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2 years ago

What is the mass, in grams, of 1.15 mol of water, H2O?

Chemistry

1 answer:

garri49 [273]2 years ago

8 0

M = n x Mr

Mr of H20 - 18.006

M = 1.15mol x 18.006g/mol
= 20.7069g
= 20.7g (3sfg)

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An oxide of nitrogen is 25.9% N by mass, has a molar mass of 108 g/mol, and contains no nitrogen-nitrogen or oxygen-oxygen bonds

Sonja [21]

Answer:

The compound is N2O4

Explanation:

We have certain important pieces of information about the compound;

1) it is an oxide (a binary compound of nitrogen and oxygen)

2) there are no N-N bonds present

3) there are no O-O bonds present

Since it contains only nitrogen and oxygen then nitrogen accounts for 25.9% of the molecule by mass then oxygen should account for (100-25.9) = 74.1% oxygen

Relative atomic mass of oxygen = 16

Relative atomic mass of nitrogen = 14

We now deduce the empirical formula

Nitrogen. Oxygen

25.9/14. 74.1/16

1.85/1.85. 4.6/1.85 (divide through by the lowest ratio)

1 2

Empirical formula is NO2

To find the molecular formula

(NO2)n = 108

(14+2(16))n= 108

46n=108

n= 108/46

n= 2

Therefore molecular formula= N2O4

4 0

2 years ago

What is 44.25 l to ml

Oliga [24]

44250

hope this helps

6 0

2 years ago

A chemist must prepare of 800.0 ml potassium hydroxide solution with a pH of 13.00 at 25Â°.

ArbitrLikvidat [17]

Answer:

4.48 grams is the mass of potassium hydroxide that the chemist must weigh out in the second step.

Explanation:

The pH of the solution = 13.00

pH + pOH = 14

pOH = 14 - pH = 14 - 13.00 = 1.00

$pOH=-\log[OH^-]$

$1.00=-\log[OH^-]$

$[OH^-]=10^{-1.00} M=0.100 M$

$KOH(aq)\rightarrow K^+(aq)+OH^-(aq)$

$[KOH]=[OH^-]=[K^+]=0.100 M$

Molariy of the KOH = 0.100 M

Volume of the KOH solution = 800 mL= 0.800 L

1 mL = 0.001 L

Moles of KOH = n

$Molarity=\frac{Moles}{Volume(L)}$

$0.100 M=\frac{n}{0.800 L}$

n = 0.0800 mol

Mass of 0.0800 moles of KOH :

0.0800 mol × 56 g/mol = 4.48 g

4.48 grams is the mass of potassium hydroxide that the chemist must weigh out in the second step.

4 0

2 years ago

The base-dissociation constant, kb, for pyridine, c5h5n, is 1.4x10-9 the acid-dissociation constant, ka, for the pyridinium ion,

sp2606 [1]

We are given the base dissociation constant, Kb, for Pyridine (C5H5N) which is 1.4x10^-9. The acid dissociation constant, Ka for the Pyridium ion or the conjugate acid of Pyridine is to be determined. We know from our chemistry classes that:

Kw = Kb * Ka

where Kw is always equal to 1x10^-14

so, to solve for Ka of Pyridium ion, substitute Kb to the equation together with Kw and solve for Ka:

1x10^-14 = 1.4x10^-9 * Ka
solve for Ka

Ka = 7.14x10^-6

Therefore, the acid dissociation constant of Pyridinium ion is 7.14x10^-6.
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5 0

2 years ago

A car starts from rest. It takes 6.0 seconds for the car to accelerate to 5 m/s2 W. What is the final velocity?

Artyom0805 [142]

Answer: 30 m/s

Explanation:

Use the first kinematic equation for linear motion

$V=V_i+at\\V=0+(5)(6)\\v=30m/s$

5 0

2 years ago