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3 years ago

How many atoms of carbon are in 24 grams of carbon?

1 answer:

3 0

(3 grams of carbon) x (1 mole of carbon/12 grams ) =3/12 = 1/4 of a mole of carbon. Then... ( 1/4 of a mole) x (6.02 x 10^23 atoms/mole) = approximately 1.5 x 10^23 atoms.

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Calculate the mass in 4.05*10^22 molecules of calcium phosphate

Marizza181 [45]

Answer:

m = 20.9 g.

Explanation:

Hello there!

In this case, according to the given information, it turns out possible for us to solve this problem by recalling both the Avogadro's number for the calculation of the moles in the given molecules of calcium phosphate and the molar mass of this compound in order to secondly calculate the mass as shown on the following setup:

$m=4.05x10^{22}molecules*\frac{1mol}{6.022x10^{23}}*\frac{310.18g}{1mol}\\\\m=20.9g$

Regards!

8 0

2 years ago

In the Haber reaction, patented by German chemist Fritz Haber in 1908, dinitrogen gas combines with dihydrogen gas to produce ga

Reptile [31]

Answer:

The rate at which ammonia is being produced is 0.41 kg/sec.

Explanation:

$N_2+3H_2\rightarrow 2NH_3$ Haber reaction

Volume of dinitrogen consumed in a second = 505 L

Temperature at which reaction is carried out,T= 172°C = 445.15 K

Pressure at which reaction is carried out, P = 0.88 atm

Let the moles of dinitrogen be n.

Using an Ideal gas equation:

$PV=nRT$

$n=\frac{PV}{RT}=\frac{0.88 atm\times 505 L}{0.0821 atm l/mol K\times 445.15 K}=12.1597 mol$

According to reaction , 1 mol of ditnitrogen gas produces 2 moles of ammonia.

Then 12.1597 mol of dinitrogen will produce :

$\frac{2}{1}\times 12.1597 mol=24.3194 mol$ of ammonia

Mass of 24.3194 moles of ammonia =24.3194 mol × 17 g/mol

=413.43 g=0.41343 kg ≈ 0.41 kg

505 L of dinitrogen are consumed in 1 second to produce 0.41 kg of ammonia in 1 second. So the rate at which ammonia is being produced is 0.41 kg/sec.

6 0

2 years ago

Please help chemistry timed test

lakkis [162]

Answer:

Barium atomic number is 56, this means that it has 56 protons in its nucleus and put it as a period 6 element. an uncharged barium atom would have 56 electron also. However as a group 2 element barium easily loses 2 electron to form the +2 cation. the ions has 56-2=54

7 0

2 years ago

Read 2 more answers

When 2 crystalline solid barium hydroxide octahydrate and crystalline solid ammonium nitrate are mixed in a beaker at room tempe

podryga [215]

It is endothermic reaction ΔH>0 (sign is +).
Because it is spontaneous reaction ΔG<span><0 (Gibbs free energy)
</span>ΔG=ΔH-TΔS, so must be TΔS>ΔH and ΔS<span>>0 (sign +).</span>

5 0

3 years ago

A reaction has a rate constant of 1.15 x 10^−2 /s at 400K and 0.685 /s at 450K.

n200080 [17]

Answer:

a) the activation barrier = 122.3 kJ/mol

b) The rate constant at 425 K = 0.1001 /s

Explanation:

Step 1: Data given

Rate constant k1 = 1.15 * 10^−2 /s at 400K (= T1)

Rate constant k2 = 0.685 /s at 450K (=T2)

Step 2: Determine the activation barrier for the reaction.

To determine the activation energy we will use the two-point Arrhenius equation:

ln(k₂/k₁) = (Ea/R)((1/T1) - (1/T2))

⇒ with Ea = the activating energy

⇒ with R = the gas constant = 8.314 J/mol* K

⇒ with k1 = rate constant 1 = 1.15 *10^-2 /s

⇒ with T1 = Temperature 1 = 400 K

⇒ with k2 = rate constant 2 = 0.685/s

⇒ with T2 = temperature 2 = 450 K

= - (Ea/R)(T₁ - T₂)/T₁T₂

Ea = (R*ln (k2/k1)) / ((1/T1)- (1/T2))

Ea = (8.314* ln(0.685/0.0115)) / ((1/400) - (1/450))

Ea = 122327.6 = 122.3 kJ/mol

B) What is the value of the rate constant at 425 K

For rate constant at 425 K.

Substitute the value of activation energy as 122327.6 J/mol, initial temperature as 400 K, final temperature as 425 K, rate constant at 400 K

1/T1 - 1/ T3 = 1/400 - 1 /425 = 1.47*10^-4

⇒ with T1 = the initial temperature = 400 K

⇒ with k1 = the rate constant at 400 K = 1.15 * 10^-2 /s

⇒ with T3 = the nex temperature = 425 K

⇒ with k3 = the rate constant at 425 K

ln(k3/k1) = Ea/R * ((1/T1)- (1/ T3))

⇒ with k3 = the rate constant at 425 K

⇒ with T3 = 425 K

k3/k1 = e^(Ea/R * ((1/T1)- (1/ T3)))

k3 = k1* e^(Ea/R * ((1/T1)- (1/ T3)))

k3 = 0.0115 * e^(122327.6/8.314 * (1.4710^-4))

k3 = 0.0115* e^2.1643

k3 = 0.1001 /s

4 0

3 years ago