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mash [69]
3 years ago
7

After examining the graphs, the scientist could

Chemistry
1 answer:
Ivanshal [37]3 years ago
5 0
4
They need more oxygen due to higher temperatures
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Use the IUPAC nomenclature rules to give the name for this compound - CrPO4.
MrRissso [65]

Answer:

i need help

Explanation:

7 0
3 years ago
Write a half-reaction for the oxidation of the manganese in MnCO3(s) to MnO2(s) in neutral groundwater where the carbonate-conta
Sergeu [11.5K]

Answer:

The half-reaction for the oxidation of the manganese in MnCO_3(s) to MnO_2(s).:

MnCO_3+2H_2O\rightarrow MnO_2(s)+HCO_3^{-}+3H^+

Explanation:

MnCO_3+H_2O\rightarrow MnO_2(s)+HCO_3^{-}

Let us say the medium in which reaction is taking place is an acidic medium. And balancing in an acidic mediums done as;

Step 1: Balance all the atom beside oxygen and hydrogen atom;

MnCO_3+H_2O\rightarrow MnO_2(s)+HCO_3^{-}

Manganese and carbon are balanced.

Step 2: Balance oxygen atom adding water on the required side:

MnCO_3+H_2O+H_2O\rightarrow MnO_2(s)+HCO_3^{-}

MnCO_3+2H_2O\rightarrow MnO_2(s)+HCO_3^{-}

Step 3: Now balance hydrogen atom by adding hydrogen ion on the required side:

MnCO_3+2H_2O\rightarrow MnO_2(s)+HCO_3^{-}+3H^+

6 0
3 years ago
what is the concentration of each of these solutions expressed as a percent sucrose by mass? a) 17g sucrose mixing in 183g water
storchak [24]
A) mass of solute = 17 g

mass of solvent = 183 g

% = mass of solute / total   x 100

% = 17 / ( 17 + 183 ) x 100

% = ( 17 / 200 ) x 100

% = 0.085 x 100

= 8.5 %
________________________________

b) mass of solute = 30.0 g

mass of solvent =300.0 g

% = mass of solute / total   x 100

% = 30.0 / ( 30.0 + 300.0 ) x 100

% = (30.0 / 330.0 ) x 100

% = 0.090 x 100

<span>= 9.0 %
</span>__________________________________________

hope this helps!


8 0
3 years ago
The decomposition of nitrogen dioxide to nitrogen monoxide and oxygen gas is a second order process as suspected in the previous
lesya [120]

Explanation:

For the given reaction 2NO_{2} \rightarrow 2NO + O_{2}

Now, expression for half-life of a second order reaction is as follows.

                  t_{1/2} = \frac{1}{[A_{0}]k}     ....... (1)

Second half life of this reaction will be t_{1/4}. So, expression for this will be as follows.

          t_{1/4} = \frac{1}{k} [\frac{1}{[A]_{f}} - \frac{1}{[A_{0}]}]  ...(2)

where [A]_{f} is the final concentration that is, \frac{[A]_{0}}{4} here and [A]_{i} is the initial concentration.

Hence, putting these values into equation (2) formula as follows.

        t_{1/4} = \frac{1}{k} [\frac{4}{[A]_{0}} - \frac{1}{[A_{0}]}]

                      = \frac{3}{[A_{0}]k}     ...... (3)

Now, dividing equation (3) by equation (1) as follows.

           \frac{t_{1/4}}{t_{1/2}} = \frac{3}{[A_{0}]k} \times [A_{0}]k                  

                                    = 3

or,                       t_{1/4} = 3 t_{1/2}    

Thus, we can conclude that one would expect the second half-life of this reaction to be three times the first half-life of this reaction.

6 0
3 years ago
What element is CU?plz helpppp
Ulleksa [173]
CU is the element symbol for Copper.
8 0
4 years ago
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