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Anvisha [2.4K]
3 years ago
6

Differences in ocean water density are larger at low latitudes near the equator because the water temperature changes often thro

ughout the year
True or False
Chemistry
1 answer:
otez555 [7]3 years ago
4 0
Answer:

False i think

Explanation:

the water temp doesn’t fluctuate much near the equator.
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By making several measurements with the same balance, a chemist obtained values of 5.224 g, 5.235 g, and 5.259 g for the mass of
stepan [7]

Answer:

a) Measurements have a good precision.

Explanation:

Accuracy is the proximity of the data to the value considered as real, in this situation we do not know the real value and we do not know if the data is accurate or not, so we can discard options b and d.

Now, precision is the proximity of the data obtained among themselves and that is what we can observe, so the appropriate answer is the option a.

6 0
3 years ago
Question
madam [21]

Answer:

The specific heat of the metal is 2.09899 J/g℃.

Explanation:

Given,

For Metal sample,

mass = 13 grams

T = 73°C

For Water sample,

mass = 60 grams

T = 22°C.

When the metal sample and water sample are mixed,

The addition of metal increases the temperature of the water, as the metal is at higher temperature, and the  addition of water decreases the temperature of metal. Therefore, heat lost by metal is equal to the heat gained by water.

Since, heat lost by metal is equal to the heat gained by water,

Qlost = Qgain

However,

Q = (mass) (ΔT) (Cp)

(mass) (ΔT) (Cp) = (mass) (ΔT) (Cp)

After mixing both samples, their temperature changes to 27°C.

It implies that , water sample temperature changed from  22°C to 27°C and metal sample temperature changed from 73°C to 27°C.

Since, Specific heat of water = 4.184 J/g°C

Let Cp be the specific heat of the metal.

Substituting values,

(13)(73°C - 27°C)(Cp) = (60)(27°C - 22℃)(4.184)

By solving, we get Cp =

Therefore, specific heat of the metal sample is 2.09899 J/g℃.

5 0
3 years ago
Draw the alkyl bromide(s) you would use in a malonic ester synthesis of ethyl 2-methyl-4-pentenoate
Rudik [331]

The structure of the alkyl bromides used in a malonic ester synthesis of ethyl 2-methyl-4-pentenoate are as drawn in the attached image.

<h3>Ethyl 2-methyl-4-pentenoate by Malonic ester synthesis.</h3>

The malonic ester synthesis is a chemical reaction characterized by the alkylation of diethyl malonate or a similar ester of malonic acid at the carbon alpha (directly adjacent) to both carbonyl groups, and subsequently converted to a substituted acetic acid.

Hence, it follows from the structure of Ethyl 2-methyl-4-pentenoate that the alkyl bromides used are Ethyl bromide and methyl bromide.

Read more on Malonic ester synthesis;

brainly.com/question/17237043

6 0
2 years ago
Read 2 more answers
Calculate the enthalpy of formation of butane, C4H10, using the balanced chemical
melisa1 [442]

Answer:

-125.4

Explanation:

Target equation is 4C(s) + 5H2(g) = C4H10

These are the data equations for enthalpy of combustion

  1. C(s) + O2(g) =O2(g) -393.5 kJ/mol * 4
  2. H2(g) + ½O2(g) =H20(l) = 285.8 kJ/mol * 5
  3. 2CO2(g) + 3H2O(l) = 13/2O2 (g) + C4H10 - 2877.1 reverse

To get target equation multiply data equation 1 by 4; multiply equation 2 by 5; and reverse equation 3, so...

Calculate 4(-393.5) + 5(-285.8) + 2877.6 and you should get the answer.

7 0
3 years ago
If the ph of hc3h5o2 is 4.2 and the ka 1.34x10^-5, what is the equilibrium concentration
n200080 [17]
<span>CH</span>₃<span>CH</span>₂<span>COOH + H</span>₂<span>O </span>↔ <span> CH</span>₃<span>CH</span>₂<span>COO</span>⁻<span> + H</span>₃<span>O</span>⁺<span> 
</span>
pH = 0.5 pKa + 0.5 pCa
0.5 pCa = pH - 0.5 pKa
              = 4.2 - (0.5 * (-log 1.34 x 10⁻⁵)) = 1.76
pCa = 3.53
Ca = antilog - 3.52 = 3 x 10⁻⁴ 
where Ca is the acid concentration 
5 0
3 years ago
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