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2 years ago

A discovery only brings extensive change when

2 answers:

dsp732 years ago

7 0

Answer is: it comes at the right time.
Some of the chemistry discoveries that come in right time (with enough informations, knowledge and right technology to use that discovery) are: Periodic Table of the Elements, discovery of electron, proton, neutron, radioactivity, pasteurization, the Haber-Bosch process.

Ipatiy [6.2K]2 years ago

7 0

Answer: A discovery only brings extensive change when (1 point)

Explanation: it comes at the right time

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Lead oxide is formed when a lead cation that has a charge of 2+ combines with an oxygen anion that has a charge of 2-. Using the

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PbO
Not sure how to show crisscross method
Make sure adding the charges together=0 and multiply the elements when necessary to balance charge

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2 years ago

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What is the molarity of solution obtained when 5.71 g of sodium carbonate-10-water is dissolved in water and made up to 250.0 cm

disa [49]

We have to know the molarity of solution obtained when 5.71 g of Na₂CO₃.10 H₂O is dissolved in water and made up to 250 cm³ solution.

The molarity of solution obtained when 5.71 g of sodium carbonate-10-water (Na₂CO₃.10 H₂O) is dissolved in water and made up to 250.0 cm^3 solutionis: (A) 0.08 mol dm⁻³

The molarit y of solution means the number of moles of solute present in one litre of solution. Here solute is Na₂CO₃.10 H₂O and solvent is water. Volume of solution is 250 cm³.

Molar mass of Na₂CO₃.10 H₂O is 286 grams which means mass of one mole of Na₂CO₃.10 H₂O is 286 grams.

5.71 grams of Na₂CO₃.10 H₂O is equal to $\frac{5.71}{286}$ = 0.0199 moles of Na₂CO₃.10 H₂O. So, 0.0199 moles of Na₂CO₃.10 H₂O present in 250 cm³ volume of solution.

Hence, number of moles of Na₂CO₃.10 H₂O present in one litre (equal to 1000 cm³) of solution is $\frac{0.0199 X 1000}{250}$ = 0.0796 moles. So, the molarity of the solution is 0.0796 mol/dm³ ≅ 0.08 mol/dm³

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2 years ago

Tollens’ test, also known as silver-mirror test, is a qualitative laboratory test used to distinguish between an aldehyde and a

Makovka662 [10]

Explanation:

Tollens' reagent is prepared by using two-step process : -

Step 1:

Silver oxide is formed by mixing aqueous silver nitrate with base like sodium hydroxide. The reaction is shown below as:

$AgNO_3 + NaOH &\rightarrow AgOH + NHO_3 \nonumber \\ 2AgOH &\rightarrow Ag_2O + H_2O$

Step 2

Ammonia solution is drop-wise added until all the silver oxide dissolves to form the reagent. The reaction is shown below as:

$Ag_2O + 4NH_3 + H_2O \rightarrow 2Ag(NH_3)_2^+ + 2OH^-$

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3 years ago

Which of the following (with specific heat capacity provided) would show the smallest temperature change upon gaining 200.0 J of

Brut [27]

Answer: The smallest temperature change is shown by water.

Explanation:

To calculate the heat absorbed or released, we use the equation:

$q=mC\times \Delta T$ ......(1)

where,

q = heat absorbed = 200.0 J

m = mass of the substance

C = specific heat of substance

$\Delta T$ = change in temperature

As, the same amount of heat is getting absorbed in all the cases. So, the change in temperature will depend on the product of mass and specific heat.

For the given options:

Option a: 50.0 g Fe, $C_{Fe}=0.449J/g^oC$

We are given:

$m=50.0g\\C_{Fe}=0.449J/g^oC$

Putting values in equation 1, we get:

$200.0J=50.0g\times 0.449J/g^oC\times \Delta T\\\\\Delta T=\frac{200}{50\times 0.449}=8.99^oC$

Change in temperature = 8.99°C

Option b: 50.0 g water, $C_{water}=4.18J/g^oC$

We are given:

$m=50.0g\\C_{water}=4.18J/g^oC$

Putting values in equation 1, we get:

$200.0J=50.0g\times 4.18J/g^oC\times \Delta T\\\\\Delta T=\frac{200}{50\times 4.18}=0.96^oC$

Change in temperature = 0.96°C

Option b: 25.0 g Pb, $C_{Pb}=0.128J/g^oC$

We are given:

$m=50.0g\\C_{Pb}=0.128J/g^oC$

Putting values in equation 1, we get:

$200.0J=25.0g\times 0.128J/g^oC\times \Delta T\\\\\Delta T=\frac{200}{25\times 0.128}=62.5^oC$

Change in temperature = 62.5°C

Option d: 25.0 g Ag, $C_{Ag}=0.235J/g^oC$

We are given:

$m=25.0g\\C_{Ag}=0.235J/g^oC$

Putting values in equation 1, we get:

$200.0J=25.0g\times 0.235J/g^oC\times \Delta T\\\\\Delta T=\frac{200}{25\times 0.235}=34.04^oC$

Change in temperature = 34.04°C

Option e: 25.0 g granite, $C_{granite}=0.79J/g^oC$

We are given:

$m=25.0g\\C_{Fe}=0.79J/g^oC$

Putting values in equation 1, we get:

$200.0J=25.0g\times 0.79J/g^oC\times \Delta T\\\\\Delta T=\frac{200}{25\times 0.79}=10.13^oC$

Change in temperature = 10.13°C

Hence, the smallest temperature change is shown by water.

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2 years ago

for the given reaction, what volume of o2 would be required to react with 4.8 l of co , measured at the same temperature and pre

SVEN [57.7K]

We can see that 2 moles of The no react with 1 mole of O2 using this equation. 4.8 L NO x 1 L O2 / 2 L NO = 2.4 L of O2 are needed at constant pressure and temperature.

What is an example of pressure?

One can see a simple illustration of pressure by using a knife against a few fruit. If you press the flat side of the knife against the fruit, the top won't be cut. The force is spread more than a wide area (low pressure).

What are different types of pressure?

The physical pressure exerted to an object is referred to as pressure. Per unit area, a parallel force is applied to the surface of the objects. F/A (Force per Area) is the basic formula for pressure.

To know more about pressure visit:

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1 year ago