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S_A_V [24]
2 years ago
6

Which if the following sentences is not true about nitrogen gas at room temperature

Chemistry
1 answer:
Ivan2 years ago
4 0

Answer:

Bud your answer would be D.

Hope that helps.

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Gold has a molar (atomic) mass of 197 g/mol. consider a 2.47 g sample of pure gold vapor. (a) calculate the number of moles of g
adell [148]
N = given mass/ molar mass.
n = number of moles
given mass = 2.47 g
molar mass = 197 g/mol

n = 2.47 / 197 
n = 0.01253 moles.
I'm sure you wanted to ask more than this. Just put some comments in. I can do the same.
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 <em>Henderson-Hasselbach (H-H) equation</em>:

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

For acetic acid, pKa = 4.75.

We <u>calculate the original number of moles for acetic acid and acetate</u>, using the <em>given concentrations and volume</em>:

  • 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 <u>calculate the new pH</u>, by using the <em>new concentrations</em>:

  • 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
H-C C-H name of formula
Vanyuwa [196]

Answer:

Ethyne

Explanation:

7 0
3 years ago
Who can answer these?
sesenic [268]
For the first one.. the Reactant side is Zn+2HCl produces Zncl2 and H2... and it is balanced too
6 0
3 years ago
if 45.0 ml of 1.50 M Ca(OH)2 are needed to neutralize 25.0 ml of HI of unknown concentration, what is the molarity of the HI?
ipn [44]

Answer:

M of HI = 5.4 M.

Explanation:

  • We have the rule: at neutralization, the no. of millimoles of acid is equal to the no. of millimoles of the base.

<em>(XMV) acid = (XMV) base.</em>

where, X is the no. of (H) or (OH) reproducible in acid or base, respectively.

M is the molarity of the acid or base.

V is the volume of the acid or base.

<em>(XMV) HI = (XMV) Ca(OH)₂.</em>

For HI; X = 1, M = ??? M, V = 25.0 mL.

For Ca(OH)₂, X = 2, M = 1.5 M, V = 45.0 mL.

<em>∴ M of HI = (XMV) Ca(OH)₂ / (XV) HI</em> = (2)(1.5 M)(45.0 mL) / (1)(25.0 mL) = <em>5.4 M.</em>

6 0
3 years ago
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