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3 years ago

How many atoms are in 355 mol Na?

Chemistry

1 answer:

eimsori [14]3 years ago

5 0

Answer:

2.138 x 10^26 atoms Na

Explanation:

atoms Na = 355 mol Na x (6.022 x 10^23 atoms Na/1 mol Na) = 2137.81 x 10^23 atoms Na = 2.138 x 10^26 atoms Na

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A food substance kept at 0Â°C becomes rotten (as determined by a good quantitative test) in 8.3 days. The same food rots in 10.6

ZanzabumX [31]

Answer:

1. 67.2 kJ/mol

Explanation:

Using the derived expression from Arrhenius Equation

$In \ (\frac{k_2}{k_1}) = \frac{E_a}{R}(\frac{T_2-T_1}{T_2*T_1})$

Given that:

time $t_1$ = 8.3 days = (8.3 × 24 ) hours = 199.2 hours

time $t_2$ = 10.6 hours

Temperature $T_1$ = 0° C = (0+273 )K = 273 K

Temperature $T_2$ = 30° C = (30+ 273) = 303 K

Rate = 8.314 J / mol

Since $(\frac{k_2}{k_1}=\frac{t_2}{t_1})$

Then we can rewrite the above expression as:

$In \ (\frac{t_2}{t_1}) = \frac{E_a}{R}(\frac{T_2-T_1}{T_2*T_1})$

$In \ (\frac{199.2}{10.6}) = \frac{E_a}{8.314}(\frac{303-273}{273*303})$

$2.934 = \frac{E_a}{8.314}(\frac{30}{82719})$

$2.934 = \frac{30E_a}{687725.766}$

$30E_a = 2.934 *687725.766$

$E_a = \frac{2.934 *687725.766}{30}$

$E_a =67255.58 \ J/mol$

$E_a =67.2 \ kJ/mol$

7 0

2 years ago

1. ancient plants and animals animals 2. consumers transpiration 3. water loss plants 4. relationship of organisms to their envi

Lilit [14]

I don't know what ur asking

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3 years ago

Determine which of the following pairs of reactants will result in a spontaneous reaction at 25°C. None of the above pairs will

Alla [95]

Answer:

Only $Fe^{3+}+Mg$ gives spontaneous reaction.

Explanation:

A redox reaction will be spontaneous if standard reduction potential ( $E^{0}$ ) of the reaction is positive. Because it leads to negative standard gibbs free energy change ( $\Delta G^{0}$ ), which is a thermodynamic condition for spontaneity of a reaction.

$E^{0}=E^{0}(reduction)-E^{0}(oxidation)$

Where $E^{0}(reduction)$ and $E^{0}(oxidation)$ represents standard reduction potential of reduction half cell and standard reduction potential of oxidation half cell.

(1) Oxidation: $Mg-2e^{-}\rightarrow Mg^{2+}$ ; $E_{Mg^{2+}\mid Mg}^{0}=-2.38V$

Reduction: $Fe^{3+}+3e^{-}\rightarrow Fe$ ; $E_{Fe^{3+}\mid Fe}^{0}=-0.04V$

So, $E^{0}=E_{Fe^{3+}\mid Fe}^{0}-E_{Mg^{2+}\mid Mg}^{0}=(-0.04+2.38)V=2.34V$

Hence this pair will give spontaneous reaction.

(2) Similarly as above, $E^{0}=E_{Pb^{2+}\mid Pb}^{0}-E_{Au^{+}\mid Au}^{0}=(-0.13-1.69)V=-1.82 V$

Hence this pair will give non-spontaneous reaction.

(3) Similarly as above, $E^{0}=E_{Ag^{+}\mid Ag}^{0}-E_{Br_{2}\mid Br^{-}}^{0}=(0.80-1.07)V=-0.27 V$

Hence this pair will give non-spontaneous reaction.

(4) Similarly as above, $E^{0}=E_{Li^{+}\mid Li}^{0}-E_{Cr^{3+}\mid Cr}^{0}=(-3.04+0.74)V=-2.30 V$

Hence this pair will give non-spontaneous reaction.

6 0

2 years ago

The compound Xe(CF3)2 decomposes in afirst-order reaction to elemental Xe with a half-life of 30. min.If you place 7.50 mg of Xe

Irina-Kira [14]

Answer:

$t=147.24\ min$

Explanation:

Given that:

Half life = 30 min

$t_{1/2}=\frac{\ln2}{k}$

Where, k is rate constant

So,

$k=\frac{\ln2}{t_{1/2}}$

$k=\frac{\ln2}{30}\ min^{-1}$

The rate constant, k = 0.0231 min⁻¹

Using integrated rate law for first order kinetics as:

$[A_t]=[A_0]e^{-kt}$

Where,

$[A_t]$ is the concentration at time t

$[A_0]$ is the initial concentration

Given that:

The rate constant, k = 0.0231 min⁻¹

Initial concentration $[A_0]$ = 7.50 mg

Final concentration $[A_t]$ = 0.25 mg

Time = ?

Applying in the above equation, we get that:-

$0.25=7.50e^{-0.0231\times t}$

$750e^{-0.0231t}=25$

$x=\frac{\ln \left(30\right)}{0.0231}$

$t=147.24\ min$

6 0

3 years ago

Chemists can use moles to calculate: A. How much of the products are needed and how much reactant will be made. B. How much prod

damaskus [11]

Answer:

c.- How much of the reactants are needed and how much product will made.

Explanation:

The moles is the matter unit used in chemistry to simplify some calculations, instead of using grams. Also the moles are very useful because the chemical reaction can be balanced.

When a Chemical reaction is balanced, then it can be easily to calculate how many moles are necessary to add in a process to obtain a quantity of grams of a product.

5 0

3 years ago