Which type of fuel must be the most carefully contained and protected

How many liters of 1.75 M solution could be made using 35 grams of NaCl?

dolphi86 [110]

Data:
M (molarity) = 1.75 M (mol/L)
m (mass) = 35 g
MM (molar Mass) of NaCl = 58.44 g/mol
V (volume) = ? (in liters)

Formula:
$M = \frac{m}{MM*V}$

Solving:
$M = \frac{m}{MM*V}$
$1.75 = \frac{35}{58.44*V}$
$1.75*58.44V = 35$
$102.27V = 35$
$V = \frac{35}{102.27}$
$\boxed{\boxed{V \approx 0.34\:L}}\end{array}}\qquad\quad\checkmark$

3 0

3 years ago

noble gases are used in many situations where a safe and constant environment must be maintained. for example, helium is used to

Lapatulllka [165]

Answer:

Noble gases have a complete valence electron shell making them stable and nonreactive.

Explanation:

4 0

2 years ago

Which of the following molecules would exhibit only London forces?a)CH4, BH3, and CCl4b)H2O, NH3, and CCl4c)PH3, NH3, and CCl4

Alex787 [66]

Answer:

a)CH₄, BH₃, and CCl₄

Explanation:

London dispersion forces:-

The bond for example, in the molecule is F-F, which is non-polar in nature because the two fluorine atoms have same electronegativity values.

The intermolecular force acting in the molecule are induced dipole-dipole forces or London Dispersion forces / van der Waals forces which are the weakest intermolecular force.

Out of the given options, H₂O , NH₃ exhibits hydrogen bonding which is:-

Hydrogen bonding:-

Hydrogen bonding is a special type of the dipole-dipole interaction and it occurs between hydrogen atom that is bonded to highly electronegative atom which is either fluorine, oxygen or nitrogen atom.

Thus option B and C rules out.

Hence, the correct option which represents the molecules which would exhibit only London forces is:- a)CH₄, BH₃, and CCl₄

4 0

4 years ago

a water sample is found to have a cl- content of 100ppm as nacl what is the concentration of chloride in moles per liter

ladessa [460]

Answer:

The concentration of chloride ion is $2.82\times10^{-3}\;mol/L$

Explanation:

We know that 1 ppm is equal to 1 mg/L.

So, the $Cl^-$ content 100 ppm suggests the presence of 100 mg of $Cl^-$ in 1 L of solution.

The molar mass of $Cl^-$ is equal to the molar mass of Cl atom as the mass of the excess electron in $Cl^-$ is negligible as compared to the mass of Cl atom.

So, the molar mass of $Cl^-$ is 35.453 g/mol.

Number of moles = (Mass)/(Molar mass)

Hence, the number of moles (N) of $Cl^-$ present in 100 mg (0.100 g) of $Cl^-$ is calculated as shown below:

$N=\frac{0.100\;g}{35.453\;g/mol}=2.82\times 10^{-3}\;mol$

So, there is $2.82\times10^{-3}\;mol$ of $Cl^-$ present in 1 L of solution.

5 0

4 years ago

5. After a day of testing race cars, you decide to

kvasek [131]

Noice but I don’t get why was this necessary

8 0

3 years ago