based on student 1’s explanation, the same substance composes both of the samples in which of the following pairs?

What volume (in mililiters) of oxygen gas is required to react with 4.03 g of my at stp

goldenfox [79]

You must use 1880 mL of O₂ to react with 4.03 g Mg.

A_r: 24.305

2Mg + O₂ ⟶ 2MgO

Moles of Mg = 4.03 g Mg × (1 mol Mg/24.305 g Mg) = 0.1658 mol Mg

Moles of O₂ = 0.1658 mol Mg × (1 mol O₂/2 mol Mg) = 0.082 90 mol O₂

STP is 25 °C and 1 bar. At STP, 1 mol of an ideal gas has a volume of 22.71 L.

Volume of O₂ = 0.082 90 mol O₂ × (22.71 L O₂/1 mol O₂) = 1.88 L = 1880 mL

4 0

3 years ago

HHHAAAAALLLPPPP It’s due tomorrow

iris [78.8K]

Answer:

D.

Explanation:

Only 0.0035% of the electromagnetic spectrum is visible to the human eye

Hopefully this helped :)

7 0

3 years ago

Which of the following is not a clue that a chemical reaction has taken place? *

DaniilM [7]

Answer:

A pure solid is heated and turns into a pure liquid.

Explanation:

No colour change recorded, only change of state, hence this is a physical change - physical changes I.e. change of state and temperature are not chemical reactions.

7 0

3 years ago

Select the pair that has the larger atom or ion listed first.

Sedbober [7]

Answer:

Correc option: $Br^- \, , Kr$

Explanation:

size of atom : it says somthing about how many shell present in a particular atom or ion and it can also be evaluated on the basis of radius of atom.

Br^- and Kr has highest number of shell as compared to other group of species .

Na ,S , Mg ,P all are from 3rd period but Kr and Br^- in the 4th period so size of species of this group will more,

Size increases on increasring the shell number

3 0

3 years ago

A buffer solution contains 0.345 M acetic acid and 0.377 M sodium acetate . If 0.0613 moles of potassium hydroxide are added to

melamori03 [73]

Answer:

pH = 5.54

Explanation:

The pH of a buffer solution is given by the Henderson-Hasselbach (H-H) equation:

pH = pKa + log $\frac{[CH_3COO^-]}{[CH_3COOH]}$

For acetic acid, pKa = 4.75.

We calculate the original number of moles for acetic acid and acetate, using the given concentrations and volume:

CH₃COO⁻ ⇒ 0.377 M * 0.250 L = 0.0942 mol CH₃COO⁻

CH₃COOH ⇒ 0.345 M * 0.250 L = 0.0862 mol CH₃COOH

The number of CH₃COO⁻ moles will increase with the added moles of KOH while the number of CH₃COOH moles will decrease by the same amount.

Now we use the H-H equation to calculate the new pH, by using the new concentrations:

pH = 4.75 + log $\frac{(0.0942+0.0613)mol/0.250L}{(0.0862-0.0613)mol/0.250L}$ = 5.54

6 0

3 years ago