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4 years ago

Consider the following intermediate reactions.

Chemistry

2 answers:

Alja [10]4 years ago

4 0

2.1648 kg of CH4 will generate 119341 KJ of energy.

Explanation:

Write down the values given in the question

CH4(g) +2 O2 → CO2(g) +2 H20 (g)

ΔH1 = - 802 kJ

2 H2O(g)→2 H2O(I)

ΔH2= -88 kJ

The overall chemical reaction is

CH4 (g)+2 O2(g)→CO2(g)+2 H2O (I) ΔH2= -890 kJ

CH4 +2 O2 → CO2 +2 H20

(1mol)+(2mol)→(1mol+2mol)

Methane (CH4) = 16 gm/mol

oxygen (O2) =32 gm/mol

Here 1 mol CH4 ang 2mol of O2 gives 1mol of CO2 and 2 mol of 2 H2O

which generate 882 KJ /mol

Therefore to produce 119341 KJ of energy

119341/882 = 135.3 mol

to produce 119341 KJ of energy, 135.3 mol of CH4 and 270.6 mol of O2 will require

=135.3 *16

=2164.8 gm

=2.1648 kg of CH4

2.1648 kg of CH4 will generate 119341 KJ of energy

schepotkina [342]4 years ago

3 0

Answer:

The answer is A on edge

Explanation:

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The sea water has 8.0x10^-1 cg of element strontium. Assuming that all strontium could be recovered, how many grams of strontium

PilotLPTM [1.2K]

984 grams of strontium will be recovered from 9.84x10^8 cubic meter of seawater.

Explanation:

From the question data given is :

volume of strontium in sea water= 9.84x10^8 cubic meter

(1 cubic metre = 1000000 ml)

so 9 .84x10^8 cubic meter

$\frac{9 .84x10^8}{1000000}$ = 984 ml.

density of sea water = 1 gram/ml

from the formula mass of strontium can be calculated.

density = $\frac{mass}{volume}$

mass = density x volume

mass = 1 x 984

= 984 grams of strontium will be recovered.

98400 centigram of strontium will be recovered.

Strontium is an alkaline earth metal and is highly reactive.

4 0

3 years ago

Alguien me da la fórmula de : Fosfuro ácido de estroncio?

natita [175]

la formula es Sr(OH)^2

6 0

3 years ago

How many milliliters of water at 25.0°C with a density of 0.997 g/mL must be mixed with 163 mL of coffee at 97.9°C so that the r

Stolb23 [73]

Answer:

248 mL

Explanation:

According to the law of conservation of energy, the sum of the heat absorbed by water (Qw) and the heat released by the coffee (Qc) is zero.

Qw + Qc = 0

Qw = -Qc [1]

We can calculate each heat using the following expression.

Q = c × m × ΔT

where,

c: specific heat

m: mass

ΔT: change in the temperature

163 mL of coffee with a density of 0.997 g/mL have a mass of:

163 mL × 0.997 g/mL = 163 g

From [1]

Qw = -Qc

cw × mw × ΔTw = -cc × mc × ΔTc

mw × ΔTw = -mc × ΔTc

mw × (54.0°C-25.0°C) = -163 g × (54.0°C-97.9°C)

mw × 29.0°C = 163 g × 43.9°C

mw = 247 g

The volume corresponding to 247 g of water is:

247 g × (1 mL/0.997 g) = 248 mL

8 0

3 years ago

Find the empirical formula of the compound ribose, a simple sugar often used as a nutritional supplement. A 14.229 g sample of r

MakcuM [25]

Answer:

CH2O

Explanation:

Firstly, we need to convert the masses of the elements to percentage compositions. This can be done by placing the mass of each element over the total mass multiplied by 100% . We can start with carbon.

C = 5.692/14.229 * 100 = 40%

O = 7.582/14.229 * 100 = 53.29%

H = 0.955/14.229 * 100 = 6.71%

We then proceed to divide each percentage composition by their atomic mass of 12, 16 and 1 respectively.

C = 40/12 = 3.333

O = 53.29/16 = 3.33

H = 6.71/2 = 6.71

Dividing by the smaller value which is 3.33

C = 3.33/3.33 = 1

O = 3.33/3.33= 1

H = 6.71/3.33 = 2

The empirical formula of the compound ribose is CH2O

6 0

3 years ago

Read 2 more answers

A sample of xenon occupies a volume of 736 mL at 2.02 atm and 1 °C. If the volume is changed to 416 mL and the temperature is ch

kozerog [31]

Answer:

$\large \boxed{\text{4.63 atm}}$

Explanation:

To solve this problem, we can use the Combined Gas Laws:

$\dfrac{p_{1}V_{1} }{n_{1}T_{1}} = \dfrac{p_{2}V_{2} }{n_{2}T_{2}}$

Data:

p₁ = 2.02 atm; V₁ = 736 mL; n₁ = n₁; T₁ = 1 °C

p₂ = ?; V₂ = 416 mL; n₂ = n₁; T₂ = 82 °C

Calculations:

(a) Convert the temperatures to kelvins

T₁ = ( 1 + 273.15) K = 274.15 K

T₂ = (82 + 273.15) K = 355.15 K

(b) Calculate the new pressure

$\begin{array}{rcl}\dfrac{p_{1}V_{1}}{n_{1} T_{1}} & = & \dfrac{p_{2}V_{2}}{n_{2} T_{2}}\\\\\dfrac{\text{2.02 atm}\times \text{736 mL}}{n _{1}\times \text{274.15 K}} & = &\dfrac{p_{2}\times \text{416 mL}}{n _{1}\times \text{355.15 K}}\\\\\text{5.423 atm} & = &1.171{p_{2}}\\p_{2} & = & \dfrac{\text{5.423 atm}}{1.171}\\\\ & = & \textbf{4.63 atm} \\\end{array}\\\text{The new pressure will be $\large \boxed{\textbf{4.63 atm}}$}}$

6 0

3 years ago