The saltwater is a homogeneous or a heterogeneous mixture

Kipish [7]

To answer this lets first see how much 1 kg is equal to in cg.

1 kg = 100000 cg

Now lets multiply:-

100000 × 1.7 = <span>170000
</span>
So, 1.7 kg = <span>170000 cg
</span>
Hope I helped ya!!!

4 0

3 years ago

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What is one molar volume of the gas ammonia (nh3) at stp?

Sedbober [7]

<span>The molar volume of ammonia at standard temperature and pressure (STP) is 22.4L. This is the standard molar volume of any gas at these conditions.</span>

6 0

3 years ago

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How many atoms are in 766.3 grams of Li?

34kurt

1 mol of any substance is made of 6.022 x 10²³ units, these units could be atoms making up elements or molecules making up a compound.

In this case. 1 mol of Li is made of 6.022 x 10²³ atoms of Li

The molar mass of Li is 6.94 g/mol

Therefore mass of 1 mol of Li is 6.94 g

In 6.94 g of Li there are - 6.022 x 10²³ atoms

Then in 766.3 g of Li there are - 6.022 x 10²³ atoms / 6.94 g x 766.3 g = 665 x 10²³

There are 6.65 x 10^25 atoms of Li

6 0

3 years ago

The densities of crystalline co2 and nh3 at 160 k are 1.56 and 0.84 g/cm3, respectively. calculate their molar volumes.

Neporo4naja [7]

The molar volume, symbol Vm<span>, is the </span>volume occupied by one mole of a substance at a given temperature and pressure. <span>It is equal to the </span>molar<span> mass divided by the mass density. Therefore, we calculate as follows:

Vm(CO2) = 44.01 / 1.56 = 28.21 cm^3 / mol
</span>Vm(NH3) = 17.03 / 0.84 = 20.27 cm^3 / mol

7 0

3 years ago

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A compound is made up of 28 g N, 24 g C, 48 g O, and 8 g H .What is the empirical formula?

vovikov84 [41]

Answer:

$\rm C_2H_8N_2O_3$ .

Explanation:

<h3>Step One: calculate the coefficients. </h3>

Look up the relative atomic mass of these four elements on a modern periodic table:

$\rm C$ : approximately $12$ .
$\rm H$ : approximately $1$ .
$\rm N$ : approximately $14$ .
$\rm O$ : approximately $16$ .

The relative atomic mass of an element is numerically equal to the mass (in grams, $\rm g$ ,) of one mole of atoms of this element.

For example, the relative atomic mass of $\rm C$ is approximately $12$ . Therefore, each mole of $\rm C\!$ atoms would have a mass of $12\; \rm g$ .

This sample contains $24\; \rm g$ of carbon. That would correspond to approximately $\displaystyle \left(\frac{24}{12}\right)\; \rm mol = 2\; \rm mol$ of $\rm C$ atoms.

Similarly, for the other three elements:

$\displaystyle n(\mathrm{H}) \approx \frac{8\; \rm g}{1\; \rm g \cdot mol^{-1}} = 8\; \rm mol$ .

$\displaystyle n(\mathrm{N}) \approx \frac{28\; \rm g}{14\; \rm g \cdot mol^{-1}} = 2\; \rm mol$ .

$\displaystyle n(\mathrm{O}) \approx \frac{48\; \rm g}{16\; \rm g \cdot mol^{-1}} = 3\; \rm mol$ .

Hence, the ratio between these elements in this compound would be:

$n(\mathrm{C}): n(\mathrm{H}): n(\mathrm{N}):n(\mathrm{O}) = 2: 8 : 2 : 3$ .

In the empirical formula of a compound, the coefficients should represent the smallest possible integer ratio between the number of atoms of these elements.

$n(\mathrm{C}): n(\mathrm{H}): n(\mathrm{N}):n(\mathrm{O}) = 2: 8 : 2 : 3$ is indeed the smallest possible integer ratio between the number of atoms of these elements.

<h3>Step Two: arrange the elements in an appropriate order</h3>

Apply the Hill System to arrange these four elements in the empirical formula. In the Hill System:

If carbon, $\rm C$ , is present in this compound, then:

$\rm C$ (carbon) and then $\rm H$ (hydrogen) will be the first two elements listed in the formula (ignore the hydrogen if it is not in the compound.)
The other elements in this compound will be listed in alphabetical order.

If there is no carbon $\rm C$ in this compound, then list all the elements in this compound in alphabetical order.

Both $\rm C$ (carbon) and $\rm H$ (hydrogen) are found in this compound. Therefore, the first element in the list would be $\rm C\!$ . The second would be $\rm H\!$ , followed by $\rm N\!$ and then $\rm O\!$ .

Hence, the empirical formula of this compound would be $\rm C_2H_8N_2O_3$ .

6 0

3 years ago