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Mademuasel [1]
3 years ago
8

The energy on the earth

Chemistry
1 answer:
Andreas93 [3]3 years ago
3 0

Answer:

<em>The earth-atmosphere energy balance is the balance between incoming energy from the Sun and outgoing energy from the Earth. When it reaches the Earth, some is reflected back to space by clouds, some are absorbed by the atmosphere, and some is absorbed in the Earth's surface. ...</em>

Explanation:

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How many centigrams are in 0.13 oz
Paha777 [63]

Answer:

I think the answer is 368.544 centigrams

Explanation:

I hope I helped

5 0
3 years ago
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The chemical reaction 2Cu + O2 → 2CuO + 315 kJ is an _____ reaction. Endothermic, combustion Exothermic, combustion Endothermic,
makvit [3.9K]

the awnser is exothermic (combustion)


4 0
3 years ago
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Consider the perbromate anion. What is the central atom? Enter its chemical symbol. How many lone pairs are around the central a
Yanka [14]

Answer:

See explanation

Explanation:

The central atom in the perbromate ion is bromine. The chemical symbol of bromine is Br. There are no lone pairs around the central bromine atom. The ion is tetrahedral in shape hence we expect a bond angle of 109°. 27 which is the ideal tetrahedral bond angle. The actual bond angle of the prebromate ion is 109.5°. The perbromate ion is BrO4^-

The observed bond angle is very close to the ideal value because of the absence of lone pairs of electrons from the central atom in the ion.

8 0
3 years ago
This is the balances equations. C3H8 + 5O2 → 3CO2 + 4H2O How many moles of oxygen are required to produce 37.15 g CO2
AleksandrR [38]
Molar mass O2 = 31.99 g/mol

Molar mass CO2 = 44.01 g/mol

Moles ratio:

<span>C3H8 + 5 O2 = 3 CO2 + 4 H2O 
</span>
5 x 44.01 g O2 ---------------- 3 x 44.01 g CO2
( mass of O2) ------------------ 37.15 g CO2

mass of O2 = 37.15 x 5 x 44.01/ 3 x 44.01

mass of O2 = 8174.8575 / 132.03

mass of O2 = 61.916 g 

Therefore:

1 mole O2 ----------------- 31.99 g
moles O2 -------------------- 61.916

moles O2 = 61.916 x 1 / 31.99

moles = 61.916 / 31.99 => 1.935 moles of O2
4 0
3 years ago
A buffer consists of 0.120 M HNO2 and 0.150 M NaNO2 at 25°C. pka of HNO2 is 3.40. a. What is the pH of the buffer? b. What is th
Mashcka [7]

Explanation:

It is known that K_{a} of HNO_{2} = 4.5 \times 10^{-4}.

(a)  Relation between K_{a} and pK_{a} is as follows.

                       pK_{a} = -log (K_{a})

Putting the values into the above formula as follows.

                      pK_{a} = -log (K_{a})

                                    = -log(4.5 \times 10^{-4})

                                     = 3.347

Also, relation between pH and  pK_{a} is as follows.

              pH = pK_{a} + log\frac{[conjugate base]}{[acid]}

                     = 3.347+ log \frac{0.15}{0.12}

                    = 3.44

Therefore, pH of the buffer is 3.44.

(b)   No. of moles of HCl added = Molarity \times volume

                                            = 11.6 M \times 0.001 L

                                             = 0.0116 mol

In the given reaction, NO^{-}_{2} will react with H^{+} to form HNO_{2}

Hence, before the reaction:

No. of moles of NO^{-}_{2} = 0.15 M \times 1.0 L

                                           = 0.15 mol

And, no. of moles of HNO_{2} = 0.12 M \times 1.0 L

                                               = 0.12 mol

On the other hand, after the reaction :  

No. of moles of NO^{-}_{2} = moles present initially - moles added

                                          = (0.15 - 0.0116) mol

                                          = 0.1384 mol

Moles of HNO_{2} = moles present initially + moles added

                               = (0.12 + 0.0116) mol

                                = 0.1316 mol

As, K_{a} = 4.5 \times 10^{-4}

           pK_{a} = -log (K_{a})

                         = -log(4.5 \times 10^{-4})

                         = 3.347

Since, volume is both in numerator and denominator, we can use mol instead of concentration.

As, pH = pK_{a} + log \frac{[conjugate base]}{[acid]}

            = 3.347+ log {0.1384/0.1316}

            = 3.369

            = 3.37 (approx)

Thus, we can conclude that pH after the addition of 1.00 mL of 11.6 M HCl to 1.00 L of the buffer solution is 3.37.

6 0
3 years ago
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