All categories

3 years ago

Calculate the predicted density for calcium, ca, given the density of strontium, sr, (2.63g/cm3) and barium, ba, (3.65 g/cm3).

1 answer:

4 0

Answer is: <span>the predicted </span>density for calcium is 1.61 g/cm³.

d(Sr) = 2.63 g/cm³; density for strontium.
d(Ba) = 3.65 g/cm³; density for barium.
Ar(Sr) = 87.62 atomic weight for strontium.
Ar(Ba) = 137.33; atomic weight for barium.
Ar(Ca) = 40.08; atomic weight for calcium.
d(Ba) - d(Sr) = 3.65 g/cm³ - 2.63 g/cm³.
d(Ba) - d(Sr) = 1.02 g/cm³.
d(Ca) = d(Sr) - 1.02 g/cm³.
d(Ca) = 2.63 g/cm³ - 1.02 g/cm³.
d(Ca) = 1.61 g/cm³.

You might be interested in

There are two binary compounds of mercury and oxygen. heating either of them results in the decomposition of the compound, with

grandymaker [24]

$\text{Hg} \text{O}$ and $\text{Hg}_{2} \text{O}$ .

Assuming complete decomposition of both samples,

$m(\text{Hg}) = m(\text{residure})$
$m(\text{O}) = m(\text{loss})$

First compound:

$m(\text{O}) = m(\text{loss}) = 0.6498 - 0.6018 = 0.048 \; g$
$m(\text{Hg}) = m(\text{residure}) = 0.6018 \; g$

$n = m/M$ ; $0.6498 \; g$ of the first compound would contain

$n(\text{O atoms}) = 0.048 \; g / 16 \; g \cdot mol^{-1}= 0.003 \; mol$
$n(\text{Hg atoms}) = 0.6018 \; g / 200.58 \; g \cdot mol^{-1}= 0.003 \; mol$

Oxygen and mercury atoms seemingly exist in the first compound at a $1:1$ ratio; thus the empirical formula for this compound would be $\text{Hg} \text{O}$ where the subscript "1" is omitted.

Similarly, for the second compound

$m(\text{O}) = m(\text{loss}) = 0.016 \; g$
$m(\text{Hg}) = m(\text{residure}) = 0.4172 - 0.016 = 0.4012 \; g$

$n = m/M$ ; $0.4172 \; g$ of the first compound would contain

$n(\text{O atoms}) = 0.016 \; g / 16 \; g \cdot mol^{-1}= 0.001 \; mol$
$n(\text{Hg atoms}) = 0.4012 \; g / 200.58 \; g \cdot mol^{-1}= 0.002 \; mol$

$n(\text{Hg}) : n(\text{O}) \approx 2:1$ and therefore the empirical formula

$\text{Hg}_{2} \text{O}$ .

8 0

3 years ago

What are you and what do you want with my attendance

Lelu [443]

Answer:

Such a waste of points but lol

8 0

3 years ago

If you're standing on the beach looking out at the ocean as far as you can see, where's your zenith?

saveliy_v [14]

Straight above your head

4 0

3 years ago

Read 2 more answers

The average atomic mass of Eu is 151.96 amu. There are only two naturally occurring isotopes of europium, Eu with a mass of 151.

earnstyle [38]

Answer:

The percentage abundance of Eu isotopes are 52 % and 48 % .

Explanation:

The formula for the calculation of the average atomic mass is:

$Average\ atomic\ mass=(\frac {\%\ of\ the\ first\ isotope}{100}\times {Mass\ of\ the\ first\ isotope})+(\frac {\%\ of\ the\ second\ isotope}{100}\times {Mass\ of\ the\ second\ isotope})$

Given that:

Since the element has only 2 isotopes, so the let the percentage of first be x and the second is 100 -x.

For first isotope,:

% = x %

Mass = 151.0 amu

For second isotope :

% = 100 - x

Mass = 153.0 amu

Given, Average Mass = 151.96 amu

Thus,

$151.96=\frac{x}{100}\times {151.0}+\frac{100-x}{100}\times {153.0}$

Solving for x, we get that:

x = 52 %

<u>Thus percentage abundance of Eu isotopes are 52 % and 48 % .</u>

7 0

3 years ago

Consider this reaction: 2al(s) + 3 cucl2(aq) → 2alcl3(aq) + 3 cu(s) if the concentration of cucl2 drops from 1.000 m to 0.655 m

kirza4 [7]

The average rate of reaction over a given interval can be calculated by taking the difference of concentration on a particular given reactant, and dividing it by the total time. In this case, (1.00 M - 0.655 M)/30 s = 0.0115 M/s, or 0.0115 mol/L-s, and this is the final rate of reaction.

8 0

3 years ago