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Xelga [282]
4 years ago
5

Cu + AgNO3 → Ag + CuNO3

Chemistry
1 answer:
tia_tia [17]4 years ago
4 0

Answer:

26.0g of Ag

Explanation:

Equation of reaction,

Cu + AgNO3 → Ag + CuNO3

We need to find the mass of Ag produced in the reaction by comparing it with the mass of Cu reacting.

From the equation of reaction,

1 mole of Cu reacts with 1 mole of Ag

Number of mole = mass / molarmass

Molar mass of Cu = 63.55g

Molar mass of Ag = 107.87g

Number of moles = mass / molar mass

Mass = number of moles × molar mass

Mass of Cu = 1 × 63.55g

Mass of Cu = 63.55g

Similarly,

Mass of Ag = 1 × 107.87

Mass of Ag = 107.87g.

Now let's go back to the equation of reaction,

1 mole of Cu = 1 mole of Ag;

63.55g of Cu = 107.87g of Ag

15.32g of Cu = x g of Ag

X = (107.87 × 15.32) / 63.55

X = 1652.568 / 63.55

X = 26.00g

26.0g of Ag will be produced when 15.32g of Cu reacts with excess AgNO3

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For IR radiation with û = 1,130 cm 1, v=__THz
Dmitry [639]

<u>Answer:</u> The frequency of the radiation is 33.9 THz

<u>Explanation:</u>

We are given:

Wave number of the radiation, \bar{\nu}=1130cm^{-1}

Wave number is defined as the number of wavelengths per unit length.

Mathematically,

\bar{\nu}=\frac{1}{\lambda}

where,

\bar{\nu} = wave number = 1130cm^{-1}

\lambda = wavelength of the radiation = ?

Putting values in above equation, we get:

1130cm^{-1}=\farc{1}{\lambda}\\\\\lambda=\frac{1}{1130cm^{-1}}=8.850\times 10^{-4}cm

Converting this into meters, we use the conversion factor:

1 m = 100 cm

So, 8.850\times 10^{-4}cm=8.850\times 10^{-4}\times 10^{-2}=8.850\times 10^{-6}m

  • The relation between frequency and wavelength is given as:

\nu=\frac{c}{\lambda}

where,

c = the speed of light = 3\times 10^8m/s

\nu = frequency of the radiation = ?

Putting values in above equation, we get:

\nu=\frac{3\times 10^8m/s}{8.850\times 10^{-4}m}

\nu=0.339\times 10^{14}Hz

Converting this into tera Hertz, we use the conversion factor:

1THz=1\times 10^{12}Hz

So, 0.339\times 10^{14}Hz\times \frac{1THz}{1\times 10^{12}Hz}=33.9THz

Hence, the frequency of the radiation is 33.9 THz

7 0
3 years ago
For the reaction 2Fe + O2 &gt; 2FeO, how many grams of iron oxide are produced from 5.00 moles of iron?
Greeley [361]

Answer:

359 grams FeO

Explanation:

To find how many grams FeO are produced, you need to use the moles of Fe, convert it to moles of FeO (using the mole-to-mole ratio from the equation), then convert the moles of FeO to grams (using the molar mass from the periodic table).

2 Fe + O₂ --> 2 FeO

Molar Mass (FeO) = 55.845 g/mol + 16.00 g/mol

Molar Mass (FeO) = 71.845 g/mol

5.00 moles Fe           2 moles FeO            71.845 grams
----------------------   x  ------------------------  x  ----------------------  = 359 grams FeO
                                   2 moles Fe               1 mole FeO

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Calculate by (a)% weight and (b) %mole each of the elements present in sugar
Musya8 [376]

Explanation:

Molecular mass of sugar = C_{12}H_{22}O_{11} : = 432 g/mol

Atomic mass of carbon atom = 12 g/mol

Atomic mass of hydrogen atom = 1 g/mol

Atomic mass of oxygen atom = 16 g/mol

a) Percentage of an element in a compound:

\frac{\text{Number of atoms of element}\times \text{Atomic mass of element}}{\text{molecular mass of compound}}\times 100

Percentage of carbon by weight in C_{12}H_{22}O_{11}:

\frac{12\times 12 g/mol}{342 g/mol}\times 100=42.10\%

Percentage of hydrogen by weight in C_{12}H_{22}O_{11}:

\frac{22\times 1g/mol}{342 g/mol}\times 100=6.43\%

Percentage of oxygen by weight in C_{12}H_{22}O_{11}:

\frac{11\times 16g/mol}{342 g/mol}\times 100=51.46\%

b) Percentage of mole each of the elements present in sugar:

=\frac{\text{Moles of atoms of an element}}{\text{total moles of all types of atoms}}\times 100

In mole of sugar we have 12 moles of carbon atom , 22 moles of hydrogen atoms and 11 moles of oxygen atoms.

Percentage of carbon by mole in C_{12}H_{22}O_{11}:

\frac{12 mol}{45 mol}\times 100=26.66\%

Percentage of hydrogen by mole in C_{12}H_{22}O_{11}:

\frac{22 mol}{45 mol}\times 100=48.88\%

Percentage of oxygen by mole in C_{12}H_{22}O_{11}:

\frac{11 mol}{45 mol}\times 100=24.44\%

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