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3 years ago

How do you boil water less than 100

1 answer:

4 0

Answer: When pressure cooker is used steam is trapped in the pot and it increases the pressure above the water. this causes the water to boil at a temperature more than 100 C. when you pull back on the plunger, the pressure in the syringe is reduced and the liquid boils at a lower temperature

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Which temperature scale does not possess negative values? A. Fahrenheit B. Celsius C. Kelvin D. No temperature scale has negativ

dimaraw [331]

Kelvin doesn't have negativity iim pretty sure

8 0

3 years ago

Single and double replacement reactions practice WS CuCl2 + F2 » CuF2 + Cl2

Lapatulllka [165]

CuCl2+F2—>CuF2+Cl2.
This is a single replacement because there is one compound and one element. Picture Cu as ‘A’ Cl2 as ‘B’ and F2 as ‘C.’ So AB+C—>AC+B. A and B “broke up” and that resulted to A going with C to create the compound CuF2 leaving Cl2 alone.

7 0

3 years ago

When 0.620 gMngMn is combined with enough hydrochloric acid to make 100.0 mLmL of solution in a coffee-cup calorimeter, all of t

OleMash [197]

Answer:

The enthalpy change during the reaction is -199. kJ/mol.

Explanation:

$Mn(s)+2HCl(aq)\rightarrow MnCl_2(aq)+H_2(g)$

Mass of solution = m

Volume of solution = 100.0 mL

Density of solution = d = 1.00 g/mL

$m=1.00 g/mL\times 100.0 mL = 100 g$

First we have to calculate the heat gained by the solution in coffee-cup calorimeter.

$q=m\times c\times (T_{final}-T_{initial})$

where,

m = mass of solution = 100 g

q = heat gained = ?

c = specific heat = $4.18 J/^oC$

$T_{final}$ = final temperature = $23.1^oC$

$T_{initial}$ = initial temperature = $28.9^oC$

Now put all the given values in the above formula, we get:

$q=100 g \times 4.18 J/^oC\times (28.9-23.1)^oC$

$q=2,242.4 J=2.242 kJ$

Now we have to calculate the enthalpy change during the reaction.

$\Delta H=-\frac{q}{n}$

where,

$\Delta H$ = enthalpy change = ?

q = heat gained = 2.242 kJ

n = number of moles fructose = $\frac{\text{Mass of manganese}}{\text{Molar mass of manganese}}=\frac{0.620 g}{54.94 g/mol}=0.0113 mol$

$\Delta H=-\frac{2.242 kJ}{0.0113 mol }=-199. kJ/mol$

Therefore, the enthalpy change during the reaction is -199. kJ/mol.

8 0

3 years ago

what is the percent by mass of a solution that contains 30 grams of potassium nitrite in 0.5 kilograms of water?

Makovka662 [10]

Answer:

5.66 %.

Explanation:

mass percent is the ratio of the mass of the solute to the mass of the solution multiplied by 100.

mass % = (mass of solute/mass of solution) x 100.

mass of potassium nitrite = 30.0 g,

mass of the solution = mass of water + mass of potassium nitrite = 500.0 g + 30.0 g = 530.0 g.

∴ mass % = (mass of solute/mass of solution) x 100 = (30.0 g/530.0 g) x 100 = 5.66 %.

5 0

3 years ago

A sample of food containing 27 g of fat, 48 g of carbohydrates and 20 g of protein is burned in a bomb calorimeter. In a perfect

rosijanka [135]

Answer:

38.3958 °C

Explanation:

As,

1 gram of carbohydrates on burning gives 4 kilocalories of energy

1 gram of protein on burning gives 4 kilocalories of energy

1 gram of fat on burning gives 9 kilocalories of energy

Thus,

27 g of fat on burning gives 9*27 = 243 kilocalories of energy

20 g of protein on burning gives 4*20 = 80 kilocalories of energy

48 gram of carbohydrates on burning gives 4*48 = 192 kilocalories of energy

Total energy = 515 kilocalories

Using,

$Q=m_{water}\times C_{water}\times (T_f-T_i)$

Given: Volume of water = 23 L = 23×10⁻³ m³

$Density=\frac{Mass}{Volume}$

Density of water= 1000 kg/m³

So, mass of the water:

$Mass\ of\ water=Density \times {Volume\ of\ water}$

$Mass\ of\ water=1000 kg/m^3 \times {0.023\ m^3}$

Mass of water = 23 kg

Initial temperature = 16°C

Specific heat of water = 0.9998 kcal/kg°C

$515=23\times 0.9998\times (T_f-16)$

Solving for final temperature as:

Final temperature = 38.3958 °C

8 0

3 years ago

How do you boil water less than 100​

How do you boil water less than 100