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2 years ago

18. Predict the products for this single replacement reaction.

Chemistry

1 answer:

fenix001 [56]2 years ago

3 0

Answer:

3rd response

4th response

Explanation:

i lowkey feel like im doing this whole chem exam for u lol. good luck anyway! u rock!

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4. Imagine you have a radioactive sample containing both Virtualium and decayed atoms of Virtualium. After analysis, you find it

Romashka [77]

My sample would be 4000 years old because on my graph, I had about 9 Virtualium left at trial 4 so I am guessing that it would be 4000 years old.

7 0

3 years ago

A gas of unknown identity diffuses at a rate of 155 mL/s in a diffusion apparatus in which carbon dioxide diffuses at the rate o

Ostrovityanka [42]

Answer:

19.07 g mol^-1

Explanation:

The computation of the molecular mass of the unknown gas is shown below:

As we know that

$\frac{Diffusion\ rate\ of unknown\ gas }{CO_{2}\ diffusion\ rate} = \frac{\sqrt{CO_{2\ molar\ mass}} }{\sqrt{Unknown\ gas\ molercular\ mass } }$

where,

Diffusion rate of unknown gas = 155 mL/s

CO_2 diffusion rate = 102 mL/s

CO_2 molar mass = 44 g mol^-1

Unknown gas molercualr mass = M_unknown

Now placing these values to the above formula

$\frac{155mL/s}{102mL/s} = \frac{\sqrt{44 g mol^{-1}} }{\sqrt{M_{unknown}} } \\\\ 1.519 = \frac{\sqrt{44 g mol^{-1}} }{\sqrt{M_{unknown}} } \\\\ {\sqrt{M_{unknown}} } = \frac{\sqrt{44 g mol^{-1}}}{1.519} \\\\ {\sqrt{M_{unknown}} } = \frac{44 g mol^{-1}}{(1.519)^{2}}$

After solving this, the molecular mass of the unknown gas is

= 19.07 g mol^-1

4 0

3 years ago

Complete and balance the molecular equation for the reaction of aqueous sodium carbonate, Na 2 CO 3 Na2CO3 , and aqueous nickel(

antoniya [11.8K]

<u>Answer:</u> The net ionic equation is written below.

<u>Explanation:</u>

Net ionic equation of any reaction does not include any spectator ions.

Spectator ions are defined as the ions which does not get involved in a chemical equation. They are found on both the sides of the chemical reaction when it is present in ionic form.

The chemical equation for the reaction of sodium carbonate and nickel (II) chloride is given as:

$Na_2CO_3(aq.)+NiCl_2(aq.)\rightarrow 2NaCl(aq.)+NiCO_3(s)$

Ionic form of the above equation follows:

$2Na^{+}(aq.)+CO_3^{2-}(aq.)+Ni^{2+}(aq.)+2Cl^{-}(aq.)\rightarrow NiCO_3(s)+2Na^+(aq.)+2Cl^-(aq.)$

As, sodium and chloride ions are present on both the sides of the reaction. Thus, it will not be present in the net ionic equation and are spectator ions.

The net ionic equation for the above reaction follows:

$Ni^{2+}(aq.)+CO_3^{2-}(aq.)\rightarrow NiCO_3(s)$

Hence, the net ionic equation is written above.

3 0

3 years ago

How many liters of oxygen are in 8.32 moles of oxygen at STP

Jlenok [28]

Answer:

186 Liters at STP conditions

Explanation:

1 mole of any gas at STP conditions occupies 22.4 Liters.

Therefore, 8.32 moles O₂(g) = 8.32 moles x 22.4Liters/mole = 186 Liters (3 sig.figs.)

6 0

2 years ago

Calculate the percent composition for each element in each of the following compounds.

jek_recluse [69]

The answer is c because the calculations for the elements have to connect with the compounds

6 0

3 years ago