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Anna11 [10]
3 years ago
13

Pleas help fill in blank

Chemistry
1 answer:
Nana76 [90]3 years ago
3 0
First on is transport
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22. How many atoms are there in 344.75 g of gold nugget? a. 1.05 x 10 to the power of 24 atoms b. 1.05 x 10 to the power of 23 a
mixas84 [53]

Answer:

1.053×10²⁴ atoms of gold

Explanation:

Hello,

Gold nugget are usually the natural occurring gold and they contain 85% - 90% weight of pure gold.

In this question, we're required to find the number of atoms in 344.75g of a gold nugget.

We can use mole concept relationship between Avogadro's number and molar mass.

1 mole = molar mass

Molar mass of gold = 197 g/mol

1 mole = Avogadro's number = 6.022 × 10²³ atoms

Number of mole = mass / molar mass

Mass = number of mole × molar mass

Mass = 1 × 197

Mass = 197g

197g is present in 6.022×10²³ atoms

344.75g will contain x atoms

x = (344.75 × 6.022×10²³) / 197

X = 1.053×10²⁴ atoms

Therefore 344.75g of gold nugget will contain 1.053×10²⁴ atoms of gold

5 0
3 years ago
Chemistry, 50 points!!! Will also mark brainliest if you answer everything
olasank [31]

Answer:

1. 2Al + Cl2 = Al2Cl2

2 TiCl4 + 2Na = Ti + 2NaCl2

3. H2O2 = H2O + O2

4. Na2S + 2HCl = H2S + 2NaCl

5. Mg(OH)2 + 2HCl = MgCl2 + 2H2O

1. 3O2 = 2O3

6 0
2 years ago
If you are given an ideal gas with pressure (P) = 259,392.00 Pa and temperature (T) = 2.00 oC of 1 mole Argon gas in a volume of
3241004551 [841]

Answer:

 R = 0.064 dm³ atm K⁻¹ mol⁻¹

Explanation:

Answer:

Explanation:

Data Given:

volume of gas V = 8.8 dm³

no. of mole of gas (n) = 1 mole

Pressure P = 259,392.00 Pa  

Convert Pascal (Pa) to atm (atmospheric pressure)

As

101,325 Pascals = 1 atm

So,

259,392.00 Pa  = 2 atm

Then Pressure (P) = 2 atm

Temperature T = 2.00 °C  

change the temperature from °C to K

As  to convert °C to K the below formula used

                   0°C + 273.15 = 273.15K

So, for 2 °C

                    2°C  + 273.15 =  275.15 K

So,

Temperature T  = 275.15 K

ideal gas constant = ?

formula used for Ideal gases

                            PV = nRT

as we have to find R of the gas:

we will rearrange the ideal gas equation as below:

R = PV / nT ........................................... (1)

Put value in equation (1)

                 R = 2atm x 8.8 dm³ / 1 mole x 275.15 K

                  R = 17.6 atm. dm³ / 275.15 mol. K

                  R = 0.064 dm³ atm K⁻¹ mol⁻¹

So the value of R is 0.064 dm³ atm K⁻¹ mol⁻¹

and the unit of R (ideal gas constant) is dm³ atm K⁻¹ mol⁻¹

7 0
3 years ago
The rate constant of a reaction is measured at different temperatures. A plot of the natural log of the rate constant as a funct
Wittaler [7]

Answer:

The activation energy is 7.11 × 10⁴ J/mol.

Explanation:

Let's consider the Arrhenius equation.

lnk=lnA-\frac{Ea}{R} .\frac{1}{T}

where,

k is the rate constant

A is a collision factor

Ea is the activation energy

R is the ideal gas constant

T is the absolute temperature

The plot of ln k vs 1/T is a straight line with lnA as intercept and -Ea/R as slope. Then,

\frac{-Ea}{R} =-8.55 \times 10^{3} K^{-1} \\Ea= 8.55 \times 10^{3} K^{-1} \times 8.314 \frac{J}{K.mol} =7.11 \times 10^{4} J/mol

8 0
3 years ago
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