A sample of tin (II) chloride has a mass of 0.49 g. After heating, it has a mass of 0.41 g.

6th grade work help me please (:

frozen [14]

The trophosphere contains the most water vapor!

3 0

3 years ago

PLZ HELP I WILL GIVE BRAINLY AND POINTS

max2010maxim [7]

Answer:

1. Define the problem

2. Conduct a literature search

3. Propose a hypothesis

4. Devise an experiment to prove or disprove

5. State conclusions

Explanation: In order to begin an experiment, you must first define a problem or question that you will be answering. Then you must research the problem in order to form a hypothesis, or an educated guess. Then you should devise and execute an experiment to answer your question. The conclusions that you draw will either prove or disprove your hypothesis. Hope this helps!

8 0

3 years ago

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How many liters of hydrogen gas will be produced at STP from the reaction of 7.179×10^23 atoms of magnesium with 54.219g of phos

Alexeev081 [22]

Answer: The volume of hydrogen gas produced will be, 12.4 L

Explanation : Given,

Mass of $H_3PO_4$ = 54.219 g

Number of atoms of $Mg$ = $7.179\times 10^{23}$

Molar mass of $H_3PO_4$ = 98 g/mol

First we have to calculate the moles of $H_3PO_4$ and $Mg$ .

$\text{Moles of }H_3PO_4=\frac{\text{Given mass }H_3PO_4}{\text{Molar mass }H_3PO_4}$

$\text{Moles of }H_3PO_4=\frac{54.219g}{98g/mol}=0.553mol$

and,

$\text{Moles of }Mg=\frac{7.179\times 10^{23}}{6.022\times 10^{23}}=1.19mol$

Now we have to calculate the limiting and excess reagent.

The balanced chemical equation is:

$3Mg+2H_3PO_4\rightarrow Mg(PO_4)_2+3H_2$

From the balanced reaction we conclude that

As, 3 mole of $Mg$ react with 2 mole of $H_3PO_4$

So, 0.553 moles of $Mg$ react with $\frac{2}{3}\times 0.553=0.369$ moles of $H_3PO_4$

From this we conclude that, $H_3PO_4$ is an excess reagent because the given moles are greater than the required moles and $Mg$ is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of $H_2$

From the reaction, we conclude that

As, 3 mole of $Mg$ react to give 3 mole of $H_2$

So, 0.553 mole of $Mg$ react to give 0.553 mole of $H_2$

Now we have to calculate the volume of $H_2$ gas at STP.

As we know that, 1 mole of substance occupies 22.4 L volume of gas.

As, 1 mole of hydrogen gas occupies 22.4 L volume of hydrogen gas

So, 0.553 mole of hydrogen gas occupies $0.553\times 22.4=12.4L$ volume of hydrogen gas

Therefore, the volume of hydrogen gas produced will be, 12.4 L

4 0

3 years ago

Consider the reaction of 19.0 g of zinc metal with excess silver nitrate to produce silver metal and zinc nitrate. The reaction

Alexeev081 [22]

Answer:

14.5 g silver

Explanation:

This is a problem using the stoichiometry of the reaction. First thing we need is the balanced equation:

Zn + 2 AgNO3 ----------------------- 2 Ag + Zn(NO3)2

We know that 14.6 g of Zn did not reacted, then we can calculate the amount of Zn reacted and do the calculation given the above reaction.

amount Zn reacted: 19.0 -14.6 g Zn = 4.4 g Zn

atomic weight of Zn: 65.37 g/mol

mol Zn reacted: 4.4 g Zn x ( 1 mol Zn/ 65.37 g Zn) = 0.067 mol Zn

We know from the balanced equation that moles of Ag are produced from 1 mol Zn therefore the mol of Ag produced are:

0.067 mol Zn x 2 mol Ag/ 1mol Zn = 0.135 mol Ag

and the mass of silver then will be given by multiplying by the atomic weight of silver:

0.135 mol Ag x 107.9 g/mol = 14.5 g Ag

4 0

3 years ago

1) G to H.<br> 2) I to J.<br> 3) J to I.<br> 4) I to H.

prisoha [69]

3 ..............................

5 0

3 years ago