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3 years ago

How many molecules of H2O are equivalent to 98.2 g of H2O

Chemistry

1 answer:

GREYUIT [131]3 years ago

3 0

Answer:

Demo Mole Quantities

58.5g NaCl(mol/58.5g)(6.02 x 1023/mol) = 6.02 x 1023 Na

Cl21 pre-1982 pennies (after 1982 pennies are mostly zinc with copper coating)

63.5g Cu( mol/ 63.5g)(6.02 x 1023/mol) = 6.02 x 1023 Cu

19.0g Al (mol/27.0g)(6.02 x 1023/mol) = 4.24 x 1023 Al

Explanation:

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10. Carbon tetrachloride has a density of 1.60 g/mL. What is the volume of 40 grams of CCl4?

ikadub [295]

Answer:

The answer is

Explanation:

The volume of a substance when given the density and mass can be found by using the formula

$volume = \frac{mass}{density}$

From the question

mass of CCl4 = 40g

Density = 1.6 g/mL

The volume is

$volume = \frac{40}{1.6}$

We have the final answer as

Hope this helps you

5 0

3 years ago

All matter has thermal energy because atoms are constantly

ololo11 [35]

The atoms are constantly in motion

3 0

3 years ago

The formation constant for the reaction ag (aq) 2nh3(aq) ag(nh3)2 (aq) is kf = 1.7 × 107 at 25°c. what is δg° at this temperatur

anzhelika [568]

The value of ΔG° at this temperature is -18034.18 J/mol

Calculation,

Given information

formation constant (Kf)= 1.7 × $10^{7}$

Universal gas constant (R) = 8.314 J/K• mol

Temperature = 25° C = 25 °C + 273 = 300 K

Formula used:

ΔG° = -RT㏑Kf

By putting the valur of R,T, Kf we get the value of ΔG°

ΔG° = - 8.314 J/K• mol×300K㏑ 1.7 × $10^{7}$

ΔG° = -2494.2㏑ 1.7 × $10^{7}$ = -18034.18 J/mol

So, change in standard Gibbs's free energy is -18034.18 J/mol

Learn about formation constant

brainly.com/question/14011682

#SPJ4

8 0

2 years ago

Charles is given two electrical conductors – aluminium and graphite. Help him to select one for making an electric wire. Justify

Iteru [2.4K]

Answer: aluminum I think

Explanation:

6 0

3 years ago

A 125g metal block at a temperature of 93.2 degrees Celsius was immersed in 100g of water at 18.3 degrees Celsius. Given the spe

nikitadnepr [17]

Answer:

$\large \boxed{34.2\, ^{\circ}\text{C}}$

Explanation:

There are two heat transfers involved: the heat lost by the metal block and the heat gained by the water.

According to the Law of Conservation of Energy, energy can neither be destroyed nor created, so the sum of these terms must be zero.

Let the metal be Component 1 and the water be Component 2.

Data:

For the metal:

$m_{1} =\text{125 g; }T_{i} = 93.2 ^{\circ}\text{C; }\\C_{1} = 0.900 \text{ J$^{\circ}$C$^{-1}$g$^{-1}$}$

For the water:

$m_{2} =\text{100 g; }T_{i} = 18.3 ^{\circ}\text{C; }\\C_{2} = 4.184 \text{ J$^{\circ}$C$^{-1}$g$^{-1}$}$

$\begin{array}{rcl}\text{Heat lost by metal + heat gained by water} & = & 0\\q_{1} + q_{2} & = & 0\\m_{1}C_{1}\Delta T_{1} + m_{2}C_{2}\Delta T_{2} & = & 0\\\text{125 g}\times 0.900 \text{ J$^{\circ}$C$^{-1}$g$^{-1}$} \times\Delta T_{1} + \text{100 g} \times 4.184 \text{ J$^{\circ}$C$^{-1}$g$^{-1}$}\Delta \times T_{2} & = & 0\\112.5\Delta T_{1} + 418.4\Delta T_{2} & = & 0\\112.5\Delta T_{1} & = & -418.4\Delta T_{2}\\\Delta T_{1} & = & -3.719\Delta T_{2}\\\end{array}$

$\Delta T_{1} = T_{\text{f}} - 93.2 ^{\circ}\text{C}\\\Delta T_{2} = T_{\text{f}} - 18.3 ^{\circ}\text{C}$

$\begin{array}{rcl}\Delta T_{1} & = & -3.719\Delta T_{2}\\T_{\text{f}} - 93.2 ^{\circ}\text{C} & = & -3.719 (T_{\text{f}} - 18.3 ^{\circ}\text{C})\\T_{\text{f}} - 93.2 ^{\circ}\text{C} & = & -3.719T_{\text{f}} + 68.06 ^{\circ}\text{C}\\4.719T_{\text{f}} & = & 161.3 ^{\circ}\text{C}\\T_{\text{f}} & = & \mathbf{34.2 ^{\circ}}\textbf{C}\\\end{array}\\\text{The final temperature of the block and the water is $\large \boxed{\mathbf{34.2\, ^{\circ}}\textbf{C}}$}$

3 0

3 years ago