Ask question

Ask question

All categories

3 years ago

7

Find the approximate atomic mass of a water molecule (h2o). give your answer in atomic mass units rounded to the nearest whole n

umber.

2 answers:

7nadin3 [17]3 years ago

6 0

Answer: 18 amu

Explanation:

The atomic mass of hydrogen is 1.008 or 1 (rounded to the nearest whole number)

The atomic mass of oxygen is 15.999 or 16 (rounded to the nearest whole number)

There are 2 hydrogen atoms and one oxygen atom in water.

1+1+16= 18 amu

Blizzard [7]3 years ago

4 0

First step is to get the mass of water molecule in grams:
From the periodic table:
molar mass of hydrogen is 1
molar mass of oxygen is 16

molar mass of a water molecule = 2(1) + 16 = 18 gm

Now, to convert the gm into amu, all you have to do is multiply the gm you got by Avogadro's number as follows:
mass of water molecule = 18 x 6.22 x 10^23 = 1.1196 x 10^25 amu which is approximately 1 x 10^25 amu

You might be interested in

You happen to be visiting Northem California and you are driving by Suisun Bay, a notorious graveyard for old ships You notice t

umka2103 [35]

Answer:

The rusting of the metal is a chemical change.

Explanation:

8 0

3 years ago

The cell theory states that all living things are composed of one or more cells that carry out the functions needed to support l

saw5 [17]

Answer:

New cells arise from existing cells.

Explanation:

According to the cell theory, "all living things are composed of one or more cells; the cell is the basic unit of life; and new cells arise from existing cells"(Lumen Learning).

Cells are the basic unit of life. Some organisms consists of only one cell while other organisms have many cells and are called multicellular organisms.

In multicellular organisms, new cells are formed by the division of preexisting cells.

7 0

3 years ago

A student mixes four reagents together, thinking that the solutions will neutralize each other. The solutions mixed together are

vaieri [72.5K]

Answer: Resulting solution will not be neutral because the moles of $OH^-$ ions is greater. The remaining concentration of $[OH^-]$ ions =0.0058 M.

Explanation:

Given,

[HCl]=0.100 M

$[HNO_3]$ = 0.200 M

$[Ca(OH)_2]$ =0.0100 M

[RbOH] =0.100 M

Few steps are involved:

Step 1: Calculating the total moles of $H^+$ ion from both the acids

moles of $H^+$ in HCl

$HCl\rightarrow {H^+}+Cl^-$

if 1 L of $HCl$ solution =0.100 moles of HCl

then 0.05L of HCl solution= 0.05 $\times$ 0.1 moles= 0.005 moles (1L=1000mL)

moles of $H^+$ in HCl = 0.005 moles

Similarliy

moles of $H^+$ in $HNO_3$

$HNO_3\rightarrow H^++NO_3^-}$

If 1L of $HNO_3$ solution= 0.200 moles

Then 0.1L of $HNO_3$ solution= 0.1 $\times$ 0.200 moles= 0.02 moles

moles of $H^+$ in $HNO_3$ =0.02 moles

so, Total moles of $H^+$ ions = 0.005+0.02= 0.025 moles .....(1)

Step 2: Calculating the total moles of $[OH^-]$ ion from both the bases

Moles of $OH^-\text{ in }Ca(OH)_2$

$Ca(OH)_2\rightarrow Ca^2{+}+2OH^-$

1 L of $Ca(OH)_2$ = 0.0100 moles

Then in 0.5 L $Ca(OH)_2$ solution = 0.5 $\times$ 0.0100 moles = 0.005 moles

$Ca(OH)_2$ produces two moles of $OH^-$ ions

moles of $OH^-$ = 0.005 $\times$ 2= 0.01 moles

Moles of $OH^-$ in $RbOH$

$RbOH\rightarrow Rb^++OH^-$

1 L of RbOH= 0.100 moles

then 0.2 [RbOH] solution= 0.2 $\times$ 0.100 moles = 0.02 moles

Moles of $OH^-$ = 0.02 moles

so,Total moles of $OH^-$ ions = 0.01 + 0.02=0.030 moles ....(2)

Step 3: Comparing the moles of both $H^+\text{ and }OH^-$ ions

One mole of $H^+$ ions will combine with one mole of $OH^-$ ions, so

Total moles of $H^+$ ions = 0.005+0.02= 0.025 moles....(1)

Total moles of $OH^-$ ions = 0.01 + 0.02=0.030 moles.....(2)

For a solution to be neutral, we have

Total moles of $H^+$ ions = total moles of $OH^-$ ions

0.025 moles $H^+$ will neutralize the 0.025 moles of $OH^-$

Moles of $OH^-$ ions is in excess (from 1 and 2)

The remaining moles of $OH^-$ will be = 0.030 - 0.025 = 0.005 moles

So,The resulting solution will not be neutral.

Remaining Concentration of $OH^-$ ions = $\frac{\text{Moles remaining}}{\text{Total volume}}$

$[OH^-]=\frac{0.005}{0.85}=0.0058M$

6 0

4 years ago

A 52 gram sample of an unknown metal requires 714 Joules of energy to heat it from

ratelena [41]

Answer: Approximately $0.267 \frac{\text{J}}{\text{g}^{\circ}\text{C}}$

===================================================

Work Shown:

We have the following variables

Q = 714 joules = heat required
m = 52 grams = mass
c = specific heat = unknown
$\Delta t$ = 82-30.5 = 51.5 = change in temperature

note: the symbol $\Delta$ is the uppercase Greek letter delta. It represents the difference or change in a value.

Apply those values into the formula below. Solve for c.

$Q = m*c*\Delta t\\\\714 = 52*c*51.5\\\\714 = 52*51.5*c\\\\714 = 2678*c\\\\2678*c = 714\\\\c = \frac{714}{2678}\\\\c \approx 0.26661687826737\\\\c \approx 0.267\\\\$

The specific heat of the unknown metal is roughly $0.267 \frac{\text{J}}{\text{g}^{\circ}\text{C}}$

3 0

3 years ago

Sodium chlorate, NaClO3, decomposes when heated to yield sodium chloride and oxygen, a reaction used to provide oxygen for the e

netineya [11]

Sodium chlorate decomposes into sodium chloride and oxygen.
The reaction is
NaClO3 --Δ--> NaCl + O2
The Δ means that heat is used
This equation is still not balanced.
Balancing the equation
2NaClO3 --Δ--> 2NaCl + 3O2
The equation is now balanced.

3 0

3 years ago