Which is a diatomic molecule? A. Ar B. CO C. CO2 D. NaCl

The equation shows one mole of ethanol fuel being burned in oxygen. Convert the energy released into its equivalent mass. C2H5OH

solong [7]

Answer:

$\boxed{\text{46.07 g}}$

Explanation:

M_r: 46.07

C₂H₅OH(ℓ) + 3O₂(g) ⟶ 2CO₂(g) + 3H₂O(ℓ); ΔH = -1418 kJ·mol⁻¹

$\text{Mass} = \text{1418 kJ} \times \dfrac{\text{1 mol}}{\text{1418 kJ}} \times \dfrac{\text{46.07 g}}{\text{1mol}}= \textbf{46.07 g}\\\\\text{1418 kJ are equivalent to $\boxed{\textbf{46.07 g}}$ of ethanol}$

4 0

4 years ago

Help please really need help

zheka24 [161]

Answer:

2.14 × 10⁻³ molecules/RSP

3.31 × 10⁻³ molecules/ESP

Explanation:

Step 1: Calculate the number of moles of Acetaminophen per Regular Strength Pill (RSP)

A Regular Strength Pill has 1.29 × 10²¹ molecules of Acetaminophen per pill. To convert molecules to moles we will use Avogadro's number: there are 6.02 × 10²³ molecules in 1 mole of molecules.

1.29 × 10²¹ molecules/RSP × 1 mol/6.02 × 10²³ molecules = 2.14 × 10⁻³ molecules/RSP

Step 2: Calculate the number of moles of Acetaminophen per Extra Strength Pill (ESP)

An Extra Strength Pill has 1.99 × 10²¹ molecules of Acetaminophen per pill. To convert molecules to moles we will use Avogadro's number: there are 6.02 × 10²³ molecules in 1 mole of molecules.

1.99 × 10²¹ molecules/ESP × 1 mol/6.02 × 10²³ molecules = 3.31 × 10⁻³ molecules/ESP

5 0

3 years ago

Which ingredients produce the best bar for growth and repair?

boyakko [2]

Weed and water and maybe cheese sticks

8 0

3 years ago

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.

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