In the equation Na + Cl2 → 2NaCl . What is the product?

What type of ions do nonmetals naturally form?

Non-metal atoms gain an electron, or electrons, from another atom to become >negatively charged ions.

8 0

3 years ago

. Sulfur dioxide can be produced in the laboratory by the reaction of hydrochloric acid and a sulfite salt such as sodium sulfit

shutvik [7]

Answer:

Mass of SO₂ can be made from 25.0 g of Na₂SO₃ and 22 g of HCl = 12.672 g

Explanation:

SO₂( sulfur dioxide) can be produced in the lab. by the reaction of hydrochloric acid & sulphite salt such as sodium.

the balanced chemical equation is as follows

Na₂SO₃ + 2 HCl → 2 NaCl + SO₂ + H₂O

Moles of Na₂SO₃ = $\frac{Mass}{Molecular mass} =\frac{25}{126} = 0.198$

Moles of HCl = $\frac{mass}{molecular mass}=\frac{22}{36.5}= 0.6$

using mole ratio method to find limiting reagent

For sodium sulfite $\frac{mole}{stoichiometry} = \frac{0.198}{1}= 0.198$

for HCl $\frac{mole}{stoichiometry} = \frac{0.6}{2}= 0.3$

since <u>sodium sulfite</u> is <u>limiting reactant</u> for above chemical reaction

1 mole of Na₂SO₃ produce 1 mole of SO₂

0.198 mole of Na₂SO₃ produce 0.198 mole of SO₂

∴ Mass of SO₂ produce = mole x molar mass of SO₂

= 0.198 x 64

= 12.672 g

8 0

3 years ago

Calculate the volume of the acid solution and the volume of the conjugate base solution that would be needed to prepare a buffer

bogdanovich [222]

Answer:

Explanation:

This can be contradictory, depending on whether the 0.1 M

is the total species concentration or the concentration of each of the two components. I'll consider this to be the former...

VA− = 9.125 mL

VHA = 15.875 mL

The Henderson-Hasselbalch equation is:

pH = pKa + log [A−][HA]

We have a pH 4.5

solution of acetic acid and acetate, so from there we can get the ratio of weak acid to conjugate base:

[A−][HA]=10

pH − pKa = 104.5 − 4.74 = 0.5754

Now, if the total concentration is

0.10 M , then:

[HA] + [A−] 0.5754

[HA] = 0.10 M

⇒[HA] = 0.10 M 1.0000 +0.5754

= 0.0635 M

−−−−−−−−

⇒[A−] = 0.0365 M

−−−−−−−−

and these concentrations are AFTER mixing. Since the total volume is 50 mL , or 0.050 L, the mols of each component (which are constant!) are:

nA − = 0.0365 molL × 0.050L =

0.001825 mols

−−−−−−−−−−−−

nHA = 0.0635 molL × 0.050L =

0.003175 mols

−−−−−−−−−−−−

So, if both of the starting concentrations were

0.20 M, we can find the volume they each start with:

VA − = 1 L0.20mols

A− × 0.001825mols A− = 0.009125 L = 9.125 mL

−−−−−−−−

VHA = 1 L 0.20 mols HA × 0.003175

mols HA = 0.015875 L = 15.875 mL

−−−−−−−−−

And this should make sense, because the total starting volume is

25.000 mL , the total ending volume is twice as large; the total species concentration is half the concentration that both species started with.

6 0

3 years ago

A silver cube with an edge length of 2.42 cm and a gold cube with an edge length of 2.75 cm are both heated to 85.4 ∘C and place

kakasveta [241]

Answer:

Explanation:

Volume of silver cube = 2.42³ = 14.17 cm³

mass of silver cube = volume x density

= 14.17 x 10.49 = 148.64 gm

Volume of gold cube = 2.75³ = 20.8 cm³

mass of gold cube = 20.8 x 19.3 = 401.44 gm

specific heat of silver and gold are .24 and .129 J /g°C

mass of 112 mL water = 112 g

Heat absorbed = heat lost = mass x specific heat x temperature fall or rise

Heat lost by metals

= 148.64 x .24 x ( 85.4 -T) + 401.44 x .129 x ( 85.4 - T )

= (35.67 + 51.78 ) x ( 85.4 - T )

87.45 x ( 85.4 - T )

= 7468.23 - 87.45 T

Heat gained by water

= 112 x 1 x ( T - 20.5 )

= 112 T - 2296

Heat lost = heat gained

7468.23 - 87.45 T = 112 T - 2296

199.45 T = 9764.23

T = 48.95° C

7 0

3 years ago

Which compound is an exception to the octet rule? H20 HCI CCI4 CIF3

Reika [66]

Answer:

clf3

Explanation:

it occupied more than 8 valence electrons

8 0

2 years ago