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amid [387]
3 years ago
12

Gold (au) goes from an oxidation number of +3 to 0 in the reaction below. 2Au3+ + 6I- ---> 2Au + 3I2. what describes the Au3+

ion?
Chemistry
2 answers:
mamaluj [8]3 years ago
7 0
In a reduction-oxidation or better known as REDOX reaction, the substance that reduces the oxidation state is known as the substance that is REDUCED. It serves as the oxidizing agent. Thus, Au3+  in this number is considered as the oxidizing agent. 
garik1379 [7]3 years ago
4 0

Answer: it is the oxidizing agent

Explanation:

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(NEED ANSWER NOW)
Anna11 [10]

Answer:it’s C

Explanation:

A distant luminous object travels rapidly away from an observer.

4 0
2 years ago
Which of the following displays the correct change in enthalpy and best describes the reaction below? 2CsCl(aq) + Na2SO4(aq) 2Na
Bond [772]
We subtract the enthalpies of the reactants from that of the products:
2\Delta
 H(NaCl)+\Delta H(Cs_2 SO_4)-2\Delta H(CsCl)-\Delta H(Na_2 SO_4) \\ 
=2(-411)+(-1400) -2(-415)-(-1380) \\ = -12 kJ
Since this is < 0, this is an exothermic reaction.

3 0
3 years ago
In the Minnesota Department of Health set a health risk limit for acetone in groundwater of 60.0 μg/L . Suppose an analytical ch
siniylev [52]

Answer:

m = 4.7 μg

Explanation:

Given data:

density of acetone = 60.0  μg/L

Volume = 79.0 mL

Mass = ?

Solution:

Formula:

d = m/v

v = 79.0 mL × 1L /1000 mL

v = 0.079 L

Now we will put the values on formula:

d = m/v

60.0  μg/L = m/0.079 L

m = 60.0 μg/L × 0.079 L

m = 4.7 μg

So health risk limit for acetone = 4.7  μg

3 0
3 years ago
What is the molarity of a NaCl stock used to make 750 ml of a dilute 10 mM solution if 5 ml of the concentrated solution was use
ch4aika [34]

Explanation:

The given data is as follows.

          M_{1} = 10 mM = 10 \times 10^{-3} M

          V_{1} = 750 ml,           V_{2} = 5 ml

          M_{2} = ?

Therefore, calculate the molarity of given NaCl stock as follows.

                  M_{1} \times V_{1} = M_{2} \times V_{2}

                  10 \times 10^{-3} \times 750 ml = M_{2} \times 5 ml

                   M_{2} = 1.5 M

Thus, we can conclude that molarity of given NaCl stock is 1.5 M.        

7 0
3 years ago
Find the horizontal range of a projectile launched at 15 degrees to the horizontal with speed of 40m/s​
vfiekz [6]

To Find :

The horizontal range of a projectile launched at 15 degrees to the horizontal with speed of 40 m/s​.

Solution :

The horizontal range of a projectile is given by :

R = \dfrac{u^2 sin 2\theta}{g} ( Here, g is acceleration due to gravity = 10 m/s² )

Putting all value in above equation :

R = \dfrac{40^2 \times sin (2 \times 15)}{10} \ m\\\\R = \dfrac{1600 \times 1}{2\times 10} \ m\\\\R = 80 \ m

Therefore, the horizontal range of projectile is 80 m.

4 0
3 years ago
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