How many grams are in 0.500 mol of Na (molar mass of Na = 22.99 g/mol)?

why can scientists ignore the forces of attraction between particles in a gas under ordinary conditions

Galina-37 [17]

The forces of attraction between particles in their gaseous state seems to be nonexistential.
therefore scientist would care less. however another state after gas which is plasma has a lesser force of attraction than the gaseous state.

8 0

3 years ago

Example of oxidizing agents include

Ivahew [28]

Explanation:

Substances generally tends to specialize as either oxidizing or reducing agents.

An oxidizing agent is an electron acceptor which causes a co-reactant to be oxidized in a reaction.

Examples are:

Non-metals especially oxygen and the halogens.

Other examples are H₂SO₄ , HNO₃, KMnO₄, K₂Cr₂O₇

learn more:

Oxidizing and reducing agents brainly.com/question/5558762''

#learnwithBrainly

4 0

3 years ago

Leave the answer blank if no precipitate will form. (Express your answer as a chemical formula.) Formula of precipitate ZnCl2(aq

Rasek [7]

Answer:

1. Zn(OH)₂ (s)

2. Ag₂CO₃ (s)

3. Ni₃(PO₄)₂(s)

4. No reaction

5. (NH₄)₂CO₃(s)

Explanation:

Let's state the equations and we analyse some solubility and precipitation information:

ZnCl₂(aq) + 2KOH(aq) → Zn(OH)₂ (s) + 2KCl (aq)

All the salts from the halogens with group 1, are soluble.

The OH⁻ reacts to Zn cation in order to produce a precipitate. This is ok, but if the base is in excess, the Zn(OH)₂ will be soluble

K₂CO₃(aq) + 2AgNO₃(aq) → Ag₂CO₃ (s) ↓ + 2KNO₃(aq)

All salts from nitrate are soluble

All salts from carbonates are insoluble

2(NH₄)₃PO₄(aq) + 3Ni(NO₃)₂(aq) → Ni₃(PO₄)₂(s) ↓ + 6NH₄NO₃(aq)

Salts from phosphates are insoluble

All salts from nitrate are soluble

NaCl(aq) + KNO3(aq) → NO REACTION

All salts from nitrate are soluble

All the salts from the halogens with group 1, are soluble

Na₂CO₃(aq) + 2NH₄Cl(aq) → 2NaCl(aq) + (NH₄)₂CO₃(s) ↓

All salts from carbonates are insoluble

All the salts from the halogens with group 1, are soluble

7 0

2 years ago

Read 2 more answers

A student measures the mass of a 6.0 cm3 block of brown sugar to be 10.0 g. What is the density of the brown sugar?

Alex17521 [72]

Answer:

1.67g/cm3

Explanation:

The formula for density is $d=\frac{m}{v}$ . The m variable stands for mass and the v variable stands for volume.

The mass of the brown sugar is 10.0g and the volume is 6.0cm3, so we can plug those values into the equation.

$d=\frac{10g}{6cm^{3} }$

$d=1.67$ $\frac{g}{cm^{3} }$

Rounded to 3 significant figures, the density of the block of brown sugar is 1.67 g/cm3. If the mass is in grams and the volume is in cm3, the unit for the final answer is $\frac{g}{cm^{3} }$ (grams per centimetres cubed).

8 0

3 years ago

Read 2 more answers

2. What group is the atom in?<br> 2 or 2A<br> 6<br> 14 <br> 1

Juliette [100K]

Answer:

$\Large \boxed{\mathrm{Group \ 4}}$