What is the V2O5 coefficient??

Count the atoms in 717 grams of Nitrogen.

MrRissso [65]

Answer:

3.08x10^25

Explanation:

Divide the mass in grams by the molar mass of nitrogen (14.0067), then multiply the result by Avogadro's number (6.022x10^23). Then round to the correct amount of significant figures.

3 0

2 years ago

Intravenous, or IV, solutions used in medicine must exert the same osmotic pressure as blood to prevent a net flow of water into

Novosadov [1.4K]

Answer:

$\boxed{\text{7.5 bar}}$

Explanation:

1. Find the molar concentration

Assume that we have 100 mL solution.

Mass of NaCl = 0.90 g

$\text{Moles of NaCl} = \text{0.90 g} \times \dfrac{\text{1 mol}}{\text{58.44 g}} = 0.0154 \text{ mol}\\\\c = \dfrac{\text{Moles of NaCl}}{\text{Litres of solution}} = \dfrac{0.0154 \text{ mol}}{\text{0.100 L}} = \text{0.154 mol/L}$

2. Find the osmotic pressure

The formula for osmotic pressure (Π) is

Π = icRT

T = (37 + 273.15) K = 310.15 K

$\Pi = 1.88 \times \text{0.154 mol}\cdot \text{L}^{-1} \times \text{0.083 14 bar}\cdot\text{L}\cdot\text{K}^{-1}\text{mol}^{-1} \times \text{310.15 K} = \text{7.5 bar}\\\\\text{The osmotic pressure of blood at body temperature is }\boxed{\textbf{7.5 bar}}$

3 0

3 years ago

PLEASE ANSWER!! I NEED HELP :)

elena-14-01-66 [18.8K]

Explanation:

To create a p-type semiconductor, group 14 semiconductors should be d o ped by atoms from group 13.

7 0

2 years ago

How many molecules are in 6.0g of sodium phosphate?

Phantasy [73]

Answer:

2.2 x 10²² molecules.

Explanation:

Firstly, we need to calculate the no. of moles in (6.0 g) sodium phosphate:

no. of moles = mass/molar mass = (6.0 g)/(163.94 g/mol) = 0.0366 mol.

It is known that every mole of a molecule contains Avogadro's number (6.022 x 10²³) of molecules.

using cross multiplication:

1.0 mole of sodium phosphate contains → 6.022 x 10²³ molecules.

0.0366 mole of sodium phosphate contains → ??? molecules.

∴ The no. of molecules in 6.0 g of sodium phosphate = (6.022 x 10²³ molecules)(0.0366 mole)/(1.0 mole) = 2.2 x 10²² molecules.

3 0

2 years ago

1) For the following reaction, 8.44 grams of carbon (graphite) are allowed to react with 9.63 grams of oxygen gas.C(graphite)(s)

Yuki888 [10]

Answer:

The answers to the question are

(a) 13.24 g

(b) (O₂)

(c) 4.8252 g

(2) 0.662 M/L

Explanation:

(a) To solve the question we write the equation as follows

C + O₂ → CO₂

That is one mole of graphite reacts with one mole of oxygen to form one mole of carbon dioxide

number of moles of graphite = 8.44/12 = 0.703 moles

number of moles of oxygen = 9.63/32 = 0.3009

However since one mole of graphite reacts with one mole of oxygen to form one mole of carbon dioxide, therefore, 0.3009 moles of oxygen will react with 0.3009 moles of carbon to fore 0.3009 moles of CO₂

The maximum mass of carbon dioxide that can be formed = mass = moles × molar mass

= 0.3009×44 = 13.24 g

(b) The formula for the limiting reagent (O₂)

Finding the limting reagent is by checking the mole balance of the reactants available to the moles specified in th stoichiometry of the reaction and selecting the reagent with the list number of moles

(c) The mass of excess reagent = 0.703 moles - 0.3009 moles = 0.4021 moles

mass of excess reagent = 0.4021 × 12 = 4.8252 g

(2) The molarity is given by number of moles per liter of solution, therefore

molar mass of mgnesium iodide = 278.1139 g/mol, number of moles of magnesium iodide in 23 g = 23g/ 278.1139 g/mol= 8.3 × 10⁻² M

Therefore the moles in 125 mL = (8.3 × 10⁻² M)/(125 mL) = (8.3 × 10⁻² M)/(0.125 L) = 0.662 M/L

3 0

3 years ago