Which cell component could be considered the distribution center of the cell?

An empty water bottle is full of air at 15°C and standard pressure. The volume of the bottle is 0.500 liter. How many moles of a

Maslowich

Answer:- 0.0211 moles of air.

Solution:- This problem is based on an ideal gas law equation.

temperature, pressure and volume are given and asked to calculate the moles of the air. Standard pressure is 1.00 atm.

T = 15 + 273 = 288 K

P = 1.00 atm

V = 0.500 L

The ideal gas law equation is, PV = nRT

where R is the universal gas law constant and its value is 0.0821 atm.L per mol per K.

n is the number of moles and its what we are asked to calculate.

For n, the equation is rearranged as:

$n=\frac{PV}{RT}$

Let's plug in the values and do the calculations:

$n=\frac{1.00atm*0.500L}{\frac{0.0821atm.K}{mol.K}*288K}$

n = 0.0211 moles

So, there are 0.0211 moles of the air in the bottle.

8 0

3 years ago

2Na + Cl2 → 2NaCI How many grams of NaCl will be produced from 33.0 g of Na and 34.0 g of Cl.?

Mademuasel [1]

54.936 grams of NaCl will be produced from 33.0 g of Na and 34.0 g of Cl.

Explanation:

Data given:

mass of Na = 33 grams

mass of $Cl_{2}$ = 24 grams

atomic mass of Na = 23 grams/moles

atomic mass of Cl = 70 grams/mole

mass of NaCl formed =?

atomic mass of NaCl = 58.44 gram/mole

balanced chemical reaction:

2Na + $Cl_{2}$ ⇒ 2NaCl

number of moles = $\frac{mass}{atomic mass}$

putting the values in the equation:

number of moles of Na = $\frac{23}{33}$

= 0.69 moles

number of moles of chlorine gas = $\frac{34}{71}$

= 0.47 moles

limiting reagent is chlorine gas in the reaction

from the balanced chemical reaction:

1 mole of chlorine gas reacts to form 2 moles of NaCl

then 0.47 moles of chlorine gas will form x moles

$\frac{2}{1}$ = $\frac{x}{0.47}$

x = 0.94 moles of NaCl is formed.

mass = atomic mass x number of moles

mass = 58.44 x 0.94

= 54.936 grams of NaCl is produced.

4 0

3 years ago

In the background information, it was stated that CaF2 has solubility, at room temperature, of 0.00160 g per 100 g of water. How

____ [38]

Answer:

2.05*10⁻⁵ moles of CF₂ can dissolve in 100 g of water.

12.82 moles of CaF₂ will dissolve in exactly 1.00 L of solution

Explanation:

First, by definition of solubility, in 100 g of water there are 0.0016 g of CaF₂. So, to know how many moles are 0.0016 g, you must know the molar mass of the compound. For that you know:

Ca: 40 g/mole
F: 19 g/mole

So the molar mass of CaF₂ is:

CaF₂= 40 g/mole + 2*19 g/mole= 78 g/mole

Now you can apply the following rule of three: if there are 78 grams of CaF₂ in 1 mole, in 0.0016 grams of the compound how many moles are there?

$moles=\frac{0.0016 grams*1 mole}{78 grams}$

moles=2.05*10⁻⁵

2.05*10⁻⁵ moles of CF₂ can dissolve in 100 g of water.

Now, to answer the following question, you can apply the following rule of three: if by definition of density in 1 mL there is 1 g of CaF₂, in 1000 mL (where 1L = 1000mL) how much mass of the compound is there?

$mass of CaF_{2}=\frac{1000 mL*1g}{1mL}$