How many atoms are in 0.31 moles

What molecules do organism use to store energy?

olga nikolaevna [1]

glycogen and triglycerides are used to store energy in the form of covalent chemical bonds, but for living things they could possibility uses lipids

6 0

3 years ago

Chemistry, need answer ASAP!! 50 points and i'll mark brainliest if correct! ^w^

pishuonlain [190]

Answer:

615 g

Explanation:

In order to convert from moles of any given substance into grams, we have to use said substance's molar mass, as follows:

# moles * Molar mass = grams of substance.

Thus, we now calculate the molar mass of beryllium iodide, BeI₂, using the molar masses of the elements:

Molar Mass of BeI₂ = Molar Mass of Be + (Molar Mass of I)*2 = 262.821 g/mol

Finally we calculate how many grams are there in 2.34 moles of BeI₂:

2.34 mol * 262.821 g/mol = 615 g

4 0

2 years ago

What is the [h3o+] in a solution that consists of 1.0 m nh3 and 2.5 m nh4cl? [kb (nh3) = 1.8 ×10-5]a) 1.1 × 10-5

Archy [21]

Answer:

[H₃O⁺] = 1.4 × 10⁻⁹ M.

Explanation:

NH₄Cl is a salt that dissolves well in water. The 2.5 M NH₄Cl will give an initial NH₄⁺ concentration of 2.5 M.

NH₃ is a weak base. It combines with water to produce NH₄⁺ and OH⁻. The opposite process can also take place. NH₄⁺ combines with OH⁻ to produce NH₃ and H₂O. The final H₃O⁺ concentration can be found from the OH⁻ concentration. What will be the final OH⁻ concentration?

Let the increase in OH⁻ concentration be x. The initial OH⁻ concentration at room temperature is 10⁻⁷ M.

Construct a RICE table for the equilibrium between NH₃ and NH₄⁺:

$\begin{array}{c|ccccccc}\text{R}&\text{NH}_3 &+&\text{H}_2\text{O}&\rightleftharpoons &{\text{NH}_4}^{+}&+&\text{OH}^{-}\\\text{I}&1.0&&&&2.5&&1.0\times 10^{-7}\\\text{C}& -x &&&& +x &&+x\\\text{E} &1.0 - x &&&&2.5+x&&1.0\times 10^{-7}+x\end{array}$ .

The $\text{K}_b$ value for ammonia is small. The value of x will be so small that at equilibrium, $1.0 - x \approx 1.0$ and $2.5- x \approx 2.5$ .

$\displaystyle \text{K}_b = \frac{[{\text{NH}_4}^{+}]\cdot [{\text{OH}}^{-}]}{[\text{NH}_3]} \approx \frac{2.5\;(x + 1.0\times 10^{-7})}{1.0}$ .

$\displaystyle \frac{2.5\;(x + 1.0\times 10^{-7})}{1.0} = 1.8\times 10^{-5}$ .

$\displaystyle [\text{OH}^{-}] = x+1.0\times 10^{-7} = 1.8\times 10^{-5} /\left(\frac{2.5}{1.0}\right) = 7.2\times 10^{-6}\;\text{mol}\cdot\text{L}^{-}$ .

Again, $\text{K}_w = 1.0\times 10^{-14}$ at room temperature.

$\displaystyle [\text{H}_3\text{O}^{+}] = \frac{\text{K}_w}{[\text{OH}^{-}]}=\frac{1.0\times 10^{-14}}{7.2\times 10^{-6}} = 1.4\times 10^{-9} \;\text{mol}\cdot\text{L}^{-1}$

7 0

3 years ago

Which particles can be stopped by human skin? ) alpha and beta particles only alpha particles alpha and gamma particles beta and

Helga [31]

Alpha and Beta particles can be stopped by our skin. Gamma particles cannot be stopped by our skin.

7 0

3 years ago

Read 2 more answers

What is the freezing point of an aqueous solution that has 25.00 g of calcium iodide dissolved in 1250 g of water?

ozzi

Answer:

- 0.380ºC

Explanation:

The lowering of the freezing point of a solvent is a colligative property ruled by the formula:

$\Delta T_f=K_f\times m\times i$

Where:

ΔTf is the lowering of the freezing point

Kf is the molal freezing constant of the solvent: 1.86 °C/m

m is the molality of the solution

i is the van't Hoff factor: the number of particles (ions) per unit of ionic compound.

a) molality, m

m = number of moles of solute/ kg of solvent

number of moles of CaI₂ = mass in grams/ molar mass

number of moles of CaI₂ = 25.00g / 293.887 g/mol = 0.0850667mol

m = 0.0850667mol/1.25 kg = 0.068053m

b) i

Each unit of CaI₂, ideally, dissociates into 1 Ca⁺ ion and 2 I⁻ ions. Thus, i = 1 + 2 = 3

c) Freezing point lowering

ΔTf = 1.86 °C/m × 0.068053m × 3 = 0.3797ºC ≈ 0.380ºC

<h2>I have problems to upload the full answer in here, so I attach a pdf file with the whole answer.</h2>

Download pdf

6 0

3 years ago