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faltersainse [42]
3 years ago
8

Fluorine gas reacts violently with water to produce hydrogen fluoride and ozone according to the following unbalanced equation.

How many moles of o, are created assuming STP conditions are in effect, given 1.88 moles of fluorine gas were
used?
__F2+__H20—__HF+__03
Chemistry
1 answer:
jeka943 years ago
6 0

Answer:

Fluorine gas reacts violently with water to produce hydrogen fluoride and ozone according to the following unbalanced equation. How many moles of o, are created assuming STP conditions are in effect, given

Explanation:

3

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Find the area of a vegetable garden that is 4.83m long and 3.4m wide. Give your answer to the correct significant figure. (use t
sergejj [24]

Answer:

The area of vegetable garden is 16.422 m².

Explanation:

Given data:

Length = 4.83 m

Width = 3.4 m

Area = ?

Solution:

Formula:

Area = length × width

Area is measured in square unit.

By putting the values in formula:

Area = length × width

Area = 4.83 m × 3.4 m

Area = 16.422 m²

The area of vegetable garden is 16.422 m².

5 0
3 years ago
Consider the following intermediate chemical equations.
QveST [7]

Answer: 250 kJ

Explanation: According to Hess’s law of constant heat summation, the heat absorbed or evolved in a given chemical equation is the same whether the process occurs in one step or several steps.

According to Hess’s law, the chemical equation can be treated as algebraic expressions and can be added or subtracted to yield the required equation. That means the enthalpy change of the overall reaction is the sum of the enthalpy changes of the intermediate reactions.

P_4(s)+6Cl_2\rightarrow 4PCl_3  \Delta H_1=-2439kJ (1)

4PCl_5(g)\rightarrow P_4(s)+10Cl_2(g)  \Delta H_2=3438kJ (2)

Net chemical equation:

PCl_5(g)\rightarrow PCl_3(g)+Cl_2(g)  \Delta H=? (3)

Adding 1 and 2 we get,

4PCl_5(g)\rightarrow 4PCl_3(g)+4Cl_2 \Delta H_3=\Delta H_1+\Delta H_2=-2439+3438=1000kJ   (4)

Now dividing equation (4) by 4, we get

PCl_5(g)\rightarrow PCl_3(g)+Cl_2

\Delta H=\frac{\Delta H_3}{4}=\frac{1000kJ}{4}=250kJ   (4)

8 0
3 years ago
Calculate the theoretical yield of ammonia produced by the reaction of 100 g of H2 gas and 200g of N2 gas? 3H2(g) + N2(g)-----&g
Alex787 [66]
Moles of Hydrogen present: 100 / 2 = 50 moles

Moles of Nitrogen present: 200 / 28 = 7.14 moles

Hydrogen required by given amount of nitrogen = 7.14 x 3 = 21.42 moles

Hydrogen is excess so we will calculate the Ammonia produced using Nitrogen.

Molar ratio of Nitrogen : Ammonia = 1 : 2

Moles of ammonia = 7.14 x 2 = 14.28 moles
8 0
4 years ago
Read 2 more answers
Oxygen and hydrogen form the polyatomic hydroxide ion. What is its charge?
prohojiy [21]

Answer:

negative

the chage on hydroxide ions os negative

4 0
3 years ago
Read 2 more answers
An interpenetrating primitive cubic structure like that of CsCl with anions in the corners has an edge length of 664 pm. If the
san4es73 [151]

Answer:

the ionic radius of the anion r^- = 312.52 \ pm

Explanation:

From the diagram shown below :

The anion Cl^- is located at the corners

The cation Cs^+ is located at the body center

The Body diagonal length =  \sqrt{3 \ a }

∴ 2 \ r^+ \ + 2r^- \ = \sqrt{3 \ a}  \\ \\ r^+ +r^- = \frac{\sqrt{3}}{2} a

Given that :

\frac{r^+}{r^-} =0.84   (i.e the  ratio of the ionic radius of the cation to the ionic radius of

                 the anion )

0.84r^- \ + r^- \ = \frac{\sqrt{3}}{2}a  \\ \\  1.84 r^- = \frac{3}{2}a \\ \\ r^- = \frac{\sqrt{3}}{2*1.84}a

Also ; a =  664 pm

Then :

r^- = \frac{\sqrt{3} }{2*1.84}*664 \ pm\\ \\ r^- = 312.52 \ pm

Therefore,  the ionic radius of the anion r^- = 312.52 \ pm

4 0
3 years ago
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