What would you need to do to calculate the molality of 10 mol of NaCl in 200

If a liquid has a Ph of 3 what is it

kvasek [131]

Lower than 7 is acid greater than 7 is a base

6 0

3 years ago

If 25.0 g NO are produced, how many grams of nitrogen gas are used?

ivanzaharov [21]

Based on the assumption that the reaction involves N and O to produce NO, if 25.0 g of NO are produced, the amount of N gas used would be 11.66 grams

<h3>Stoichiometric calculation</h3>

From the equation of the reaction:

N + O ---------> NO

Mole ratio of N to NO is 1:1

Mole of 25.0 g of NO = 25/30.01 = 0.833 moles

Equivalent mole of N = 0.833 moles

Mass of 0.833 moles N = 0.833 x 14 = 11.66 grams

More on stoichiometric calculations can be found here: brainly.com/question/8062886

7 0

2 years ago

The most abundant elements in the universe are hydrogen and helium, but there are also small but significant amounts of heavier

Stells [14]

Answer:

in nuclear fusion deep in the interiors of stars

Explanation:

Nuclear fusion -

It is the type of reaction , where two or more atomic nuclei of the atom merges together to release two or more different nuclei along with some subatomic particles , is referred to as a nuclear fusion reaction .

The reaction can very well be done on stars , because of very high energy .

Hence , a nuclear fusion occurs deep inside the stars .

7 0

3 years ago

A compound contains 1.2 g of carbon, 3.2 g of oxygen and 0.2g of hydrogen. Find the formula of the compound

Karolina [17]

Answer:

The empirical formula of the compound is $C_{0.504}HO_{1.008}$ .

Explanation:

We need to determine the empirical formula in its simplest form, where hydrogen ( $H$ ) is scaled up to a mole, since it has the molar mass, and both carbon ( $C$ ) and oxygen ( $O$ ) are also scaled up in the same magnitude. The empirical formula is of the form:

$C_{x}HO_{y}$

Where $x$ , $y$ are the number of moles of the carbon and oxygen, respectively.

The scale factor ( $r$ ), no unit, is calculated by the following formula:

$r = \frac{M_{H}}{m_{H}}$ (1)

Where:

$m_{H}$ - Mass of hydrogen, in grams.

$M_{H}$ - Molar mass of hydrogen, in grams per mole.

If we know that $M_{H} = 1.008\,\frac{g}{mol}$ and $m_{H} = 0.2\,g$ , then the scale factor is:

$r = \frac{1.008}{0.2}$

$r = 5.04$

The molar masses of carbon ( $M_{C}$ ) and oxygen ( $M_{O}$ ) are $12.011\,\frac{g}{mol}$ and $15.999\,\frac{g}{mol}$ , then, the respective numbers of moles are: ( $r = 5.04$ , $m_{C} = 1.2\,g$ , $m_{O} = 3.2\,g$ )

Carbon

$n_{C} = \frac{r\cdot m_{C}}{M_{C}}$ (2)

$n_{C} = \frac{(5.04)\cdot (1.2\,g)}{12.011\,\frac{g}{mol} }$

$n_{C} = 0.504\,moles$

Oxygen

$n_{O} = \frac{r\cdot m_{O}}{M_{O}}$ (3)

$n_{O} = \frac{(5.04)\cdot (3.2\,g)}{15.999\,\frac{g}{mol} }$

$n_{O} = 1.008\,moles$

Hence, the empirical formula of the compound is $C_{0.504}HO_{1.008}$ .

3 0

3 years ago

Plz help u might find it easy

PIT_PIT [208]

The melting point of pure lead : 327 °C

<h3>Further explanation</h3>

Given

Amount of tin and melting point of solder

Required

The melting point

Solution

The composition of solder = tin and lead

So if it is 100% tin, 0% lead or 0% tin, 100% lead

From the table it is shown that when the position is 100% tin in the solder, the melting point of the solder is 232 °C, so it shows that the melting point of pure lead is obtained when% tin in solder = 0 (100% lead in solder), so that the melting point is obtained. : 327 °C

3 0

2 years ago