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zlopas [31]
2 years ago
10

Define the following terms. a. Capillary action b. Chromatogram

Chemistry
1 answer:
jonny [76]2 years ago
6 0

Answer:

<em>1</em><em>.</em><em>Capillary action is important for moving water (and all of the things that are dissolved in it) around. </em>

<em>2</em><em>.</em><em>the pattern formed on an adsorbent medium by the layers of components separated by chromatography. </em>

Explanation:

hope this help you

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Answer:

when cell is exposed to radiation we get brain tumour or cancer.

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2 years ago
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Round to 4 significant figures: 42,561
arsen [322]

6 is the fourth significant figures

if the number behind it is 5 or more then 5, you must add 1 to the number and ALL the number behind it will turn into 0

so that the answer is 42560

7 0
3 years ago
When 0.200 grams of Al reacts with 15.00 mL of a 0.500 M copper(II) chloride solution, how many moles of solid Cu would be produ
Naddik [55]
When The balanced equation is:
2Al + 3CuCl2 ⇒3 Cu + 2AlCl3
So, we want to find the limiting reactant:
1- no. of moles of 2Al = MV/n = (Wt * V )/ (M.Wt*n*V) = Wt / (M.Wt *n)
        
where M= molarity, V= volume per liter and n = number of moles in the balanced equation.
by substitute: 
∴ no. of moles of 2Al = 0.2 / (26.98 * 2)= 0.003706 moles.
2- no.of moles of 3CuCl2= M*v / n = (0.5*(15/1000)) / 3= 0.0025 moles.
So, CuCl2 is determining the no.of moles of the products.
∴The no. of moles of 3Cu = 0.0025 moles.
∴The no.of moles of Cu= 3*0.0025=  0.0075 moles.
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 = 0.0075 * 63.546 =0.477 g


3 0
2 years ago
149.3 g of H2O at 95 ◦C is poured over 412 g Fe at 5 ◦C in an insulated vessel. What is the final temperature? The specific heat
abruzzese [7]
You are correct, but you needn't worry about the signs so much. Just remember that the negative sign is used to denote a loss of energy; since the water is hotter, it will be losing energy (-Q) and the iron will gain energy (Q). Now, we substitute the values:
-149.3 * 4.184 * (T - 95) = 412 * 0.44 * (T - 5)
Solving this equation for T,
T = 74.8 °C
6 0
2 years ago
A helium filled ballon had a volume of 8.50 L on the ground at 20.0 C and a pressure of 750.0 Torr. After the ballon was release
Marrrta [24]

Answer:

V_2=12.1L

Explanation:

Hello!

In this case, according to the given data of volume, pressure and temperature, it is possible to infer this problem can be solved via the combined gas law:

\frac{P_1V_1}{T_1} =\frac{P_2V_2}{T_2}

Thus, regarding the question, we evidence we need V2, but first we make sure the temperatures are in Kelvins:

T_1=20+273=293K\\\\T_2=-40+273=233K

Then, we obtain:

V_2=\frac{P_1V_1T_2}{T_1P_2}\\\\V_2=\frac{0.987atm*8.50L*233K}{293K*0.550atm}\\\\V_2=12.1L

Best regards!

5 0
2 years ago
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