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MrRissso [65]
2 years ago
7

If the concentration of a reactant is tripled (all other things remain constant), and the reaction rate increases nine times, wh

at is the reaction order with respect to the tripled reactant? Enter your answer as a number.If the concentration of a reactant is increased 1.5 times (all other things remain constant), and the reaction rate increases 2.25 times, what is the reaction order with respect to the reactant? Enter your answer as a number.If the concentration of a reactant is tripled (all other things remain constant), and the reaction rate remains constant, what is the reaction order with respect to the tripled reactant? Enter your answer as a number.When 29.0 mL of 0.220 M KIO3 is combined with 38.0 mL of H2SO3 and 50.0 mL of water, what is the resulting concentration of KIO3?
Chemistry
1 answer:
Nikitich [7]2 years ago
7 0
<h3>Answer:</h3><h3>\sqrt{ \times  { -  {2 { \times x3 \times \frac{?}{?} }^{2} }^{2}  \times \frac{?}{?} }^{2}  \times \frac{?}{?} }  \\</h3>

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the stability of atomic nuclei is related to the _____. ratio of protons to electrons ratio of neutrons to protons number of pro
jenyasd209 [6]
<span>ratio of neutrons to protons</span>
6 0
2 years ago
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If your lungs were filled with air containing this level of lead, how many lead atoms would be in your lungs? (Assume a total lu
kolbaska11 [484]

Answer:

1.505×10^23 atoms of lead

Explanation:

Volume of lead in the lungs = total volume of lungs = 5.60L

1 mole = 22.4L

5.6L of lead = 5.6/22.4 = 0.25 mole

From Avogadro's law

1 mole of lead contains 6.02×10^23 atoms of lead

0.25 mole of lead = 0.25×6.02×10^23 = 1.505×10^23 atoms of lead

6 0
2 years ago
84. Heavy water, D2O (molar mass = 20.03 g mol–1), can be separated from ordinary water, H2O (molar mass = 18.01), as a result o
ipn [44]

Answer:

relative rate of diffusion is 1.05

Explanation:

According to Graham's law of difussion:

Rate of diffusion is inversely proportional to the square root of molecular weight of a molecule.

For two given molecules:

\frac{(rate)_{1}}{(rate)_{2}}=\frac{\sqrt{M_{2}} }{\sqrt{M_{1}}}

The given molecules are

Water = 18.01

Heavy water =20.03

Thus the relative rate of diffusion will be:

\frac{(rate)_{water}}{(rate)_{heavywater}}=\sqrt{ \frac{20.03}{18.01}}=1.05

8 0
2 years ago
How many grams of oxygen are in 56 g of c2h2o2?
jolli1 [7]
First, find percent of oxygen: atom/molecule... there are 2 atoms of Oxygen so: O2/C2H2O2 which is: 32g O2 / 58g C2H2O2 =32/58. 

<span>Next, multiply this by the total mass (56g C2H2O2) and the units will cancel out (g*g/g -> g) leaving you with the mass of Oxygen: </span>

<span>56g C2H2O2 * 32g O2/58g C2H2O2 = 56*32/58= 31g</span>
3 0
2 years ago
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N2(g) + 3h2(g) ⇌ 2nh3(g) the equilibrium constant kc at 375°c is 1.2. starting with [h2]0 = 0.76 m, [n2]0 = 0.60 m and [nh3]0 =
nekit [7.7K]
N2(g)   +  3  H2(g) =  2NH3(g)

Qc =  (NH3^2)   / { (N2)(H)^3)}

Qc=  0.48^2  /{ ( 0.60) (0.760^3) }=  0.875

Qc < Kc  therefore  the  equilibrium  will   shift     to  the  right.  This  implies  that  Nh3  concentration  will    increase    
8 0
2 years ago
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