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3 years ago

I have no idea how to do this help plz!

Chemistry

2 answers:

Wittaler [7]3 years ago

8 0

The answes are
C)
Then it is
)A
Im doing this for class too bruh

PSYCHO15rus [73]3 years ago

5 0

C A and B D . Sorry if I’m wrong

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What is the frequency of 3.98 x 10^-77

LekaFEV [45]

Answer:

What is the frequency of a 6.9 x 10-13 m wave? 3.00 x 108 = 6.9x10-13 mly). GAMMA. V = 4.35 x 10 20 5-11. 3. What is the wavelength of a 2.99 Hz wave?

Missing: 3.98 ‎77

Explanation:

5 0

3 years ago

3: Consider molecules of hydrogen (tiny ones) and oxygen (bigger ones) in a gas mixture. If they have the same average kinetic e

mezya [45]

Answer: Hydrogen molecules will have greatest average speed.

Explanation:

The formula for average speed is :

$\mu_{av}={\sqrt{\frac{8RT}{\pi\times M}}}$

R = gas constant

T = temperature

M = Molecular Mass

Now putting all the values:

$\frac{\nu_{H_2}}{\nu_{O_2}}=\sqrt{\frac{8RT}{\pi\times M_{H_2}}}/{\sqrt{\frac{8RT}{\pi\times M_{O_2}}}$

$\frac{\nu_{H_2}}{\nu_{O_2}}=\sqrt{\frac{16}{2}}$

$\frac{\nu_{av}_{H_2}}{\nu_{av}_{O_2}}=8$

Thus average speed of hydrogen is 8 times the average speed of oxygen. Thus hydrogen molecules will have greatest average speed

8 0

3 years ago

When you have finished adding the necessary amount of NaOH to your beaker, what color will the litmus paper turn?

Svetllana [295]

Answer: The red litmus paper turns blue on dipping in NaOH solution.

Explanation:

Litmus paper is the indicator that detects the nature of the solution, whether it is acidic or basic.

There are 2 types of litmus paper:

Red litmus paper: This paper will turn blue if it is dipped in basic solution and will remain as such if it is dipped in acidic solution.
Blue litmus paper: This paper will turn red if it is dipped in acidic solution and will remain as such if it is dipped in basic solution.

NaOH is a strong base, so when a red litmus paper is dipped in the beaker having necessary amount of NaOH, the red litmus paper turns into blue.

7 0

3 years ago

Match the following aqueous solutions with the appropriate letter from the column on the right.1. 0.19 m AgNO3 2. 0.17 m CrSO4 3

vichka [17]

Answer:

0.13 m of $Mn(NO_3)_2$ → Highest boiling point

0.19 m of $AgNO_3$ → Second Highest boiling point

0.17 m of $CrSO_4$ → Third highest boiling point

0.31 m Sucrose (nonelectrolyte) → Lowest boiling point

Explanation:

Elevation in boiling is given by :

$\Delta T_b=i\times k_b\times m$

Where :

i = van't Hoff factor

$k_b$ = Molal Elevation constant of solvent

m = molaity of the solution

1) 0.19 m of $AgNO_3$

$AgNO_3\rightarrow Ag^++NO_3^{-}$

i = 2 (electrolyte)

Molality of the solution = 0.19

Elevation is boiling point of solution:

$\Delta T_b=2\times k_b\times 0.19 m$

$\Delta T_b=0.38 m\times k_b$

2) 0.17 m of $CrSO_4$

$CrSO_4\rightarrow Cr^{2+}+SO_4^{2-}$

i = 2 (electrolyte)

Molality of the solution = 0.17

Elevation is boiling point solution :

$\Delta T_b=2\times k_b\times 0.17 m$

$\Delta T_b=0.34 m\times k_b$

3) 0.13 m of $Mn(NO_3)_2$

$Mn(NO_3)_2\rightarrow Mn^{2+}+2NO_3^{-}$

i = 3 (electrolyte)

Molality of the solution = 0.13

Elevation is boiling point solution :

$\Delta T_b=3\times k_b\times 0.13 m$

$\Delta T_b=0.39 m\times k_b$

4) 0.31 m Sucrose (nonelectrolyte)

i = 1 ( non electrolyte)

Molality of the solution = 0.31 m

Elevation is boiling point solution :

$\Delta T_b=1\times k_b\times 0.31 m$

$\Delta T_b=0.31 m\times k_b$

Higher the value of elevation in temperature higher will be the boiling point of the solution .

The decreasing order of solution from highest boiling point to lowest boiling point is :

$0.39 m\times k_b>0.38 m\times k_b>0.34 m\times k_b>0.31 m\times k_b$

0.13 m of $Mn(NO_3)_2$ → Highest boiling point

0.19 m of $AgNO_3$ → Second Highest boiling point

0.17 m of $CrSO_4$ → Third highest boiling point

0.31 m Sucrose (nonelectrolyte) → Lowest boiling point

6 0

3 years ago

Suppose 2.8 moles of methane are allowed to react with 5 moles of oxygen.

Ronch [10]

Answer : The limiting reagent is $O_2$

Solution : Given,

Moles of methane = 2.8 moles

Moles of $O_2$ = 5 moles

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

$CH_4+2O_2\rightarrow CO_2+2H_2O$

From the balanced reaction we conclude that

As, 2 mole of $O_2$ react with 1 mole of $CH_4$

So, 5 moles of $O_2$ react with $\frac{5}{2}=2.5$ moles of $CH_4$

From this we conclude that, $CH_4$ is an excess reagent because the given moles are greater than the required moles and $O_2$ is a limiting reagent and it limits the formation of product.

Hence, the limiting reagent is $O_2$

8 0

3 years ago