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2 years ago

What solution turns litmus paper blue? A. acidic B. basic C. neutral D. none

2 answers:

8 0

The correct answer is B. Blue litmus paper turns<span> red under acidic conditions and red </span>litmus paper turns blue<span> under basic conditions.

Happy studying ^-^

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Over [174]2 years ago

6 0

Litmus paper turns blue in a basic solution
so B

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Which of the following changes would have no effect on the equilibrium position of the reaction below? 2 NOBR (g) 2 NO (g)+ Br2

nasty-shy [4]

Answer:

D) cutting the concentrations of both NOBr and NO in half

Explanation:

The equilibrium reaction given in the question is as follows -

2NOBr ( g ) ↔ 2NO ( g ) + Br₂ ( g )

The equilibrium constant for the above reaction can be written as -

K = [ NO ]² [ Br₂ ] / [ NOBr ] ²

Therefore from the condition given in the question , the changes that will not affect the equilibrium will be , reducing the concentration of both NOBr and NO to half ,

Hence ,

the new concentrations are as follows -

[ NoBr ] ' = 1/2 [ NoBr ]

[ NO ] ' = 1/2 [ NO ]

Hence the new equilibrium constant equation can be written as -

K ' = [ NO ] ' ² [ Br₂ ] / [ NOBr ] ' ²

Substituting the new concentration terms ,

K ' = 1/2 [ NO ] ² [ Br₂ ] / 1/2 [ NoBr ] ²

K ' = 1/4 [ NO ] ² [ Br₂ ] / 1/4 [ NoBr ] ²

The value of 1 / 4 in the numerator and the denominator is cancelled -

K ' = [ NO ] ² [ Br₂ ] / [ NoBr ] ²

Hence ,

K' = K

5 0

3 years ago

Examples of noble gases

LekaFEV [45]

Natural gas, desolate

4 0

2 years ago

Which of these shows a volume of 1.25 liters expressed in milliliters

stellarik [79]

I would use a conversation calculator

3 0

2 years ago

Read 2 more answers

__C2H4+__O2-->__CO2+__H2O

mihalych1998 [28]

Answer:

Here is your answer mate :D

3 0

2 years ago

Express each of the following answers in the IUPAC format. Do not include any capitals or spaces in your name. Separate multiple

lorasvet [3.4K]

The full question is shown in the image attached

Answer:

See explanation

Explanation:

In naming an alkane, the first thing we do is to obtain the parent chain by counting the number of carbon atoms in the chain.

When we obtain that, then we identify the substituents and number them in such a way that they have the lowest numbers. The compounds shown have the following names according to the order in which the structures appear in the image attached;

1. 2-methyl propane

2. 2,4-dimethyl heptane

3. 2,2,3,3-tetramethyl butane

4. 5-ethyl-2,4-dimethyl octane

5 0

2 years ago