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3 years ago

The most familiar elements are typically the most

Chemistry

2 answers:

egoroff_w [7]3 years ago

5 0

Explanation:

The most familiar elements are the ones most abundant in the earth and therefore the most economic ones. Good examples of this elements are oxygen, carbon, nitrogen silicium.

rodikova [14]3 years ago

4 0

-Are the most used in daily life and basic human needs

(such as oxygen,nitrogen,carbon, etc)

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What is the third alkali metal

andrey2020 [161]

Answer:

potassium

The third alkali metal is K (potassium). The atomic number of K (potassium) is 19. Thus, the atomic number of third alkali metal is 19

Explanation:

3 0

3 years ago

Read 2 more answers

Does changing the number of neutrons change the identity of the element you have built ?

worty [1.4K]

Answer:

If you change the number of neutrons somehow, nothing will happen because it carry's no charge at all.

Explanation:

4 0

3 years ago

A lead mass is heated and placed in a foam cup calorimeter containing 40.0 mL of water at 17.0°C. The water reaches a temperatur

lbvjy [14]

Answer: 502 Joules

Explanation:

To calculate the mass of water, we use the equation:

$\text{Density of substance}=\frac{\text{Mass of substance}}{\text{Volume of substance}}$

Density of water = 1 g/mL

Volume of water = 40.0 mL

Putting values in above equation, we get:

$1g/mL=\frac{\text{Mass of water}}{40.0mL}\\\\\text{Mass of water}=(1g/mL\times 40.0mL)=40.0g$

When metal is dipped in water, the amount of heat released by lead will be equal to the amount of heat absorbed by water.

$Heat_{\text{absorbed}}=Heat_{\text{released}}$

The equation used to calculate heat released or absorbed follows:

$q=m\times c\times \Delta T$

q = heat absorbed by water

$m$ = mass of water = 40.0 g

$T_{final}$ = final temperature of water = 20.0°C

$T_{initial$ = initial temperature of water = 17.0°C

$c$ = specific heat of water= 4.186 J/g°C

Putting values in equation 1, we get:

$q=40.0\times 4.186\times (20.0-17.0)]$

$q=502J$

Hence, the joules of heat were re-leased by the lead is 502

5 0

3 years ago

How many moles of sodium carbonate are contained by 57.3g of sodium carbonate

Lady_Fox [76]

Answer:

$\boxed {\boxed {\sf 0.541 \ mol \ Na_2CO_3}}$

Explanation:

We are asked to find how many moles of sodium carbonate are in 57.3 grams of the substance.

Carbonate is CO₃ and has an oxidation number of -2. Sodium is Na and has an oxidation number of +1. There must be 2 moles of sodium so the charge of the sodium balances the charge of the carbonate. The formula is Na₂CO₃.

We will convert grams to moles using the molar mass or the mass of 1 mole of a substance. They are found on the Periodic Table as the atomic masses, but the units are grams per mole instead of atomic mass units. Look up the molar masses of the individual elements.

Na: 22.9897693 g/mol
C: 12.011 g/mol
O: 15.999 g/mol

Remember the formula contains subscripts. There are multiple moles of some elements in 1 mole of the compound. We multiply the element's molar mass by the subscript after it, then add everything together.

Na₂ = 22.9897693 * 2= 45.9795386 g/mol
O₃ = 15.999 * 3= 47.997 g/mol
Na₂CO₃= 45.9795386 + 12.011 + 47.997 =105.9875386 g/mol

We will convert using dimensional analysis. Set up a ratio using the molar mass.

$\frac {105.9875386 \ g \ Na_2CO_3}{1 \ mol \ Na_2CO_3}$

We are converting 57.3 grams to moles, so we multiply by this value.

$57.3 \ g \ Na_2CO_3} *\frac {105.9875386 \ g \ Na_2CO_3}{1 \ mol \ Na_2CO_3}$

Flip the ratio so the units of grams of sodium carbonate cancel.

$57.3 \ g \ Na_2CO_3} *\frac {1 \ mol \ Na_2CO_3}{105.9875386 \ g \ Na_2CO_3}$

$57.3 } *\frac {1 \ mol \ Na_2CO_3}{105.9875386 }$

$\frac {57.3 }{105.9875386 } \ mol \ Na_2CO_3$

$0.5406295944 \ mol \ Na_2CO_3$

The original measurement of moles has 3 significant figures, so our answer must have the same. For the number we found that is the thousandth place. The 6 in the ten-thousandth place to the right tells us to round the 0 up to a 1.

$0.541 \ mol \ Na_2CO_3$

There are approximately <u>0.541 moles of sodium carbonate</u> in 57.3 grams.

6 0

2 years ago

How many moles of hydrogen gas can be produced if 0.78 moles of hydrochloric acid reacts with excess solid zinc according to the

White raven [17]

Answer:

The answer to your question is It will be formed 0.39 moles of H₂

Explanation:

Data

moles of H₂ = ?

moles of HCl = 0.78

moles of Zinc = excess

Balanced chemical reaction

2 HCl + Zn ⇒ 1 H₂ + ZnCl₂

Process

1.- Use proportions to solve this problem. Consider the coefficients of the balanced reaction.

2 moles of HCl ---------------------- 1 mol of H₂

0.78 moles of HCl ----------------- x

x = (0.78 x 1) / 2

- Simplification

x = 0.78 / 2

- Result

x = 0.39 moles of H₂

3 0

3 years ago