Titration of the newly freed EDTA required 28.2 mL of 0.0310 M Ca2+

Step 2: Calculate mole ratio

The reaction of EDTA and any metal ion is 1:1, the number of mol of Cd2+ and Mn2+ in the mixture equals the total number of mol of EDTA minus the number of mol of EDTA consumed in the back titration with Ca2+:

Step 3: Calculate total mol of EDTA

Total EDTA = (56.3 mL EDTA)(0.0500 M EDTA) = 0.002815 mol EDTA

Consumed EDTA = 0.002815 mol – (13.4 mL Ca2+)(0.0310 M Ca2+) = 0.002815 - 0.0004154 = 0.0023996 mol EDTA

Step 4: Calculate total moles of CD2+ and Mn2+

So, the total moles of Cd2+ and Mn2+ must be 0.0023996 mol

Step 5: Calculate remaining moles of Cd2+

The quantity of cadmium must be the same as the quantity of EDTA freed after the reaction with cyanide:

Moles Cd2+ = (28.2 mL Ca2+)(0.0310 M Ca2+) = 0.0008742 mol Cd2+.

Step 6: Calculate remaining moles of Mn2+

The remaining moles must be Mn2+: 0.0023996 - 0.0008742 = 0.0015254 moles Mn2+

Step 7: Calculate initial concentrations

The initial concentrations must have been:

(0.0008742 mol Cd2+)/(50.0 mL) = 0.0175 M Cd2+

(0.0015254 mol Mn2+)/(50.0 mL) = 0.0305 M Mn2+

The concentration of Cd2+ is 0.0175 M

The concentration of Mn2+ is 0.0305 M

4 0

3 years ago

HELP ASAP

Vika [28.1K]

Equilibrium constant of a reaction is the ratio of concentrations of the products and the reactants when the reaction is in equilibrium. This value is independent of the concentrations since the conditions are at equilibrium instead it depends on ionic strength and temperature.

First, we write the equilibrium expression.

K = [H2S]^2 / [H2]^2 x [S2]
K = (0.725^2) / [(0.208^2) (1.13 x 10^-6)]
K = 10751545.56 or 1.08 x 10^7

Thus, the answer is A.

4 0

3 years ago

Which type of cell is a rose thorn? prokaryotic eukaryotic plant eukaryotic animal chloroplastic

Ede4ka [16]

Plant cells are eukaryotic cells. Prokaryotic cells do not contain a membrane bound nucleus, mitochondria or other membrane bound cell structures (organelles), the DNA of prokaryotic cells are located in the cytoplasm of the cell. ... Plant cells are eukaryotic because they have a nuclear membrane.
so therefore, A rose thorn is a eukaryotic plant cell.

5 0

4 years ago

Read 2 more answers

The percent composition of carbon in C6H12O6 is:

nata0808 [166]

To calculate percent composition, you first need to find the molar mass of C (carbon), H (hydrogen) and O (oxygen).
C is 12.01
H is 1.00
O is 16

Then multiply each by the number of atoms of each element in the formula (the number that comes after each element in the equation for example C6 means 6 carbon atoms.

C: 12.01 x 6= 72.06
H: 1x12= 12
O: 16x6= 96

Then add them up.
72.06+ 12+ 96= 180.06

Now find the percent composition of carbon.

72.06/ 180.06 x 100= 40.01%

So the answer is C 40%.

8 0

4 years ago