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2 years ago

What is density?

Chemistry

2 answers:

kirza4 [7]2 years ago

7 0

Answer: 3, the amount per unit volume of a particular material

Alenkinab [10]2 years ago

6 0

3 is the correct answer

You might be interested in

The density (g/L) of CO2 gas at STP ( 273 K, 1.00 atm) is

Triss [41]

The density : 1.96 g/L

<h3>Further explanation</h3>

Ideal gas Law

Pv=nRT

where

P = pressure, atm , N/m²

V = volume, liter

n = number of moles

R = gas constant = 0.082 l.atm / mol K (P= atm, v= liter),or 8,314 J/mol K (P=Pa or N/m², v= m³)

T = temperature, Kelvin

so for density,

MW CO₂ = 44 g/mol

$\tt \rho=\dfrac{P\times MW}{RT}\\\\\rho=\dfrac{1\times 44}{0.082\times 273}\\\\\rho=1.96~g/L$

3 0

2 years ago

N2(g) + 3H2(g) → 2NH3(g) How many grams of N2 are required to produce 240.0g NH3?

just olya [345]

Answer:

$\large \boxed{\text{197.4 g}}$

Explanation:

We will need a chemical equation with masses and molar masses, so, let's gather all the information in one place.

Mᵣ: 28.01 17.03

N₂ + 3H₂ ⟶ 2NH₃

m/g: 240.0

(a) Moles of NH₃

$\text{Moles of NH}_{3} = \text{240.0 g NH}_{3}\times \dfrac{\text{1 mol NH}_{3}}{\text{17.03 g NH}_{3}}= \text{14.09 mol NH}_{3}$

(b) Moles of N₂

$\text{Moles of N$_{2}$} = \text{14.09 mol NH}_{3} \times \dfrac{\text{1 mol N$_{2}$}}{\text{2 mol NH}_{3}} = \text{7.046 mol N$_{2}$}$

(c) Mass of N₂

$\text{Mass of N$_{2}$} =\text{7.046 mol N$_{2}$} \times \dfrac{\text{28.01 g N$_{2}$}}{\text{1 mol N$_{2}$}} = \textbf{197.4 g N$_{2}$}\\\\\text{The reaction requires $\large \boxed{\textbf{197.4 g}}$ of N$_{2}$}$

7 0

3 years ago

Read 2 more answers

Would someone please help?

sveta [45]

The season that is starting is winter.
The answer to 21 is (4)

3 0

3 years ago

in a typical person, the level of glucose is about 85 mg/ 100 mL of blood. if an average body contains about 11 pt of blood, how

Vanyuwa [196]

Let's note that 1 pint = 473.1765 mL, so 11 pints should be 5204.9415 mL.

We make a proportion out of the word problem

(85 mg glucose/ 100 mL) times (1 g/ 1000 mg) = 4.4242 grams of glucose

5 0

3 years ago

2AgNO3 + CaCl2 → 2AgCl + Ca(NO3)2

Harrizon [31]

It can be found that 337.5 g of AgCl formed from 100 g of silver nitrate and 258.4 g of AgCl from 100 g of CaCl₂.

<u>Explanation:</u>

2AgNO₃ + CaCl₂ → 2 AgCl + Ca(NO₃)₂

We have to find the amount of AgCl formed from 100 g of Silver nitrate by writing the expression.

$100 g \text { of } A g N O_{3} \times \frac{2 \text { mol } A g N O_{3}}{169.87 g A g N O_{3}} \times \frac{2 \text { mol } A g C l}{1 \text { mol } A g N O_{3}} \times \frac{143.32 g A g C l}{1 \text { mol } A g C l}$

= 337.5 g AgCl

In the same way, we can find the amount of silver chloride produced from 100 g of Calcium chloride.

It can be found as 258.4 g of AgCl produced from 100 g of Calcium chloride.

4 0

3 years ago