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weqwewe [10]
3 years ago
14

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

Chemistry
1 answer:
Makovka662 [10]3 years ago
8 0
It will be the second one
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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
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