551 cal of heat is added to 5.00 g ice at –20.0 °c. what is the final temperature of the water?

1 answer:

5 0

This item can be answered using the equation,

h = mcp(dT) + mHv

where h is the heat, m is the mass of the substance, cp is the specific heat, and dT is the temperature difference, Hv is the latent heat of fusion. Substituting the known values from the given above,

551 cal = (5 g)(1 cal/g°C)(T - -20) + (5 g)(80 cal/g)

The value of T from the equation is 10.2.

Answer: 10.2°C

You might be interested in

Elements with similar properties are located

Nataly [62]

Those elements with similar properties are in the same column.

4 0

3 years ago

Determine the type of reaction for the following equation: C2H6 + O2à H2O + CO2

valkas [14]

Answer:

The type of reaction for the following equation is combustion equation.

Explanation:

Combustion reaction is defined as the chemical reaction in which a hydrocarbon reacts with oxygen gas to produce carbon dioxide gas and water molecule.

$\text{hydrocarbon}+O_2\rightarrow CO_2+H_2O$

The reaction given to us:

$C_2H_6 + \frac{7}{2}O_2\rightarrow 3H_2O + 2CO_2$

When 1 mole of ethane reacts with 7/2 moles of oxygen gas it gives 3 moles of water and 2 moles of carbon dioxide gas.

The type of reaction for the following equation is combustion equation.

4 0

3 years ago

A 7.36 g sample of copper is contaminated with a additional 0.51 g sample of zinc. suppose an atomic mass measurement was perfor

OleMash [197]

The average molecular weight of the mixture can be calculated using this formula:
MWav = x1MW1 + x2MW2

Where x is the mass fraction of the components of the mixture, in this case, copper (63.546 g/mol) and zinc (65.38 g/mol).

x1 = 7.36 / (7.36+0.51)=0.935
x2 = 0.51 / (7.36+0.51)=0.065

So,
MWav = 0.935(63.546) + 0.065(65.38) = 63.665 g/mol

3 0

3 years ago

Given the two reactions H2S(aq)⇌HS−(aq)+H+(aq), K1 = 9.57×10−8, and HS−(aq)⇌S2−(aq)+H+(aq), K2 = 1.46×10−19, what is the equilib

Dvinal [7]

Answer: The value of $K_c$ for the final reaction is $7.16\times 10^{25}$

Explanation:

The given chemical equations follows:

Equation 1: $H_2S(aq.)\rightleftharpoons HS^-(aq.)+H^(aq.);K_1$

Equation 2: $HS^-(aq.)\rightleftharpoons S^{2-}(aq.)+H^(aq.);K_2$

The net equation follows:

$S^{2-}(aq.)+2H^+(aq.)\rightleftharpoons H_2S(aq.);K_c$

As, the net reaction is the result of the addition of reverse of first equation and the reverse of second equation. So, the equilibrium constant for the net reaction will be the multiplication of inverse of first equilibrium constant and the inverse of second equilibrium constant.

The value of equilibrium constant for net reaction is:

$K_c=\frac{1}{K_1}\times \frac{1}{K_2}$