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3 years ago

Part 2 please and explain please

Chemistry

1 answer:

vekshin13 years ago

6 0

1) 7.15 * 10^6

since the exponent of the scientific notation is positive, move 6 places to the right.

so, 7.15 * 10^6 = 7150000

2) 3.03 * 10^-8

Since the exponent of the scientific notation is negative, move the decimal point 8 places to the left.

so, 3.03 * 10^-8 = 0.0000000303

3) 4.9 * 10^-1

since the exponent of the scientific notation is negative, move 1 decimal place to the left.

so, 4.9 * 10^-1 = 0.49

4) 2.886 * 10^5

since the exponent of the scientific notation is positive, move 5 decimal places to the right.

so, 2.886 * 10^5 = 288600

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When H2(g) reacts with Cl2(g) to form HCl(g) , 185 kJ of energy are evolved for each mole of H2(g) that reacts. Write a balanced

Katyanochek1 [597]

Answer:

H₂(g) + Cl₂(g) → 2HCl(g) + 185kJ

Explanation:

In a chemical reaction, enthalpy of reaction ΔH is a thermodynamic constant that gives information if the reaction is exothermic (Produce heat if reacts) or endothermic (Consume heat if reacts).

In the reaction:

H₂(g) + Cl₂(g) → 2HCl(g) ΔH = -185kJ

As ΔH <0, the reaction is exothermic, that means, produce heat, writing a balanced thermochemical equation:

H₂(g) + Cl₂(g) → 2HCl(g) + 185kJ

The enthalpy is as a product beacause an exothermic reaction produces heat.

I hope it helps!

7 0

3 years ago

The metabolic oxidation of glucose, C6H12O6, in our bodies produces CO2, which is expelled from our lungs as a gas.

enot [183]

Answer:

$\large \boxed{\text{21.6 L}}$

Explanation:

We must do the conversions

mass of C₆H₁₂O₆ ⟶ moles of C₆H₁₂O₆ ⟶ moles of CO₂ ⟶ volume of CO₂

We will need a chemical equation with masses and molar masses, so, let's gather all the information in one place.

Mᵣ: 180.16

C₆H₁₂O₆ + 6O₂ ⟶ 6CO₂ + 6H₂O

m/g: 24.5

(a) Moles of C₆H₁₂O₆

$\text{Moles of C$_{6}$H$_{12}$O}_{6} = \text{24.5 g C$_{6}$H$_{12}$O}_{6}\times \dfrac{\text{1 mol C$_{6}$H$_{12}$O}_{6}}{\text{180.16 g C$_{6}$H$_{12}$O}_{6}}\\\\= \text{0.1360 mol C$_{6}$H$_{12}$O}_{6}$

(b) Moles of CO₂

$\text{Moles of CO}_{2} =\text{0.1360 mol C$_{6}$H$_{12}$O}_{6} \times \dfrac{\text{6 mol CO}_{2}}{\text{1 mol C$_{6}$H$_{12}$O}_{6}} = \text{0.8159 mol CO}_{2}$

(c) Volume of CO₂

We can use the Ideal Gas Law.

pV = nRT

Data:

p = 0.960 atm

n = 0.8159 mol

T = 37 °C

(i) Convert the temperature to kelvins

T = (37 + 273.15) K= 310.15 K

(ii) Calculate the volume

$\begin{array}{rcl}pV &=& nRT\\\text{0.960 atm} \times V & = & \text{0.8159 mol} \times \text{0.082 06 L}\cdot\text{atm}\cdot\text{K}^{-1}\text{mol}^{-1} \times \text{310.15 K}\\0.960V & = & \text{20.77 L}\\V & = & \textbf{21.6 L} \\\end{array}\\\text{The volume of carbon dioxide is $\large \boxed{\textbf{21.6 L}}$}$