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3 years ago

When copper ii chloride reacts with sodium nitrate copper ii nitrate and sodium chloride are formed?

1 answer:

5 0

The balanced chemical reaction would be written as follows:

CuCl2 + 2NaNO3 = Cu(NO3)2 + 2NaCl

This is an example of a double displacement type of reaction where the ions of the reactants displaces each other forming two new substances. Hope this answers the question.

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Which option is a mixture?<br> A. potassium<br> B. sugar<br> C. water<br> D. salad

PSYCHO15rus [73]

Answer:

D.Salad

Explanation:

A mixture is a substance that combines two or more substance that are physically combined. A mixture can be easily seperated by a physical means . A mixture is the combination of various substances that are not chemically combined .Examples of mixtures are cake and salad . The cake contains various substances like egg, flour and sugar. This mixture can be easily separated into it individual entities by a physical means .

Base on the option the only mixture among the option is salad . Salad is a a mixtures that combines substances like tomatoes , fruits and vegetable. The individual substances that makes up the salad can be easily separated.

Sugar is a compound . Water is a compound as it contains two elements and potassium is an element.

6 0

3 years ago

Read 2 more answers

The activation energy for a reaction is changed from 184 kJ/mol to 59.0 kJ/mol at 600. K by the introduction of a catalyst. If t

11111nata11111 [884]

Answer:

The catalyzed reaction will take 2.85 seconds to occur.

Explanation:

The activation energy of a reaction is given by:

$k = Ae^{-\frac{E_{a}}{RT}}$

For the reaction without catalyst we have:

$k_{1} = Ae^{-\frac{E_{a_{1}}}{RT}}$ (1)

And for the reaction with the catalyst:

$k_{2} = Ae^{-\frac{E_{a_{2}}}{RT}}$ (2)

Assuming that frequency factor (A) and the temperature (T) are constant, by dividing equation (1) with equation (2) we have:

$\frac{k_{1}}{k_{2}} = \frac{Ae^{-\frac{E_{a_{1}}}{RT}}}{Ae^{-\frac{E_{a_{2}}}{RT}}}$

$\frac{k_{1}}{k_{2}} = e^{\frac{E_{a_{2}} - E_{a_{1}}}{RT}$

$\frac{k_{1}}{k_{2}} = e^{\frac{59.0 \cdot 10^{3}J/mol - 184 \cdot 10^{3} J/mol}{8.314 J/Kmol*600 K} = 1.31 \cdot 10^{-11}$

Since the reaction rate is related to the time as follow:

$k = \frac{\Delta [R]}{t}$

And assuming that the initial concentrations ([R]) are the same, we have:

$\frac{k_{1}}{k_{2}} = \frac{\Delta [R]/t_{1}}{\Delta [R]/t_{2}}$

$\frac{k_{1}}{k_{2}} = \frac{t_{2}}{t_{1}}$

$t_{2} = t_{1}\frac{k_{1}}{k_{2}} = 6900 y*1.31 \cdot 10^{-11} = 9.04 \cdot 10^{-8} y*\frac{365 d}{1 y}*\frac{24 h}{1 d}*\frac{3600 s}{1 h} = 2.85 s$

Therefore, the catalyzed reaction will take 2.85 seconds to occur.

I hope it helps you!

4 0

3 years ago

Given the balanced equation, 2 H2 + O2 -> 2 H2O, answer the following

9966 [12]

Answer: $O_2$ is the limiting reagent

Explanation:

To calculate the moles :

$\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}$

$\text{Moles of} H_2=\frac{10.0g}{2g/mol}=5.00moles$

$\text{Moles of} O_2=\frac{20.0g}{32g/mol}=0.625moles$

The balanced chemical reaction is :

$2H_2+O_2\rightarrow 2H_2O$

According to stoichiometry :

1 moles of $O_2$ require = 2 moles of $H_2$

Thus 0.625 moles of $O_2$ will require= $\frac{2}{1}\times 0.625=1.25moles$ of $H_2$

Thus $O_2$ is the limiting reagent as it limits the formation of product and $H_2$ is the excess reagent as it is present more than the required amount.

8 0

3 years ago

Set the Room temp. to -10.0 °C and observe. What is happening?

Lady bird [3.3K]

Answer:

its getting colder

Explanation:

6 0

3 years ago

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Is energy used during an explosion

Whitepunk [10]