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3 years ago

What term best describes ammonia

2 answers:

7 0

Answer: Description: Ammonia is a colourless alkaline gas and is one of the most abundant nitrogen-containing compounds in the atmosphere. It is an irritant with a characteristic pungent odor that is widely used in industry. ... Ammonia is both a neurotoxin and a metabotoxin.

Explanation:

olga2289 [7]3 years ago

7 0

Ammonia is a colorless alkaline gas. It is both a neurotoxin and metabotoxin

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How many moles of KCl are in <br> 28g of KCl?

eduard

Answer:

0.38 moles KCl

Explanation:

(28 g KCl) / (74.55 g/mol KCl) = 0.38 moles KCl

7 0

3 years ago

Read 2 more answers

Which answer is an isotope of 40/20x? <br><br> Question Pictured here:

Nesterboy [21]

An isotope of ₂₀⁴⁰X ⇒₂₀⁴⁴X

<h3>Further explanation </h3>

The elements in nature have several types of isotopes

Isotopes are atoms that have the same number of protons and a different number of neutrons.

So Isotopes are elements that have the same Atomic Number (Proton)

In the following element notation,

$\large {{{A} \atop {Z}} \right X}$

X = symbol of elemental atom

A = mass number

= number of protons + number of neutrons

Z = atomic number

= number of protons = number of electrons, on neutral elements

For symbol of isotope X :

₂₀X⁴⁰ : atomic number = 20, mass number = 40

A. ₄₀²⁰X : atomic number = 40, mass number = 20, neutrons=20

B. ₂₀⁴⁰X²⁺ : atomic number = 20, mass number = 40, neutrons=20

C. ₂₁⁴⁰X : atomic number = 21, mass number = 40, neutrons=19

D. ₂₀⁴⁴X : atomic number = 20, mass number = 44, neutrons=22

7 0

3 years ago

Which of the following provide information about the shape of an orbital

Kobotan [32]

S, P. S is a sphere P is an infinity

7 0

4 years ago

What volume of 1.25 M solution is needed to make 250 mL of 0.75 M solution?

zysi [14]

Answer:

V1= 150ml

Explanation:

C1= 1.25, V1=?, C2= 0.75, V2= 250ml

Applying dilution formula

C1V1= C2V2

V1= (0.75×250)/1.25 = 150ml

6 0

3 years ago

A 500 mL hypertonic saline solution is labeled as consisting of 5.21 % w/w NaCl. Given that the density of salt water is 1.02 g/

kicyunya [14]

<u>Answer:</u> The molarity of saline solution is 0.909 M

<u>Explanation:</u>

We are given:

5.21 w/w % NaCl

This means that 5.21 grams of NaCl is present in 100 grams of solution

To calculate mass of a substance, we use the equation:

$\text{Density of substance}=\frac{\text{Mass of substance}}{\text{Volume of substance}}$

Density of solution = 1.02 g/mL

Mass of solution = 100 g

Putting values in above equation, we get:

$1.02g/mL=\frac{100g}{\text{Volume of solution}}\\\\\text{Volume of solution}=\frac{100g}{1.02g/mL}=98.04mL$

To calculate the moalrity of solution, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}$

Given mass of NaCl = 5.21 g

Molar mass of NaCl = 58.44 g/mol

Volume of solution = 98.04 mL

Putting values in above equation, we get:

$\text{Molarity of solution}=\frac{5.21\times 1000}{58.44g/mol\times 98.04}\\\\\text{Molarity of solution}=0.909M$

Hence, the molarity of saline solution is 0.909 M

3 0

3 years ago