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WARRIOR [948]
3 years ago
12

3,010,000 into scientific notation

Chemistry
2 answers:
sdas [7]3 years ago
6 0

Answer:

3.01 x 10^6

Explanation:

We start at the farthest right of this number, then shifting it 6 places to the left until we get only one whole digit in our answer. In this case are whole digit would be the number "3" from the millions place written out as 3.01 x 10^6. Because we started at the right of the number and shifted to the left 6 times, we get a positive exponent of "6."

Here is an example of a negative exponent in scientific notation:

skelet666 [1.2K]3 years ago
3 0

Answer:

3.01 x 10 to the power of 6

Explanation:

Step 1

To find a, take the number and move a decimal place to the right one position.

Original Number: 3,010,000

New Number: 3.010000

Step 2

Now, to find b, count how many places to the right of the decimal.

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ki77a [65]
E.g. in H3PO4 (O, -2).
8. The sum of the oxidation states of all the atoms in a species must be equal to net charge on the species. e.g. Net Charge of HClO4 = 0, i.e. [+1(H)+7(Cl)-2<span>*4(O)] = 0.</span>
4 0
3 years ago
If 8.00 g NH4NO3 is dissolved in 1000 g of water, the water decreases in temperature from 21.00 degrees Celsius to 20.39 degrees
telo118 [61]

Answer:

25.7 kJ/mol

Explanation:

There are two heats involved.

heat of solution of NH₄NO₃ + heat from water = 0

q₁  +  q₂  =  0

n  =  moles of NH₄NO₃  =  8.00 g NH₄NO₃  ×  1 mol NH₄NO₃/80.0 g NH₄NO₃          

∴ n =   0.100 mol NH₄NO₃

q₁ = n * ΔHsoln = 0.100 mol * ΔHsoln

m  =  mass of solution  =  1000.0 g + 8.00 g  =  1008.0 g

q₂  =  mcΔT  = 58.0 g  ×  4.184 J°C⁻¹  g⁻¹  × ((20.39-21)°C) = -2570.19 J

q₁  +  q₂  =  0.100 mol  ×ΔHsoln  – 2570.19 J  =  0

ΔHsoln  =  +2570.19 J  /0.100 mol  =  +25702 J/mol  =  +25.7 kJ/mol

7 0
3 years ago
The half-reaction occurring at the cathode in the balanced reaction shown below is __________?2Ag(s) + 1/2 O2(g) + 2H+(aq) → 2Ag
Shtirlitz [24]

Answer:

Correct choice are C and D (they are both, the same).

Explanation:

Cathode is the positive(+) electrode where a reduction occurs.

Reduction is the chemical reaction where the oxidation state is reduced.

2Ag(s) + 1/2 O2(g) + 2H+(aq) → 2Ag+(aq) + H2O (l)

A. 2H2O (l) → O2 (g) + 4H+ (aq) + 4e-

B. 2Ag (s) → 2Ag+ (aq) + 2e-

C. 1/2 O2 (g) + 2H+ (aq) + 2e- → H2O (l)

D. 1/2 O2 (g) + 2H+ (aq) + 2e- → H2O (l)

C or D, are ok. They are the same equation.

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3 0
4 years ago
Consider the chemical equations shown here.
vichka [17]

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Explanation:

The overall equation for the reaction that produces  P₄0₁₀ is :

P₄0₆_{s} + 20₂_{g}   ⇒    P₄0₁₀_{s}

Now let us derive this equation:

Given equations:

   P₄_{s} + 30₂_{g}  ⇒ P₄0₆_{s}  equation 1;

   P₄_{s} + 50₂_{g} ⇒  P₄0₁₀_{s}  equation 2;

To get the overall combined equation, the equation 1 must be reversed and added to equation 2:

            P₄0₆_{s} ⇒ P₄_{s} + 30₂_{g}   equation 3

                      +

            equation 2:

 P₄_{s} + 50₂_{g}  +    P₄0₆_{s}  ⇒  P₄0₁₀_{s}  +  P₄_{s} + 30₂_{g}  

cancelling specie that appears on both sides and removing excess oxygen gas on the reactant side gives;

   

                  P₄0₆_{s} + 20₂_{g}   ⇒    P₄0₁₀_{s}

learn more:

Net equation brainly.com/question/2947744

#learnwithBrainly

5 0
3 years ago
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8 0
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