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tangare [24]
2 years ago
6

Find the mass in kilograms of the liquid air that is required to produce 600L of oxygen. In normal condition, 1L of liquid air h

as 1.3g​
Chemistry
2 answers:
Step2247 [10]2 years ago
8 0

Explanation:

1L is 1.3 g so 600L is 780 g

1kg= 1000g so

780 g = 0.78kg

I GUESS

IgorC [24]2 years ago
3 0

Mass in kilograms of liquid air required = 0.78 kg

<u>Given that </u>

1 Litre of liquid air contains 1.3 grams of oxygen ( air )

<u />

<u> Determine the a</u><u>mount of liquid air</u><u> in Kg</u>

volume of air given = 600 L

mass of liquid air required = x

1 litre = 1.3 grams

600 L =  x

∴ x ( mass of liquid air ) = 1.3 * 600

                                       = 780 g  = 0.78 kg

Hence we can conclude that Mass in kilograms of liquid air required = 0.78 kg

Learn more about liquid air : brainly.com/question/636295

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Which of the following is the Arrhenius equation?<br>​
natka813 [3]

Answer:

A is the answer the reason why is because I got it right when I was doing my test

3 0
2 years ago
Consider the reaction of magnesium metal with hydrochloric acid to produce magnesium chloride and hydrogen gas. If 3.29 mol of m
gregori [183]

Answer:

1.645 moles of excess reactant that is of magnesium metal are left over.

Explanation:

Moles of magnesium metal = 3.29 mol

Moles of HCl = 3.29 mol

Mg(s)+2HCl(aq)\rightarrow MgCl_2(aq)+H_2(g)

According to recation, 2 moles of HCl reacts with 1 mol of magnesium metal, then 3.29 moles of HCl will react with :

\frac{1}{2}\times 3.29 mol=1.645 mol of magnesium metal

Moles of HCl left = 3.29mol - 3.29 mol = 0

Moles of magnesium metal left = 3.29 mol - 1.645 mol = 1.645 mol

1.645 moles of excess reactant that is of magnesium metal are left over.

7 0
3 years ago
Two equilibrium reactions of nitrogen with oxygen, with their corresponding equilibrium constants (Kc) at a certain temperature,
7nadin3 [17]

Answer:

Kc = 1.54e - 31 / 2.61e - 24

Explanation:

1 )   N_{2}(gas) + O_{2}(gas)\rightarrow 2NO(gas)  ; Kc = 1.54e - 31

2)   N_{2}(gas) + 1/2O_{2}(gas)\rightarrow N_{2}O(gas)  ; Kc = 2.16e - 24

   upon reversing  ( 2 )  equation

     N_{2}O(gas)\rightarrow N_{2}(gas) + 1/2O_{2}(gas)   Kc = 1/2.16e - 24  

    now adding 1 and reversed equation (2)

       N_{2}(gas) + O_{2}(gas)\rightarrow 2NO(gas)

      N_{2}O(gas)\rightarrow N_{2}(gas) + 1/2O_{2}(gas)

   we get ,

                  N_{2}O(gas) + 1/2O_{2}(gas)\rightarrow 2NO(gas)  Kc = 1.54e-31 × 1/2.61e - 24

       equilibrium constant of equation (3) is -

            Kc = 1.54e - 31 / 2.61e - 24

3 0
3 years ago
A compound is found to have 55.7% hafnium+and+44.3%+chlorine.+what+is+the+empirical+formula?
uranmaximum [27]

The empirical formula of a compound found to have 55.7% hafnium and 44.3% chlorine is HfCl4.

<h3>How to calculate empirical formula?</h3>

The empirical formula of a compound is a notation indicating the ratios of the various elements present in a compound, without regard to the actual numbers.

The empirical formula of the given compound can be calculated as follows:

  • Hafnium = 55.7% = 55.7g
  • Chlorine = 44.3% = 44.3g

First, we convert mass values to moles by dividing by the molar mass of each element

  • Hafnium = 55.7g ÷ 178.49g/mol = 0.312mol
  • Chlorine = 44.3g ÷ 35.5g/mol = 1.25mol

Next, we divide each mole value by the smallest

  • Hafnium = 0.312 ÷ 0.312 = 1
  • Chlorine = 1.25 ÷ 0.312 = 4

Therefore, the empirical formula of a compound found to have 55.7% hafnium and 44.3% chlorine is HfCl4.

Learn more about empirical formula at: brainly.com/question/14044066

#SPJ1

7 0
1 year ago
What are some problems chemist must consider when developing new technologies?
Ratling [72]
The study of chemistry helps us understand the nature of the world around us. Chemistry is always developing to keep up with any phenomenon that appears in nature.
Therefore, scientists and chemists are always developing new technologies. However, chemists must very careful when developing these new technologies. They should consider any bad chemical reactions that might occur and also chemicals that harmful to either the individuals or the environment.
3 0
3 years ago
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