In a reaction to produce sulfuric acid, the theoretical yield is 300.g. What is the percent yield if the actual yield is 280.g?

One meter is equal to 10 9 nanometers. How long in meters is 670 nanometers? 6.70 × 10 –7 m

Mila [183]

Answer:

The answer to your question is: the first option

Explanation:

Meter is a unit of length in the SI, international system of units. Its symbol is m.

Nanometer is equivalent to 10⁻⁹ meters, as its name says it is used to measure extremely small particles.

-To solve this problem we use a rule of three.

1 meter ⇒ 10⁹ nanometers

x meters ⇒ 670 nanometers

Simplify

x = (670 x 1)/( 10⁹)

x = 6.7 x 10 ⁻⁷ meters

6 0

3 years ago

Find the elements sodium, oxygen, and phosphorus on the periodic table. Describe the monoatomic ions each would form. Then, give

miskamm [114]

Sodium would form Na+

Oxygen would form O-2

Phosphorus would form -3 or +5

Sodium and oxygen would combine to form ionic bonds because one is a metal and the other is non-metal, while Phosporus and oxygen would combine to form covalent bonds because they are both non-metals.

4 0

3 years ago

Read 2 more answers

Empirical formula of c3h12o6

morpeh [17]

Answer:

$CH_{4} O_{2}$

Explanation:

The empirical formula is the simplest whole-number ratio of the elements in a compound to one another. To find the empirical formula given a molecular formula, divide all the subscripts by their greatest common factor.

3, 12, and 6 can all be divided by 3, which makes the empirical formula $CH_{4} O_{2}$

5 0

3 years ago

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 $\rm O\!$ .