From the following heats of reaction,

Use words to describe basic cell division

ryzh [129]

Cell division<span> is the process by which a parent </span>cell<span> divides into two or more daughter cells</span>

6 0

3 years ago

In a solution made by dissolving 50 grams of calcium chloride in 1000 grams of water, which component is considered to be the so

yKpoI14uk [10]

Calcium chloride is the solute because it was dissolved in water.
Therefore water is the solvent.

4 0

4 years ago

Read 2 more answers

The normal boiling point of ethanol (C2H5OH) is 78.3 oC and its molar enthalpy of vaporization is 38.56 kJ/mol. What is the chan

Sloan [31]

Answer:

The change in entropy in the system is -231.5 J/K

Explanation:

Step 1: Data given

The normal boiling point of ethanol (C2H5OH) is 78.3 °C = 351.45 K

The molar enthalpy of vaporization of ethanol is 38.56 kJ/mol = 38560 J/mol

Mass of ethanol = 97.2 grams

Pressure = 1 atm

Step 2: Calculate the entropy change of vaporization

The entropy change of vaporization = molar enthalpy of vaporization of ethanol / temperature

The entropy change of vaporization = 38560 J/mol / 351.45 K

The entropy change of vaporization = -109.72 J/k*mol

Step 3: Calculate moles of ethanol

Moles ethanol = mass / molar mass ethanol

Moles ethanol = 97.2 grams / 46.07 g/mol

Moles ethanol = 2.11 moles

Step 4: Calculate change in entropy

For 1 mol = -109.72 J/K*mol

For 2.11 moles = -231.5 J/K

The change in entropy in the system is -231.5 J/K

7 0

3 years ago

The Henry's law constant (kH) for O2 in water at 20°C is 1.28e-3 mol/l atm. How many grams of O2 will dissolve in 3.5 L of H2O t

Sophie [7]

Answer : The mass of $O_2$ dissolved will be, 0.2365 grams

Explanation :

First we have to calculate the concentration of $O_2$ .

As we know that,

$C_{O_2}=k_H\times p_{O_2}$

where,

$C_{O_2}$ = concentration of $O_2$ = ?

$p_{O_2}$ = partial pressure of $O_2$ = 1.65 atm

$k_H$ = Henry's law constant = $1.28\times 10^{-3}mole/L.atm$

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

$C_{O_2}=(1.28\times 10^{-3}mole/L.atm)\times (1.65atm)$

$C_{O_2}=2.112\times 10^{-3}mole/L$

The concentration of $O_2$ = $2.112\times 10^{-3}mole/L$

Now we have to calculate the moles of $O_2$

$\text{Moles of }O_2=\text{Concentration of }O_2\times \text{volume of solution}$

$\text{Moles of }O_2=(2.112\times 10^{-3}mole/L)\times (3.5L)=7.392\times 10^{-3}mole$

Now we have to calculate the mass of $O_2$

$\text{Mass of }O_2=\text{Moles of }O_2\times \text{Molar mass of }O_2$

$\text{Mass of }O_2=(7.392\times 10^{-3}mole)\times (32g/mole)=0.2365g$

Therefore, the mass of $O_2$ dissolved will be, 0.2365 grams

4 0

3 years ago

The ratio of oxygen to carbon by mass in carbon monoxide is 1.33:1.00. part a find the formula of an oxide of carbon in which th

zheka24 [161]

To calculate the molecular formula, convert the mass ratio into molar ratio as follows:

mass ratio of O:C=2:1

molar mass of carbon is 12 g/mol and that of oxygen is 16 g/mol thus, number of moles can be calculated as follows:

n=\frac{m}{M}

calculating ratio,

O:C=\frac{2 g}{16 g/mol}:\frac{1 g}{12 g/mol}=\frac{1}{8}:\frac{1}{12}=12:8=3:2

thus, molecular formula will be C_{2}O_{3}

5 0

3 years ago