When 18.0 g H20 is mixed with 33.5 g Fe, which is the limiting reactant?

Chemistry

1 answer:

solmaris [256]3 years ago

7 0

Answer:

si.mple fe

Explanation:

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If the concentration of the HCl used in your titration was 1.00 M and you used 23.68 mL to reach the endpoint, calculate the con

drek231 [11]

Answer: The concentration of the OH-, CB = 0.473 M.

Explanation:

The balanced equation of reaction is:

2HCl + Ca(OH)2 ===> CaCl2 + 2H2O

Using titration equation of formula

CAVA/CBVB = NA/NB

Where NA is the number of mole of acid = 2 (from the balanced equation of reaction)

NB is the number of mole of base = 1 (from the balanced equation of reaction)

CA is the concentration of acid = 1M

CB is the concentration of base = to be calculated

VA is the volume of acid = 23.65 ml

VB is the volume of base = 25mL

Substituting

1×23.65/CB×25 = 2/1

Therefore CB =1×23.65×1/25×2

CB = 0.473 M.

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3 years ago

Suppose a mercury thermometer contains 3.250g of mercury and has a capillary that is 0.180mm in diameter.. How far does the merc

Helga [31]

1) You need to get volume of both temperatures by using first attached formula V= Mass/Density $V_1 = (3.250 g)/(13.596 g/(cm^3)) = .2390 cm^3
At 25 degrees C (V_2e have:

V_2 = (3.250 g)/(13.534 g/(cm^3)) = .2401 cm^3$

2) Using the second formula you get the height of 0 degree
$h = V/(pi * r^2)


h_1 =(.2390 cm^3)/(pi*(.009 cm)^2) = 939.2 cm$

(radius in cm is $(0.180 mm) / 2 * (1 cm)/(10 mm) or .009 cm$

3) Then with h1 you can easily get the height of 25 degrees
$h_2 =(.2401 cm^3)/(pi*(.009 cm)^2) = 943.5 cm$

Subtract 943.5 cm - 939.2 cm, and obtain a rise in mercury height of 4.3 cm

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How many atoms are in 17.0 moles of Pb?

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No. of atoms=mols*avagadros no.
N=n*No
N=17 * 6.022 *10^23
No. Of atoms=(17) (6.022*10^23)

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