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son4ous [18]
4 years ago
8

This week and last, we've be talking a lot about liquid properties. For me, these always tie in to cooking. Often times when we

are boiling water for pasta or some other food, we add salt. Adding salt has a number of effects. It changes the boiling point as well as lowers the specific heat capacity. What are the implications of this in your cooking?
Chemistry
2 answers:
-Dominant- [34]4 years ago
6 0

Answer:

It makes the pasta to get hot faster and boil quicker.

Explanation:

Adding salt to water actually raises the boiling point of the water, due to a phenomenon called boiling point elevation. Essentially, adding any non-volatile solute such as salt to a liquid causes a decrease in the liquid’s vapour pressure. A liquid boils when the vapour pressure above it equals atmospheric pressure, so a lower vapour pressure means you need a higher temperature to boil the water. The reason salt makes water boil faster has to do with specific heat capacities, or the energy it takes to raise the temperature of a substance. Salt ions dissolved in water bind to water molecules, holding them stable and making it harder for them to move around. As a result, the non-salt bound water molecules receive more of the energy provided by the stove, and therefore they get hot faster and boil quicker.

Nady [450]4 years ago
4 0

Answer:Adding salt in boiling water lowers water's specific heat capacity however it also raises its boiling point and therefore increases the temperature at which the water takes to boils (and hence increases the time taken to cook food)

Explanation:When salt is added to boiling water,the specific heat capacity of water(which is generally high) decreases and hence less energy is required to heat the water to same temperature.However salt water has a high boiling point which means once the salt is added to boiling water,its boiling temperature actually rises.The salt water needs a higher temperature before it actually starts to boil.So in short salt water boils at a higher temperature than an unsalty water.

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Aqueous sulfuric acid reacts with solid sodium hydroxide to produce aqueous sodium sulfate and liquid water . If of water is pro
sergeinik [125]

The question is incomplete, here is the complete question:

Aqueous sulfuric acid reacts with solid sodium hydroxide to produce aqueous sodium sulfate and liquid water. If 12.5 g of water is produced from the reaction of 72.6 g of sulfuric acid and 77.0 g of sodium hydroxide, calculate the percent yield of water. Be sure your answer has the correct number of significant digits in it.

<u>Answer:</u> The percent yield of water in the reaction is 46.85 %.

<u>Explanation:</u>

To calculate the number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}     .....(1)

  • <u>For NaOH:</u>

Given mass of NaOH = 77.0 g

Molar mass of NaOH = 40 g/mol

Putting values in equation 1, we get:

\text{Moles of NaOH}=\frac{77.0g}{40g/mol}=1.925mol

  • <u>For sulfuric acid:</u>

Given mass of sulfuric acid = 72.6 g

Molar mass of sulfuric acid = 98 g/mol

Putting values in equation 1, we get:

\text{Moles of sulfuric acid}=\frac{72.6g}{98g/mol}=0.741mol

The chemical equation for the reaction of NaOH and sulfuric acid follows:

H_2SO_4+2NaOH\rightarrow Na_2SO_4+2H_2O

By Stoichiometry of the reaction:

1 mole of sulfuric acid reacts with 2 moles of NaOH

So, 0.741 moles of sulfuric acid will react with = \frac{2}{1}\times 0.741=1.482mol of NaOH

As, given amount of NaOH is more than the required amount. So, it is considered as an excess reagent.

Thus, sulfuric acid is considered as a limiting reagent because it limits the formation of product.

By Stoichiometry of the reaction:

1 mole of sulfuric acid produces 2 moles of water

So, 0.741 moles of sulfuric acid will produce = \frac{2}{1}\times 0.741=1.482moles of water

Now, calculating the mass of water from equation 1, we get:

Molar mass of water = 18 g/mol

Moles of water = 1.482 moles

Putting values in equation 1, we get:

1.482mol=\frac{\text{Mass of water}}{18g/mol}\\\\\text{Mass of water}=(1.482mol\times 18g/mol)=26.68g

To calculate the percentage yield of water, we use the equation:

\%\text{ yield}=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100

Experimental yield of water = 12.5 g

Theoretical yield of water = 26.68 g

Putting values in above equation, we get:

\%\text{ yield of water}=\frac{12.5g}{26.68g}\times 100\\\\\% \text{yield of water}=46.85\%

Hence, the percent yield of water in the reaction is 46.85 %.

8 0
3 years ago
What is the radioactivity of the sample of sodium -24 after 30h?
statuscvo [17]
Half life is 15 hours
so 30 hours would be 2 half lives

after 15 hours it would be at 50%
after 15 more hours it would be at 50% of 50%

so .5 x .5 = .25

it would be 25%
5 0
3 years ago
Read 2 more answers
How many moles of NH3 are produced when 3.00 mol of H2 react completely in N2+3G2 2NH3
EastWind [94]

N2 + 3H2 -----> 2NH3

3 mol 2 mol


When 3 mol H2 react completely 2 mol NH3 are formed.

3 0
4 years ago
Ammonium perchlorate is the solid rocket fuel used by the U.S. Space Shuttle. It reacts with itself to produce nitrogen gas , ch
SpyIntel [72]

Answer:

2.94 g of water are produced by this decomposition

Explanation:

We need to determine the reaction to solve this problem:

The reactant is the ammonium perchlorate

The products are N₂, Cl₂, O₂ and H₂O

The reaction is: 2NH₄ClO₄ →  N₂ + Cl₂ + 2O₂ + 4H₂O

We convert the mass of the reactant to moles:

9.6 g . 1 mol / 117.45 g = 0.0817 moles

Ratio is 2:4. Let's make a rule of three:

2 moles of perchlorate produce 4 moles of water.

Then, 0.0817 moles of perchlorate will produce (0.0817 . 4)/2 = 0.163 moles

We convert the moles of produced water, to mass (g)

0.613 mol . 18 g / 1mol = 2.94 g

8 0
3 years ago
How much water needs to be added
Olegator [25]

Answer:

165 ml

Explanation:

We are given;

Initial volume; V_a = 55 ml

Initial molarity; M_a = 3 M

Molarity of desired solution; M_b = 0.75 M

Volume of desired solution; V_b = (55 + x) ml

Where x is the volume of water to be added.

To solve for V_b, we will use the equation ;

M_a•V_a = M_b•V_b

V_b = (M_a•V_a)/M_b

V_b = (3 × 55)/0.75

V_b = 220 mL

Thus;

(55 + x) = 220

x = 220 - 55

x = 165 mL

8 0
3 years ago
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