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2 years ago

How many moles are in 2.3x1024 g of Ag? Plz help!!

Chemistry

1 answer:

ziro4ka [17]2 years ago

7 0

Answer:

2.13 × 10²² moles

Explanation:

Relation between Mole and Mass is,

Mole = Mass / M.Mass

Putting values,

Mole = 2.3 × 10²⁴ g / 107.87 g/mol

Mole = 2.13 × 10²² moles

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Radiation fog forms at night when the skies are clear and the relative humidity

LenKa [72]

Answer: A. True

Explanation:

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3 years ago

If the temperature of a gas confined to such a cylinder is raised from 287°C to 587°C, what is the ratio of the initial volume t

jenyasd209 [6]

Answer:

1 : 2

Explanation:

From the question given, we obtained the following information:

T1 = 287°C = 287 +273 = 560K

T1 = 587°C = 587 +273 =860K

V1 = V

V2 =?

V1 /T1 = V2 /T2

V/560 = V2/860

Cross multiply

V2 x 560 = 860 x V

Divide both side by 560

V2 = (860 x V) / 560

V2 = 1.54V = 2V

The ratio of initial to final volume =

V1 : V2

= V : 2V

= 1 : 2

8 0

3 years ago

How to do thw part where it says list the elements to period 2<br>and the metals

marissa [1.9K]

A)2nd period - Li, Be, B,C,N,O,F,Ne
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7 0

3 years ago

2C3H7OH + 9O2 -> 6CO2 + 8H2O

MAVERICK [17]

Answer:

Up to $18\; \rm mol$ $\rm CO_2$ .

Explanation:

The equation for this reaction: $2\; \rm C_3 H_7 OH + 9\; O_2 \to 6\; CO_2 + 8\; H_2O$ is indeed balanced.

Notice that the coefficient of $\rm C_3 H_7 OH$ is $2$ while that of $\rm CO_2$ is $6$ .

Therefore, each unit of this reaction would consume $2$ units of $\rm C_3H_7OH$ while producing $6$ units of $\rm CO_2$ . In other words, this reaction would produce three times as much $\rm CO_2\!$ as it consumes $\rm C_3H_7OH\!$ .

If all that $6\; \rm mol$ of $\rm C_3H_7OH$ molecules are actually consumed through this reaction, three times as many $\rm CO_2$ molecules would be produced. That corresponds to $3 \times 6\; \rm mol = 18\; \rm mol$ of $\rm CO_2\!$ molecules.

6 0

3 years ago

A student titrated 50 cm of sulphuric acid (H2SO4)of an unknown concentration

Talja [164]

No. of mol. of NaOH used: (30/1000)*0.5= 0.015mol
Mole ratio of NaOH:H2SO4= 2:1
No. of mol. of H2SO4: 0.015/2= 0.0075mol
Conc. of acid= 0.0075/(50/1000)= 0.15moldm^-3

Hope it helps, drop a thanks

3 0

2 years ago