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soldi70 [24.7K]
2 years ago
12

Determine the enthalpy change when 23.5 g of carbon is reacted with oxygen according to the reaction C(s) + O2(g) CO2(g) delta H

= –394 kJ/mol
Chemistry
1 answer:
xeze [42]2 years ago
8 0
We are provided with the amount of energy released when one mole of carbon reacts. We mus first convert the given mass of carbon to moles and then compute the energy released for the given amount.

Moles = mass / atomic mass
Moles = 23.5 / 12
Moles = 1.96 moles


One mole releases 394 kJ/mol
1.96 moles will release:
394*1.96
= 772.24

The enthalpy change of the reaction will be -772.24 kJ
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The diameter of carbon atom is.000000000154m. What is this expressed in scientific notation.
Margarita [4]

1.54 x 10^-10

thats the answer

5 0
3 years ago
Calculate the mass of magnesium carbonate ( MgCO3), in grams, required to produce 110.0 g of carbon dioxide using the following
bearhunter [10]

Answer:

210.7~g~MgCO_3

Explanation:

We have to start with the <u>reaction</u>:

MgCO_3~->~MgO~+~CO_2

We have the same amount of atoms on both sides, so, we can continue. The next step is to find the <u>number of moles</u> that we have in the 110.0 g of carbon dioxide, to this, we have to know the <u>atomic mass of each atom</u>:

C: 12 g/mol

O: 16 g/mol

Mg: 23.3 g/mol

If we take into account the number of atoms in the formula, we can calculate the <u>molar mass</u> of carbon dioxide:

(12*1)+(16*2)=44~g/mol

In other words: 1~mol~CO_2=~44~g~CO_2. With this in mind, we can calculate the moles:

110~g~CO_2\frac{1~mol~CO_2}{44~g~CO_2}=25~mol~CO_2

Now, the <u>molar ratio</u> between carbon dioxide and magnesium carbonate is 1:1, so:

2.5~mol~CO_2=2.5~mol~MgCO_3

With the molar mass of MgCO_3 ((23.3*1)+(12*1)+(16*3)=84.3~g/mol. With this in mind, we can calculate the <u>grams of magnesium carbonate</u>:

2.5~mol~MgCO_3\frac{84.3~g~MgCO_3}{1~mol~MgCO_3}=210.7~g~MgCO_3

I hope it helps!

8 0
3 years ago
Ba2+(aq)+SO42−(aq)→BaSO4(s)
adelina 88 [10]
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4 0
3 years ago
An environmental scientist developed a new analytical method for the determination of cadmium (cd^2+) in mussels. To validate th
Anettt [7]

Answer:

The method is accurate  in the calculation of the Cu^+2

Explanation:

As a first step we have to calculate the <u>average concentration </u>of Cu^+2 find it by the method.

\frac{0.782+0.762+0.825+0.838+0.761 }{5} =0.79 ppm

Then we have to find the<u> standard deviation:</u>

s=\sqrt{\frac{1}{N-1}\sum_{i=1}^N(x_i-\bar{x})^2}=0.0359

For the confidence interval we have to use the formula:

μ=Average±\frac{t*s}{\sqrt{n} }

Where:

t=t student constant with 95 % of confidence and 5 data=2.78

μ= 0.79  ±  \frac{2.78*0.0359}{\sqrt{5} }

upper limit:  0.84

lower limit: 0.75

If we compare the limits of the value obtanied by the method (Figure 1 Red line) with the reference material (Figure 1 blue line) we can see that the values obtained by the method are within the values suggested by the reference material. So, it's method is accurate.

7 0
3 years ago
How many grams of water vapor (H2O) are in a 10.2 liter sample at 0.98 atmospheres and 26ÁC? Show all work used to solve this pr
klio [65]
The answer is 7.33 g.

<span>To calculate this, we will use the the ideal gas law:
PV = nRT
where
P - pressure of the gas,
V - volume of the gas,
n - amount of substance of gas,
R - gas constant,
T - temperature of the gas.</span>

Since the amount of substance of gas (n) can be expressed as mass (m) divided by molar mass (M), then:

PV = RTm/M

It is given:

P = 0.98 atm

V = 10.2 l

T = 26°C = 299.15 K 

R = 0.082 l atm/Kmol (gas constant)

M (H2O) = 2Ar(H) + Ar(O) = 2*1 + 16 = 2 + 16 = 18g

m = ?

Since PV = RTm/M, then:

m = PVM/RT

m = 0.98 · 10.2 · 18 / 0.082 · 299.15 = 179.928/24.5303 = 7.33 g

6 0
3 years ago
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