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alexgriva [62]
2 years ago
11

A candle is placed between a concave mirror and its focal point.

Chemistry
2 answers:
natali 33 [55]2 years ago
6 0

Virtual and enlarged..............

What grade are you in btw ?

jarptica [38.1K]2 years ago
3 0

Answer: C AND D

Explanation:

You might be interested in
When aluminium and chlorine compared , which has smaller atomic size ? Why ?​
Zielflug [23.3K]
Chlorine has a smaller atomic size.
Explanation: As you move towards right of the periodic table, the atomic size decreases. This is because the number of protons is increasing towards the right of the periodic table, which applies a greater inward force on the electrons. that is why the elements on the right of the periodic table have a smaller atomic size when compared to the elements on the left. Since chlorine is on the right side of aluminium, it has a smaller atomic size.
7 0
2 years ago
Read 2 more answers
Devise a way to separate a mixture of sand, salt, and iron filings. Write the procedure using a step-by-step method and explain
aev [14]
1. Iron fillings are magnetic, so use a magnet to pull the iron fillings out of the mix.
2. Then you can put the salt and sand mixture into water, since salt is soluble, and the salt will dissolve, leaving you with sand.
3 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
2 years ago
Everyday examples of combined gas laws
Kruka [31]
Idk but look it up on google
5 0
3 years ago
Hydroxylamine is a weak molecular base with kb = 6.6 x 10-9. what is the ph of a 0.0500 m solution of hydroxylamine?
7nadin3 [17]
Answer is: pH of hydroxylamine solution is 9,23.

Kb(NH₂OH) = 1,8·10⁻⁵<span>.
c</span>₀(NH₂OH)<span> = 0,0500 M = 0,05 mol/L.
c(NH</span>₂⁺) = c(OH⁻) = x.

c(NH₂OH<span>) = 0,05 mol/L - x.
Kb = c(NH</span>₂⁺) · c(OH⁻) / c(NH₂OH).

0,0000000066 = x² /  (0,05 mol/L - x). 

solve quadratic equation: x = c(OH⁻) = 0,000018 mol/L.<span>
pOH = -log(</span>0,000018 mol/L) = 4,74.<span>
pH = 14 - 4,74 = 9,23.</span>


4 0
3 years ago
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