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sesenic [268]
2 years ago
7

Add 92.2 km to 9426 m and report the answer in

Chemistry
1 answer:
Mashcka [7]2 years ago
3 0

to convert m into km we have to divide it by 1000.so,

9426/1000=9.426km

92.2km+9.426km=101.626km.

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A 0.500-g sample of KCl is added to 50.0g of water in a calprimeter (Figure 5.12) If the temperature decreases by 1.05C. what is
Sauron [17]

Answer : The reaction is endothermic.

Explanation :

Formula used :

Q=m\times c\times \Delta T

where,

\Delta T = change in temperature = 1.05^oC

Q = heat involved in the dissolution of KCl = ?

m = mass = 0.500 + 50.0 = 50.5 g

c = specific heat of resulting solution = 4.18J/g^oC

Now put all the given value in the above formula, we get:

Q=50.5g\times 4.18J/g^oC\times 1.05^oC

Q=+221.64J

The heat involved in the dissolution of KCl is positive that means as the change in temperature decreases then the reaction is endothermic and as the change in temperature increases then the reaction is exothermic.

Hence, the reaction is endothermic.

8 0
3 years ago
Calculate the molality of isoborneol in the product if, a) the melting point of pure camphor is 179°C and the melting point take
tresset_1 [31]

Answer:

The molality of isoborneol in camphor is 0.53 mol/kg.

Explanation:

Melting point of pure camphor= T =179°C

Melting point of sample = T_f = 165°C

Depression in freezing point = \Delta T_f=?

\Delta T_f=T- T_f=179^oC-165^oC=14^oC

Depression in freezing point  is also given by formula:

\Delta T_f=i\times K_f\times m

K_f = The freezing point depression constant

m = molality of the sample

i = van't Hoff factor

We have: K_f = 40°C kg/mol

i = 1 (  organic compounds)

\Delta T_f=14^oC

14^oC=1\times 40^oC kg/mol\times m

m=\frac{14^oC}{1\times 40^oC kg/mol}=0.35 mol/kg

The molality of isoborneol in camphor is 0.53 mol/kg.

8 0
3 years ago
A sample of propane(c3h8)has 3.84x10^24 H atoms.
deff fn [24]

Answer:

A) 14. 25 × 10²³ Carbon atoms

B) 34.72 grams

Explanation:

1 molecule of Propane has 3 atoms of Carbon and 8 atoms of Hydrogen.

The sample has 3.84 × 10²⁴ H atoms.

If 8 atoms of Hydrogrn are present in 1 molecule of propane.

3.84 × 10²⁴ H atoms are present in

\mathfrak{ \frac{3.8 }{8} \times 10 ^{24}}

<u>= 4.75 × 10²³ molecules of Propane</u>.

- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -

No. of Carbon atoms in 1 molecule of propane = 3

=> C atoms in 4.75× 10²³ molecules of Propane = 3 × 4.75 × 10²³

<u>= 14.25 × 10²³ </u>

<u>________________________________________</u>

<u>Gram</u><u> </u><u>Molecular</u><u> </u><u>Mass</u><u> </u><u>of</u><u> </u><u>Propane</u><u>(</u><u>C3H8</u><u>)</u>

= 3 × 12 + 8 × 1

= 36 + 8

= 44 g

1 mole of propane weighs 44g and has 6.02× 10²³ molecules of Propane.

=> 6.02 × 10²³ molecules of Propane weigh = 44 g

=> 4. 75 × 10²³ molecules of Propane weigh =

\mathsf{ \frac{44 }{6.02 \times  {10}^{23} } \times 4.75 \times  {10}^{23}  }

\mathsf{  = \frac{44 }{6.02 \times   \cancel{{10}^{23} }} \times 4.75 \times \cancel{ {10}^{23}}  }

\mathsf{  = \frac{44 }{6.02 } \times 4.75   }

<u>= 34.72 g</u>

8 0
2 years ago
What happens when you drop iodine on a leaf?
insens350 [35]
When you drop iodine on a leaf you may observe a colour change of orange/brown to a blue/black complex.

This is because in the leaf there are starch molecules that form a blue/black complex with the starch molecules.

hope that helps :)
6 0
2 years ago
What is the mass of 0.100 mole of neon? (Watch sf’s)
dimaraw [331]

Answer:

The answer to your question is letter D. 2.02 g

Explanation:

Data

moles of Ne = 0.100

atomic mass of Neon = 20.18 g

Process

1.- Use proportions to find the answer

                   20.18 g of Ne ------------------  1 mol of Ne

                        x                 ------------------  0.1 moles

                        x = (0.1 x 20.18)/1

                        x = 2.018

2.- Consider the significant figures

      0.100 has three significant figures so the answer must be  2.02 g

4 0
3 years ago
Read 2 more answers
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