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frosja888 [35]
3 years ago
13

The transfer of thermal energy from a warmer object to a cooler object is called

Chemistry
1 answer:
NISA [10]3 years ago
3 0

Answer:

c

Explanation:

somehow i didnt have to look this up to help u lol i learned this is 6th grade

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According to the below equation, how many moles of SO2 are required to generate 1.43×1024 water molecules?
podryga [215]
Make sure the equation is always balanced first. (It is balanced for this question already) 6.022 x 10^23 is Avogadro’s number. In one mole of anything there is always 6.022 x 10^23 molecules, formula units, atoms. For one mol of an element/ compound use molar mass (grams).

Multiply everything on the top = 8.61x10^47
Multiple everything on bottom= 1.20x10^24
Divide top and bottom = 7.15x10^23

Answer: 7.15x10^23 mol SO2
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3 years ago
Consider the formula for glucose: C6H12O6. What does this indicate about the relationship of the reactants to glucose?
stellarik [79]

The correct option is D.

The reactants that combine together to form glucose are carbon dioxide, water and energy from the sun. Six molecules of carbon dioxide combine with six molecules of water in the presence of sunlight to form glucose. The chemical equation for the reaction is given below"

6CO2 + 6H2O + Sunlight = C6H12O6.

6 0
2 years ago
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An object has a volume of 2 milliliters and a mass of 10 grams, calculate the density of the object
MArishka [77]
5000 kilogram/cubic meter
5 0
3 years ago
A piece of lithium is placed on the surface of some water in a beaker. hydrogen is given off. Lithium hydroxide is formed. What
Gala2k [10]
<span>A piece of lithium is placed on the surface of some water in a beaker. Hydrogen is given off and Lithium hydroxide is formed. The word equation would be as follows:
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lithium<span> + </span>water<span>→ </span>lithium hydroxide<span> + </span>hydrogen

Hope this answers the question. Have a nice day.
8 0
2 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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