How many elements are present in the compound

The decomposition of nitrogen dioxide is described by the following chemical equation:

kow [346]

Answer:

The balanced chemical equation will be " $NO_{2}+O\rightarrow NO+O_{2}$ ".

Explanation:

The given equation is:

$2NO_{2}\rightarrow 2NO+O_{2}$

Step 1:

$2NO_{2}\rightarrow NO+O$ ...(equation 1)

Step 2:

$NO_{2}+O\rightarrow NO+O_{2}$ ...(equation 2)

On adding "equation 1" and "equation 2", we get

⇒ $NO_{2}+NO_{2}+O\rightarrow NO+O+NO+O_{2}$

⇒ $2NO_{2}\rightarrow 2NO+O_{2}$

The second step:

⇒ $NO_{2}+O\rightarrow NO+O_{2}$

3 0

3 years ago

Make your escape balancing chemical equations

Anettt [7]

Answer:

dsfewrewr

Explanation:

https://newsengin.zendesk.com/hc/vew/community/posts/360056216292-%D9%85%D9%88%D9%82%D8%B9-%D8%A7%D9%84%D9%86%D9%88%D8%B1-%D9%85%D8%B3%D9%84%D8%B3%D9%84-%D8%A7%D9%84%D9%85%D8%A4%D8%B3%D8%B3-%D8%B9%D8%AB%D9%85%D8%A7%D9%86-%D8%A7%D9%84%D8%AD%D9%84%D9%82%D8%A9-8-%D8%A7%D9%84%D8%AB%D8%A7%D9%85%D9%86%D8%A9-%D9%83%D8%A7%D9%85%D9%84%D8%A9-%D9%85%D8%AA%D8%B1%D8%AC%D9%85%D8%A9-%D9%82%D9%8A%D8%A7%D9%85%D8%A9-%D8%B9%D8%AB%D9%85%D8%A7%D9%86-%D8%A7%D9%84%D8%AD%D9%84%D9%82%D8%A9-8-%D9%83%D8%A7%D9%85%D9%84%D8%A9-%D9%85%D8%AA%D8%B1%D8%AC%D9%85%D8%A9

https://newsengin.zendesk.com/hc/pst/community/posts/360056216252-Serial-TV-%D9%85%D8%B3%D9%84%D8%B3%D9%84-%D8%A7%D9%84%D9%85%D8%A4%D8%B3%D8%B3-%D8%B9%D8%AB%D9%85%D8%A7%D9%86-%D8%A7%D9%84%D8%AD%D9%84%D9%82%D8%A9-7-%D8%A7%D9%84%D8%B3%D8%A7%D8%A8%D8%B9%D8%A9-%D8%A7%D9%88%D9%86%D9%84%D8%A7%D9%8A%D9%86-%D8%A7%D9%84%D9%85%D8%A4%D8%B3%D8%B3-%D8%B9%D8%AB%D9%85%D8%A7%D9%86-%D8%A7%D9%84%D8%AD%D9%84%D9%82%D8%A9-7-%D9%83%D8%A7%D9%85%D9%84%D8%A9-%D9%85%D8%AA%D8%B1%D8%AC%D9%85%D8%A9-HD

https://newsengin.zendesk.com/hc/vew/community/posts/360056216292-%D9%85%D9%88%D9%82%D8%B9-%D8%A7%D9%84%D9%86%D9%88%D8%B1-%D9%85%D8%B3%D9%84%D8%B3%D9%84-%D8%A7%D9%84%D9%85%D8%A4%D8%B3%D8%B3-%D8%B9%D8%AB%D9%85%D8%A7%D9%86-%D8%A7%D9%84%D8%AD%D9%84%D9%82%D8%A9-8-%D8%A7%D9%84%D8%AB%D8%A7%D9%85%D9%86%D8%A9-%D9%83%D8%A7%D9%85%D9%84%D8%A9-%D9%85%D8%AA%D8%B1%D8%AC%D9%85%D8%A9-%D9%82%D9%8A%D8%A7%D9%85%D8%A9-%D8%B9%D8%AB%D9%85%D8%A7%D9%86-%D8%A7%D9%84%D8%AD%D9%84%D9%82%D8%A9-8-%D9%83%D8%A7%D9%85%D9%84%D8%A9-%D9%85%D8%AA%D8%B1%D8%AC%D9%85%D8%A9

7 0

3 years ago

A mixture of CS2(g) and excess O2(g) is placed in a 10 L reaction vessel at 100.0 ∘C and a pressure of 3.10 atm . A spark causes

ziro4ka [17]

Answer:

PCO2 = 0.6 25 atm

PSO2 = 1.2 75 atm

PO2 = 0.6 atm

Explanation:

Step 1: Data given

Volume = 10.0 L

Temperature = 100.0 °C

Pressure = 3.10 °C

After reaction, the temperature returns to 100.0 ∘C, and the mixture of product gases (CO2, SO2, and unreacted O2) is found to have a pressure of 2.50 atm

Step 2: The balanced equation

CS2(g)+3O2(g)→CO2(g)+2SO2(g)

Step 3: Name the reactants and products

a = CS2

b = O2 before reaction

c = CO2

d = SO2

e = nS O2 after reaction with n = the number of moles

Step 4: Calculate moles before reaction

PV = nRT

n = PV/(RT)

(na + nb) = (3.10atm) * (10.0L) / ((0.08206 Latm/moleK) * (373.15K))

(na + nb) = 1.0124

Step 5: Calculate moles after reaction

PV = nRT

n = PV/(RT)

nc + nd + ne) = PV/(RT) = (2.50 atm)*(10.0L) / ((0.08206 Latm/moleK)*(373.15K))

(nc + nd + ne) = 0.816 moles

Step 6: Calculate mol fraction

For 1 mole CS2 we need 3 moles O2 to produce 1 mole of CO2 and 2 moles of SO2

moles O2 remaining = ne = nb - 3na

moles CO2 produced = nc = na

moles SO2 producted = nd = 2na

(nc + nd + ne) = 0.816 moles = nb - 3na + na + 2na = 0.816

nb = 0.816

. (na + nb) = 1.0124

na = 1.0124 moles - 0.816 moles = 0.208

which leads to

nc = na = 0.208

nd = 2na = 2*0.208 = 0.416

ne = 0.816 - 3*0.208 = 0.192

mole fraction CO2 = 0.208 / (0.208 + 0.416 + 0.192) = 0.25

mole fraction SO2 = 0.416 / (0.208 + 0.416 + 0.192) = 0.5 1

mole fraction O2 = 0.192 /(0.208 + 0.416 + 0.192) = 0.24

Step 6: Calculate partial pressure

PCO2 = 0.25 * 2.50 atm = 0.6 25 atm

PSO2 = 0.51 * 2.50 atm = 1.2 75 atm

PO2 = 0.24 * 2.50 atm = 0.6 atm

Step 7: Control results

now let's verify a couple of things

PV = nRT

P = nRT/V

before rxn

P = (0.208 + 0.816) * (0.08206 L*atm/mole*K) * (373.15K) / (10.0L) ≈ 3.10 atm

after rxn

P = ((0.208 +0.416+0.192) * (0.08206 L*atm/mole*K) * (373.15K) / (10.0L) ≈ 2.50 atm

8 0

3 years ago

Suppose of copper(II) acetate is dissolved in of a aqueous solution of sodium chromate. Calculate the final molarity of acetate

uranmaximum [27]

Answer:

0.0714 M for the given variables

Explanation:

The question is missing some data, but one of the original questions regarding this problem provides the following data:

Mass of copper(II) acetate: $m_{(AcO)_2Cu} = 0.972 g$

Volume of the sodium chromate solution: $V_{Na_2CrO_4} = 150.0 mL$

Molarity of the sodium chromate solution: $c_{Na_2CrO_4} = 0.0400 M$

Now, when copper(II) acetate reacts with sodium chromate, an insoluble copper(II) chromate is formed:

$(CH_3COO)_2Cu (aq) + Na_2CrO_4 (aq)\rightarrow 2 CH_3COONa (aq) + CuCrO_4 (s)$

Find moles of each reactant. or copper(II) acetate, divide its mass by the molar mass:

$n_{(AcO)_2Cu} = \frac{0.972 g}{181.63 g/mol} = 0.0053515 mol$

Moles of the sodium chromate solution would be found by multiplying its volume by molarity:

$n_{Na_2CrO_4} = 0.0400 M\cdot 0.1500 L = 0.00600 mol$

Find the limiting reactant. Notice that stoichiometry of this reaction is 1 : 1, so we can compare moles directly. Moles of copper(II) acetate are lower than moles of sodium chromate, so copper(II) acetate is our limiting reactant.

Write the net ionic equation for this reaction:

$Cu^{2+} (aq) + CrO_4^{2-} (aq)\rightarrow CuCrO_4 (s)$