The amount of heat energy needed to heat 200 g of water from 15 °C to its boiling point, and boil it, is

What layer is made of metals and is a solid?<br> A.Mesosphere<br> B.inner core<br> C.outer core

Nookie1986 [14]

B because inner core has metal and it’s solid

7 0

3 years ago

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7. A small stone added to boiling liquids to make them boil more evenly without sudden violent releases of

Rzqust [24]

Answer:

A boiling chip, boiling stone, porous bit or anti-bumping granule is a tiny, unevenly shaped piece of substance added to liquids to make them boil more calmly.

These help in making the liquid boil more easily

8 0

3 years ago

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A 32.5 g iron rod, initially at 22.4 ∘C, is submerged into an unknown mass of water at 63.0 ∘C, in an insulated container. The f

Allisa [31]

Answer:

The mass of water $m_{w}$ = 39.18 gm

Explanation:

Mass of iron $m_{iron}$ = 32.5 gm

Initial temperature of iron $T_{1}$ = 22.4°c = 295.4 K

Specific heat of iron $C_{iron}$ = 0.448 $\frac{KJ}{kg K}$

Mass of water = $m_{w}$

Specific heat of water $C_{w} = 4.2 \frac{KJ}{kg K}$

Initial temperature of water $T_{2}$ = 336 K

Final temperature after equilibrium $T_{f}$ = 59.7°c = 332.7 K

When iron rod is submerged into water then

Heat lost by water = Heat gain by iron rod

$m_{w}$ $C_{w}$ ( $T_{2}$ - $T_{f}$ ) = $m_{iron}$ $C_{iron}$ ( $T_{f}$ - $T_{1}$ )

Put all the values in above formula we get

$m_{w}$ × 4.2 × ( 336 - 332.7 ) = 32.5 × 0.448 × ( 332.7 - 295.4 )

$m_{w}$ = 39.18 gm

Therefore the mass of water $m_{w}$ = 39.18 gm

8 0

3 years ago

What is the percent by volume of ethanol (C2H60, or ethyl alcohol) in the final solution when 90 mL of ethanol is diluted to a v

r-ruslan [8.4K]

Answer:

16.0%.

Explanation:

Volume percent of a substance is the ratio of the substance volume to the solution volume multiplied by 100.

V % of ethanol = (volume of ethanol / volume of the solution) x 100.

volume of ethanol = 90.0 mL, volume of the solution = 550.0 mL.

∴ V % of ethanol = (90.0 mL / 550.0 mL) x 100 = 16.36% ≅ 16.0%.

8 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