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Colt1911 [192]
4 years ago
9

For you last required journal entry of the quarter, but on a separate piece of paper, create a poster which compares the mass of

a mole of candy to something interesting. For example, “The mass of a mole of M&M’s is equal to the mass of 1,000 Jupiters” (This may not be true, I made it up.) Show the math neatly in one corner of the poster.
Chemistry
1 answer:
Yuki888 [10]4 years ago
8 0
With a mole of skittles placed side by side, you can go around the earth <span>191.17 trillion times. 

I hope this helps! If it didn't please reply. Please mark it the brainliest if it did!

</span>(6.022x10^23) x .5" = 3.011x10^23 inches.
<span>3.011x10^23 / 12 in per ft(=.251)
</span><span>.251 /5280 ft per mile = 4.75x10^18 miles
</span><span>4.75x10^18 miles / 24850 mile around the earth = 1.9117x10^14 time around the earth.
</span><span>191.17 trillion times around the earth</span>
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Prospectors are considering searching for gold on a plot of land that contains 2.45 g of gold per bucket of soil. If the volume
makvit [3.9K]

Answer:

13.2 g of gold

Explanation:

We'll begin by converting 5.25 L to ft³.

This can be obtained as follow:

Recall:

1 L = 0.0353 ft³

Therefore,

5.25 L = 5.25 × 0.0353

5.25 L = 1.85×10¯¹ ft³

From the question given above,

2.45 g of gold is present in 5.25 L ( i.e 1.85×10¯¹ ft³) of soil.

Therefore, Xg of gold will be present in 1 ft³ of soil i.e

Xg of gold = 2.45/1.85×10¯¹

Xg of gold = 13.2 g

Therefore, 13.2 g of gold is present in 1 ft³ of the soil.

8 0
4 years ago
What is the molarity of HCl found in a titration where 50 ml of HCl is titrated with 50 ml of .1 M NaOH? Needs to be answered as
Aneli [31]

0.1 mol/L . The concentration of the HCl is 0.1 mol/L

a) Write the <em>balanced chemical equation </em>

HCl + NaOH → NaCl + H2O

b) Calculate the <em>moles of NaOH </em>

Moles of NaOH = 0.050 L NaOH x (0.1 mol NaOH/1 L NaOH)

= 0.0050 mol NaOH

c) Calculate the <em>moles of HCl </em>

Moles of HCl = 0.0050 mol NaOH x (1 mol HCl/1 mol NaOH)

= 0.0050 mol HCl

d) Calculate the <em>molar concentration</em> of the HCl

<em>c</em> = moles/litres = 0.0050 mol/0.050 L = 0.1 mol/L

7 0
4 years ago
In your body, carbon dioxide, CO2, dissolves in the blood to form carbonic acid, H2CO3 as follows:
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Answer:

STOP USING THIS SIGHT STOP

Explanation:

not good for ua quart of stain covers 100 square ft how many quarts should you buy to stain the ramp assume you dont have to stain the bottem its lenth is 7 ft 1 side is 35ft the other 35 1/6 ft and the hight is 35 ft

a quart of stain covers 100 square ft how many quarts should you buy to stain the ramp assume you dont have to stain the bottem its lenth is 7 ft 1 side is 35ft the other 35 1/6 ft and the hight is 35 ft

a quart of stain covers 100 square ft how many quarts should you buy to stain the ramp assume you dont have to stain the bottem its lenth is 7 ft 1 side is 35ft the other 35 1/6 ft and the hight is 35 ft

a quart of stain covers 100 square ft how many quarts should you buy to stain the ramp assume you dont have to stain the bottem its lenth is 7 ft 1 side is 35ft the other 35 1/6 ft and the hight is 35 ft

a quart of stain covers 100 square ft how many quarts should you buy to stain the ramp assume you dont have to stain the bottem its lenth is 7 ft 1 side is 35ft the other 35 1/6 ft and the hight is 35 ft

a quart of stain covers 100 square ft how many quarts should you buy to stain the ramp assume you dont have to stain the bottem its lenth is 7 ft 1 side is 35ft the other 35 1/6 ft and the hight is 35 ft

a quart of stain covers 100 square ft how many quarts should you buy to stain the ramp assume you dont have to stain the bottem its lenth is 7 ft 1 side is 35ft the other 35 1/6 ft and the hight is 35 ft

a quart of stain covers 100 square ft how many quarts should you buy to stain the ramp assume you dont have to stain the bottem its lenth is 7 ft 1 side is 35ft the other 35 1/6 ft and the hight is 35 ft

a quart of stain covers 100 square ft how many quarts should you buy to stain the ramp assume you dont have to stain the bottem its lenth is 7 ft 1 side is 35ft the other 35 1/6 ft and the hight is 35 ft

a quart of stain covers 100 square ft how many quarts should you buy to stain the ramp assume you dont have to stain the bottem its lenth is 7 ft 1 side is 35ft the other 35 1/6 ft and the hight is 35 ft

a quart of stain covers 100 square ft how many quarts should you buy to stain the ramp assume you dont have to stain the bottem its lenth is 7 ft 1 side is 35ft the other 35 1/6 ft and the hight is 35 ft

3 0
3 years ago
Which property of water helps it move upward by capillary action
zhannawk [14.2K]
It’s the water method that helps it
8 0
3 years ago
Carbon tetrachloride can be produced by the following reaction: Suppose 1.20 mol of and 3.60 mol of were placed in a 1.00-L flas
hjlf

The given question is incomplete. The complete question is :

Carbon tetrachloride can be produced by the following reaction:

CS_2(g)+3Cl_2(g)\rightleftharpoons S_2Cl_2(g)+CCl_4(g)

Suppose 1.20 mol CS_2(g) of and 3.60 mol of Cl_2(g)  were placed in a 1.00-L flask at an unknown temperature. After equilibrium has been achieved, the mixture contains 0.72 mol  of CCl_4. Calculate equilibrium constant at the unknown temperature.

Answer: The equilibrium constant at unknown temperature is 0.36

Explanation:

Moles of  CS_2 = 1.20 mole

Moles of  Cl_2 = 3.60 mole

Volume of solution = 1.00  L

Initial concentration of CS_2 = \frac{moles}{volume}=\frac{1.20mol}{1L}=1.20M

Initial concentration of Cl_2 = \frac{moles}{volume}=\frac{3.60mol}{1L}=3.60M

The given balanced equilibrium reaction is,

                 CS_2(g)+3Cl_2(g)\rightleftharpoons S_2Cl_2(g)+CCl_4(g)

Initial conc.         1.20 M        3.60 M                  0                  0

At eqm. conc.     (1.20-x) M   (3.60-3x) M   (x) M        (x) M

The expression for equilibrium constant for this reaction will be,

K_c=\frac{[S_2Cl_2]\times [CCl_4]}{[Cl_2]^3[CS_2]}

Now put all the given values in this expression, we get :

K_c=\frac{(x)\times (x)}{(3.60-3x)^3\times (1.20-x)}

Given :Equilibrium concentration of CCl_4 , x = \frac{moles}{volume}=\frac{0.72mol}{1L}=0.72M

K_c=\frac{(0.72)\times (0.72)}{(3.60-3\times 0.72)^3\times (1.20-0.72)}

K_c=0.36

Thus equilibrium constant at unknown temperature is 0.36

4 0
3 years ago
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