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3 years ago

Hydrogen may not be advantagous as a fuel because-------

Chemistry

1 answer:

sergeinik [125]3 years ago

7 0

Answer:

Hydrogen may not be advantageous as a fuel because...

- Its expensive

- Its difficult to store

- Its highly flammable

- Its dependent on fossil fuels

Explanation:

Its expensive - Not only is hydrogen gas expensive, but it also takes a lot of work to free from other elements. It is both expensive and time-consuming to produce.

Its difficult to store - Moving hydrogen is not an easy task. Moving anything more than small amounts of hydrogen was also very expensive, making it impractical.

Its highly flammable - When exposed to the atmosphere, hydrogen could potentially form explosive mixtures.

Its dependent on fossil fuels - Hydrogen energy itself is renewable. However, the process of separating it from oxygen uses non-renewable sources such as coal and oil.

~Hope this Helps!~

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what is the molarity of an HCI solution if 25.0 ml of 0.185 M NaOH is required to neutralize 0.0200 L of HCI?

butalik [34]

Answer:

Molarity of HCl solution = 0.25 M

Explanation:

Given data:

Volume of NaOH= V₁ = 25.0 mL (25/1000 = 0.025 L)

Molarity of NaOH solution=M₁ = 0.185 M

Volume of HCl solution = V₂ = 0.0200 L

Molarity of HCl solution = M₂= ?

Solution:

M₁V₁ = M₂V₂

0.185 M ×0.025 L = M₂ × 0.0200 L

M₂ = 0.185 M ×0.025 L / 0.0200 L

M₂ = 0.005M.L /0.0200 L

M₂ = 0.25 M

7 0

3 years ago

Hurry i only have till today to finsh this!

Leona [35]

Answer:

do it yourself

Explanation:

5 0

3 years ago

Read 2 more answers

How many atoms are in 25.00 g of B?

klio [65]

Answer:

There are 1.393 x 10²⁴ atoms in 25.00 g of B.

Explanation:

Hey there!

We are given a value, in grams, that we need to convert to a number of atoms.

We can convert grams to atoms by using Avogadro's Number ( $N_A$ ). This number is equivalent to $6.022 \times 10^{23}$ .

This number can be used to convert any values to:

atoms
molecules
formula units
moles

In order to do this problem, we will need to use dimensional analysis (DA). This process allows us to convert from grams to atoms.

We need to set up our ratios in order to work this out. We can use a periodic table to help us through this next part of the problem.

1. Locating the number of moles of B in the sample

We first need to find the amount of moles of boron (B) there are in the sample.

Checking a periodic table, the atomic mass in atomic mass units (amu) is 10.81 amu.

Atomic mass units can easily be converted to grams and these units can be used interchangeably.

Therefore, for each atom of boron, it weighs 10.81 grams to us. This is equivalent to the mass of one mole of boron.

To find the number of moles, we have two possible ratios we can use:

$\displaystyle \frac{1 \ mole \ B}{10.81 \ grams \ B}$

$\displaystyle \frac{10.81 \ grams \ B}{1 \ mole \ B}$

These ratios mean the same thing, but we need to convert our final unit to moles.

We are given a sample in grams, and when dividing our units, we need to keep moles.

Since the first portion of our expression is in grams, we need to have grams in the bottom of our expression.

$\displaystyle 25.00 \ \text{grams B} \ \times \frac{1 \text{mole B}}{10.81 \ \text{grams B}}$

We can now simplify the expression. Our grams B unit will cancel out, so we are therefore left with moles B remaining.

2. Locating the number of atoms in the sample

Now with our equation, we can convert our number of moles that would be solved if we stopped with the above. However, we need to convert to atoms.

We use Avogadro's number and create a ratio with that of moles.

$\displaystyle \frac{6.022 \times 10^{23}\text{atoms}}{1 \text{mole B}}$

$\displaystyle \frac{1 \text{mole B}}{6.022 \times 10^{23} \text{atoms}}$

We need to cancel out our moles and end with atoms, so we must have moles in the denominator. Therefore, we use the first ratio.

Using our previous expression, we multiply by this new ratio and solve the expression.

$\displaystyle 25.00 \ \text{grams B} \ \times \frac{1 \text{mole B}}{10.81 \ \text{grams B}} \ \times \frac{6.022 \times 10^{23}\text{atoms}}{1 \text{mole B}}$

This expression can now be operated. You will need a calculator to perform this calculation.

Our numerator is:

$[(25.00 \times 1 \times (6.022 \times 10^{23})]$

Plugging this into a calculator, we get:

$1.5055 \times 10^{25}$

Our denominator is:

$(1 \times 10.81 \times 1)$

This simplifies to:

$10.81$

Dividing our numerator and denominator:

$\displaystyle \frac{1.5055 \times 10^{25}}{10.81}$

Plugging this into a calculator, we get:

$1.392691952 \times 10^{24}$

3. Simplifying with significant figures

Now, we need to take into account that we have significant figures. We are given this original value:

$25.00$

This value has four significant figures, which means we need to round our value we received above to four significant figures.

$\approx 1.393$

Our units are added as well as our scientific notation:

$1.393 \times 10^{24} \ \text{atoms of B}$

Therefore, our final answer is choice A.

8 0

2 years ago

How do you calculate metal density?

nekit [7.7K]

Answer:

You can divide the mass by the volume to calculate the density of the metal

Explanation:

7 0

2 years ago

Which electron transition represents a gain of energy

Helen [10]

Answer:

The transition from lower energy level to higher energy level require a gain of energy.

Explanation:

When transition occur from lower energy level to higher energy level require a gain of energy. Electron could not jump unto higher energy level without gaining thew energy.

When electron jump into lower energy level from high energy level it loses the energy.

For example electron when jumped from 2nd to 3rd shell it gain energy and when in return back to 2nd shell from 3rd shell it loses energy.

The process is called excitation and de-excitation.

Excitation:

When the energy is provided to the atom the electrons by absorbing the energy jump to the higher energy levels. This process is called excitation. The amount of energy absorbed by the electron is exactly equal to the energy difference of orbits.

De-excitation:

When the excited electron fall back to the lower energy levels the energy is released in the form of radiations. this energy is exactly equal to the energy difference between the orbits. The characteristics bright colors are due to the these emitted radiations. These emitted radiations can be seen if they are fall in the visible region of spectrum.

3 0

3 years ago