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4 years ago

Which aspects of Dalton's atomic theory are disproven by Thomson's discovery of the

Chemistry

1 answer:

abruzzese [7]4 years ago

4 0

Answer:

Dalton’s theory suggests that an atom is indivisible. However, Thomson discovered the existence of subatomic particles called electrons.

Explanation:

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15. Ammonium nitrate, NH4NO3, and aluminum powder react explosively producing nitrogen gas, water vapor and aluminum oxide. Writ

Lunna [17]

Answer:

$2Al_(_S_) ~+~3NH_4NO_3_(_a_q_) ->3N_2_(_g_) ~+~6H_2O_(_g_) ~+~Al_2O_3_(_S_)$

$Entalpy=-2861.9~KJ$

Explanation:

In this case, we have to start with the reagents:

$Al~+~NH_4NO_3$

The compounds given by the problem are:

-) Nitrogen gas = $N_2$

-) Water vapor = $H_2O$

-) Aluminum oxide = $Al_2O_3$

Now, we can put the products in the reaction:

$Al_(_S_) ~+~NH_4NO_3_(_aq_) ->N_2_(_g_) ~+~H_2O_(_g_) ~+~Al_2O_3_(_S_)$

When we balance the reaction we will obtain:

$2Al_(_S_) ~+~3NH_4NO_3_(_a_q_) ->3N_2_(_g_) ~+~6H_2O_(_g_) ~+~Al_2O_3_(_S_)$

Now, for the enthalpy change, we have to find the standard enthalpy values:

$Al_(_S_)=0~KJ/mol$

$NH_4NO_3_(_a_q_)=-132.0~KJ/mol$

$N_2_(_g_)=0~KJ/mol$

$H_2O_(_g_)=~-~241.8~KJ/mol$

$Al_2O_3_(_S_)=~-~1675.7~KJ/mol$

With this in mind, if we multiply the number of moles (in the balanced reaction) by the standard enthalpy value, we can calculate the energy of the reagents:

$(0*2)~+~(-132*3)=~-396~KJ$

And the products:

$(0*3)~+~(-241.8*6)~+~(-1675.7*1)=-3125.9~KJ$

Finally, for the total enthalpy we have to subtract products by reagents :

$(-3125.9~KJ)-(-396~KJ)=-2729.9~KJ$

I hope it helps!

8 0

3 years ago

The hardest known mineral is _____. topaz talc diamond graphite

vagabundo [1.1K]

The hardest known mineral is diamond.

5 0

3 years ago

Read 2 more answers

NaCl has a Delta. Hfus = 30. 2 kJ/mol. What is the mass of a sample of NaCl that needs 732. 6 kJ of heat to melt completely? Use

tia_tia [17]

The mass of NaCl sample has been 24.3 g. Thus, option A is correct.

The heat of fusion has been the amount of heat required to convert 1 mole of substance into solid to liquid state.

The heat required has been given as:

$Q=m\Delta H$

<h3>Computation for the mass of NaCl</h3>

The given solution has heat of fusion, $\Delta H_{fus}=30.2\rm kJ/mol$

The heat required to melt the sample has been, $Q=732.6\;\rm kJ/mol$

Substituting the values for the mass of NaCl

$\rm 732.6\;kJ/mol=Mass\;\times\;30.2\;kJ/mol\\\\
Mass=\dfrac{732.6}{30.2}\;g\\\\
 Mass=24.3\;g\\$

The mass of NaCl sample has been 24.3 g. Thus, option A is correct.

Learn more about heat of fusion, here:

brainly.com/question/87248

5 0

3 years ago

The flask contains 10.0 mL of HCl and a few drops of phenolphthalein indicator. The buret contains 0.160 M NaOH. It requires 18.

olchik [2.2K]

Answer:

Approximately $0.291\; \rm M$ (rounded to two significant figures.)

Explanation:

The unit of concentration $\rm M$ is the same as $\rm mol \cdot L^{-1}$ (moles per liter.) On the other hand, the volume of both the $\rm NaOH$ solution and the original $\rm HCl$ solution here are in milliliters. Convert these two volumes to liters:

$V(\mathrm{NaOH}) = 18.2\; \rm mL = 18.2 \times 10^{-3}\; \rm L = 0.0182\; \rm L$ .
$V(\text{$\mathrm{HCl}$, original}) = 10.0\; \rm mL = 10.0\times 10^{-3}\; \rm L = 0.0100\; \rm L$ .

Calculate the number of moles of $\rm NaOH$ in that $0.0182\; \rm L$ of $0.160\; \rm M$ solution:

$\begin{aligned} n(\mathrm{NaOH}) &= c(\mathrm{NaOH})\cdot V(\mathrm{NaOH})\\ &= 0.160\; \rm mol \cdot L^{-1} \times 0.0182\; \rm L \approx 0.00291\; \rm mol\end{aligned}$ .

$\rm HCl$ reacts with $\rm NaOH$ at a one-to-one ratio:

$\rm HCl\; (aq) + NaOH\; (aq) \to NaCl\; (aq) + H_2O\; (l)$ .

Coefficient ratio:

$\displaystyle \frac{n(\mathrm{HCl})}{n(\mathrm{NaOH})} = 1$ .

In other words, one mole of $\rm NaOH$ would neutralize exactly one mole of $\rm HCl$ . In this titration, $0.291\; \rm mol$ of $\rm NaOH\!$ was required. Therefore, the same amount of $\rm HC$ should be present in the original solution:

$\begin{aligned}&n(\text{$\mathrm{HCl}$, original})\\ &= n(\mathrm{NaOH})\cdot \frac{n(\mathrm{HCl})}{n(\mathrm{NaOH})} \\ &\approx 0.00291\; \rm mol \times 1 = 0.00291\; \rm mol\end{aligned}$ .

Calculate the concentration of the original $\rm HCl$ solution:

$\displaystyle c(\text{$\mathrm{HCl}$, original}) = \frac{n(\text{$\mathrm{HCl}$, original})}{V(\text{$\mathrm{HCl}$, original})} \approx \frac{0.00291\; \rm mol}{0.0100\; \rm L} \approx 0.291\; \rm M$ .

5 0

3 years ago

6th grade science . :((

Nadusha1986 [10]

Answer:

C- private

Explanation:

This event seems to be exclusive. That is why it is private.

3 0

3 years ago