I need help!! i need to make it like 3:2

What is the freezing point of a solution of 465 g of sucrose c12h22o11 dissolved in 575 ml of water?

Amanda [17]

Answer:

The freezing point of the solution is - 4.39 °C.

Explanation:

We can solve this problem using the relation:

ΔTf = (Kf)(m),

where, ΔTf is the depression in the freezing point.

Kf is the molal freezing point depression constant of water = -1.86 °C/m,

density of water = 1 g/mL.

So, the mass of 575 mL is 575 g = 0.575 kg.

m is the molality of the solution (m = moles of solute / kg of solvent = (465 g / 342.3 g/mol)/(0.575 kg) = 2.36 m.

∴ ΔTf = (Kf)(m) = (-1.86 °C/m)(2.36 m) = - 4.39 °C.

∵ The freezing point if water is 0.0 °C and it is depressed by - 4.39 °C.

∴ The freezing point of the solution is - 4.39 °C.

6 0

3 years ago

Describe what half-life is. answer in a paragraph please.

dangina [55]

Half-life is defined as the quantity to reduce to half of its initial value.

Explanation:

The term half-life is generally used in nuclear physics which describes how long a stable atom can survive a radioactive decay or how quickly an unstable stable atom can undergo radioactive decay. Half-life is a constant and does not have any units.

The formula to calculate half-life:

N(t) = $\bold{N_{0} e^{-\lambda t}}$

Here N(t) is the quantity which is “not decayed”.

$N_0$ is the “initial quantity” of the substance.

λ is the “decay constant”

4 0

3 years ago

If an ocean wave has a frequency of 2 hz and a speed of 4 m/s , what is the wavelength

Anastaziya [24]

Answer:

2 m

Explanation:

6 0

3 years ago

Under what pressure did 10L of gas change to if the original pressure was 4 atm on 3L?

cluponka [151]

Answer:

11111

Explanation:

1110001001

3 0

3 years ago

Methanol, ethanol, and n−propanol are three common alcohols. When 1.00 g of each of these alcohols is burned in air, heat is lib

KengaRu [80]

Answer:

For methanol: Heat of combustion = -22.6 kJ / 0.0312 moles = -724.3590 kJ/mol (negative sign signifies release of heat)

For ethanol: Heat of combustion = -29.7 kJ / 0.0217 moles = -1368.6636 kJ/mol (negative sign signifies release of heat)

For propanol: Heat of combustion = -33.4 kJ / 0.0166 moles = -2012.0482 kJ/mol (negative sign signifies release of heat)

Explanation:

Given:

Mass of Methanol = 1.0 g

Mass of ethanol = 1.00 g

Mass of n-propanol = 1.00 g

For methanol:

2 CH₃OH + 3 O₂ ----> 2 CO₂ + 4 H₂O, ∆H₀ = -22.6 kJ/g (negative sign signifies release of heat)

1 g of methanol on combustion gives 22.6 kJ of energy

Calculation of moles of methanol:

$moles=\frac{Mass(m)}{Molar\ mass (M)}$

Molar mass of methanol = 32.04 g/mol

Thus moles of methanol = 1 g/ (32.04 g/mol) = 0.0312 moles

Hence energy in kJ/mol:

Heat of combustion = -22.6 kJ / 0.0312 moles = -724.3590 kJ/mol (negative sign signifies release of heat)

For ethanol:

C₂H₅OH + 3 O₂ ----> 2 CO₂ + 3 H₂O, ∆H₀ = -29.7 kJ/g (negative sign signifies release of heat)

1 g of ethanol on combustion gives 29.7 kJ of energy

Calculation of moles of ethanol:

$moles=\frac{Mass(m)}{Molar\ mass (M)}$

Molar mass of ethanol = 46.07 g/mol

Thus moles of ethanol = 1 g/ (46.07 g/mol) = 0.0217 moles

Hence energy in kJ/mol:

Heat of combustion = -29.7 kJ / 0.0217 moles = -1368.6636 kJ/mol (negative sign signifies release of heat)

For propanol:

2 C₃H₇OH + 9 O₂ ----> 6 CO₂ + 8 H₂O, ∆H₀ = -33.4 kJ/g , (negative sign signifies release of heat)

1 g of methanol on combustion gives 33.4 kJ of energy

Calculation of moles of methanol:

$moles=\frac{Mass(m)}{Molar\ mass (M)}$

Molar mass of methanol = 60.09 g/mol

Thus moles of methanol = 1 g/ (60.09 g/mol) = 0.0166 moles

Hence energy in kJ/mol:

Heat of combustion = -33.4 kJ / 0.0166 moles = -2012.0482 kJ/mol (negative sign signifies release of heat)

5 0

3 years ago