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3 years ago

How many times of o are there?

Chemistry

2 answers:

GaryK [48]3 years ago

7 0

4! your welcome
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lions [1.4K]3 years ago

3 0

I would say 4 but you can figure it out or if you have a parent near by to

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Luke and Sian want to plant a vegetable garden in their yard. A soil testing kit measures the soil pH at 5.0, but the lettuce th

dexar [7]

Answer:

So, Luke and Sian has to increase the pH of the soil by adding base to it.

Explanation:

The pH is defined as the negative logarithm of the hydrogen ion concentration in their aqueous solution.

$pH=-\log[H^+]$

With increase in hydrogen ion concentration the pH value decreases.
With decrease in hydrogen ion concentration the pH value increases.

The pH of the soil after testing it on a kit comes out be 5.0, but they both need pH of the soil to 6.5.

Comparison of pH of soil:

= 5.0 < 6.5

= High hydrogen ion concentration > High hydrogen ion concentration

So, Luke and Sian has to increase the pH of the soil by adding base .Doing so will decrease the hydrogen ion concentration in the soil (where as addition of acid lower the pH of soil).

4 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 $\large \boxed{\textbf{21.6 L}}$}$

7 0

3 years ago

What tests can you do to identify gases, based on their chemical properties?

GuDViN [60]

Tests for gases
Hydrogen, oxygen, carbon dioxide, ammonia and chlorine can be identified using different tests.
Hydrogen. A lighted wooden splint makes a popping sound in a test tube of hydrogen.
Oxygen. A glowing wooden splint relights in a test tube of oxygen.

5 0

3 years ago

VA 19.75-g sample was heated by 12.35 calories. The specific heat of the sample is 0.125 cal/g°C. What was the initial temperatu

SOVA2 [1]

Answer:

31.9 °C

Explanation:

The formula for the heat q absorbed by an object is

q = mCΔT where ΔT = (T₂ - T₁)

Data:

q = 12.35 cal

m = 19.75 g

C = 0.125 cal°C⁻¹g⁻¹

T₂ = 37.0 °C

Calculations

(a) Calculate ΔT

q = mCΔT

12.35 cal = 19.25 g × 0.125 cal°C⁻¹g⁻¹ × ΔT

12.35 = 2.406ΔT °C⁻¹

ΔT = 12.35/(2.406 °C⁻¹) = 5.13 °C

(b) Calculate T₂

ΔT = T₂ - T₁

T₁ = T₂ - ΔT = 37.0 °C - 5.13 °C = 31.9 °C

The original temperature was 31.9 °C.

6 0

3 years ago

The use of evaporation and condensation to separate the partsof a mixture is called _____

Keith_Richards [23]

Answer: Evaporation Condensation is used to seperate the parts of a mixture

Explanation: This is because it is the first order of the distilation process

5 0

2 years ago