The moon makes light the same way that the sun does.True or False

The names and chemical formulae of some chemical compounds are written in the first two columns of the table below. Each compoun

charle [14.2K]

Answer:

See below

Explanation:

Name Formula Major species

Zinc iodide ZnI₂ H₂O(ℓ), I⁻(aq), Zn²⁺(aq),

Nitrogen(I) oxide N₂O H₂O(ℓ), N₂O(aq)

Sodium nitrite NaNO₂ H₂O(ℓ), Na⁺(aq), NO₂⁻(aq)

Glucose C₆H₁₂O₆ H₂O(ℓ), C₆H₁₂O₆(aq)

Nickel(II) iodide NiI₂ H₂O(ℓ), I⁻(aq), Ni²⁺(aq)

Glucose and nitrogen(I) oxide are covalent compounds. They do not dissociate in solution.
The compounds containing metals are ionic. They produce ions in solution.
ZnI₂ and NiI₂ produce twice as many iodide ions as metal ions.

6 0

3 years ago

How did oxygen first enter Earth's atmosphere?

forsale [732]

Answer:

I think c biological processes

8 0

2 years ago

I just need to know the answer for six please and thank you!

nydimaria [60]

Answer:

Explanation:

3(NH₄)₂Cr₂O₇

8 0

2 years ago

A gas has a volume of 1.75L at -23°C and 150.0 kPa. At what temperature would the gas occupy 1.30L at 210.0 kPa?

Nastasia [14]

Answer:

At -13 $^{0}\textrm{C}$ , the gas would occupy 1.30L at 210.0 kPa.

Explanation:

Let's assume the gas behaves ideally.

As amount of gas remains constant in both state therefore in accordance with combined gas law for an ideal gas-

$\frac{P_{1}V_{1}}{T_{1}}=\frac{P_{2}V_{2}}{T_{2}}$

where $P_{1}$ and $P_{2}$ are initial and final pressure respectively.

$V_{1}$ and $V_{2}$ are initial and final volume respectively.

$T_{1}$ and $T_{2}$ are initial and final temperature in kelvin scale respectively.

Here $P_{1}=150.0kPa$ , $V_{1}=1.75L$ , $T_{1}=(273-23)K=250K$ , $P_{2}=210.0kPa$ and $V_{2}=1.30L$

Hence $T_{2}=\frac{P_{2}V_{2}T_{1}}{P_{1}V_{1}}$

$\Rightarrow T_{2}=\frac{(210.0kPa)\times (1.30L)\times (250K)}{(150.0kPa)\times (1.75L)}$

$\Rightarrow T_{2}=260K$

$\Rightarrow T_{2}=(260-273)^{0}\textrm{C}=-13^{0}\textrm{C}$

So at -13 $^{0}\textrm{C}$ , the gas would occupy 1.30L at 210.0 kPa.

5 0

3 years ago

A chemist must dilute 47.2 mL of 150. mM aqueous sodium nitrate solution until the concentration falls to . He'll do this by add

erma4kov [3.2K]

Answer:

0.295 L

Explanation:

It seems your question lacks the final concentration value. But an internet search tells me this might be the complete question:

" A chemist must dilute 47.2 mL of 150. mM aqueous sodium nitrate solution until the concentration falls to 24.0 mM. He'll do this by adding distilled water to the solution until it reaches a certain final volume. Calculate this final volume, in liters. Be sure your answer has the correct number of significant digits. "

Keep in mind that if your value is different, the answer will be different as well. However the methodology will remain the same.

To solve this problem we can use the formula C₁V₁=C₂V₂

Where the subscript 1 refers to the concentrated solution and the subscript 2 to the diluted one.

47.2 mL * 150 mM = 24.0 mM * V₂

V₂ = 295 mL

And converting into L becomes:

295 mL * $\frac{1 L}{1000mL}$ = 0.295 L

6 0

3 years ago