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Nimfa-mama [501]
3 years ago
6

How many particles are in 1 mol of carbon? 1 mol of lithium? 1 mol of eggs? will 1 mol of each of these substances have the same

mass?
Chemistry
1 answer:
andrezito [222]3 years ago
8 0
<span>One mole of a substance contains Avogadro's number of atoms/molecules/the like. This would mean that all of the items described would have approximately 6.022 * 10^23 atoms, even though their masses would differ. This would be due to the molar mass of each substance being different because of the constituent elements in the substance.</span>
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The Gateway Arch in St. Louis, MO is approximately 630 ft tall. How many U.S. dimes would be in a stack of the same
miskamm [114]

Answer:

142240

Explanation:

We are told in the question:

Height of Gateway Arch in St. Louis, MO = 630ft

We are asked, how many U.S. dimes would be in a stack of the same

height when 1 dime is 1.35 mm thick.

Step 1

Convert height in ft to mm

1 ft = 304.8 mm

630ft =

Cross Multiply

630ft × 304.8mm/1ft

= 192024 mm.

Step 2

To find how many US dimes would be in a stack of the same height

= Total thickness/ Thickness of 1 US dime

= 192024 mm/1.35mm

= 142240

Therefore, the number of dimes that would be in a stack of the same

height is 142240

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3 years ago
100 POINTS BRAINLIEST PICTURE BELOW
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Answer:

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Explanation:

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8 0
1 year ago
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What else can copper react with?
FinnZ [79.3K]

Answer:

copper

Explanation:

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3 years ago
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What are some ways population size is limited within an ecosystem?
Vsevolod [243]
<span>Population growth is limited in ecosystems because of lack of resources. If there were unlimited resources, a population would grow exponentially. For example, if a single bacterium were placed in a petri dish and was given unlimited resources, it would multiply to cover the entire world within several days. Unfortunately, resources are limited, so populations are too.</span>
6 0
3 years ago
Complete and balance the chemical equations for the precipitation reactions, if any, between the following pairs of reactants, a
Crank

Explanation:

a. Pb(NO_3)_2(aq) + Na_2SO_4(aq) → ?

Pb(NO_3)_2(aq) + Na_2SO_4(aq)\rightarrow PbSO_4(s)+2NaNO_3(aq)

Pb(NO_3)_2(aq)\rightarrow Pb^{2+}(aq)+2NO_3^{-}(aq)

Na_2SO_4(aq)\rightarrow 2Na^++SO_4^{2-}(aq)

Pb^{2+}(aq)+2NO_3^{-}(aq)+2Na^++SO_4^{2-}(aq)\rightarrow PbSO_4(s)+2Na^++2NO_3^{-}(aq)

Removing common ions from both sides, we get the net ionic equation:

Pb^{2+}(aq)+SO_4^{2-}(aq)\rightarrow PbSO_4(s)

b. NiCl_2(aq) + NH_4NO_3(aq) →

NiCl_2(aq) + NH4NO_3(aq) \rightarrow Ni(NO_3)_2+NH_4Cl(aq)

No precipitation is occuring.

c. Fe_Cl2(aq) + Na_2S(aq) →

FeCl_2(aq) + Na_2S(aq)\rightarrow FeS(s)+2NaCl(aq)

FeCl_2(aq)\rightarrow Fe^{2+}(aq)+2Cl^{-}(aq)

Na_2S(aq)\rightarrow 2Na^++S{2-}(aq)

Fe^{2+}(aq)+2Cl^{-}(aq)+2Na^++S^{2-}(aq)\rightarrow FeS(s)+2Na^++2Cl^{-}(aq)

Removing common ions from both sides, we get the net ionic equation:

Fe^{2+}(aq)+S^{2-}(aq)\rightarrow FeS(s)

d.MgSO_4(aq) + BaCl_2(aq) →

MgSO4(aq) + BaCl2(aq)\rightarrow BaSO_4(s)+MgCl_2

MgSO_4(aq)\rightarrow Mg^{2+}(aq)+SO_4^{2-}(aq)

BaCl_2(aq)\rightarrow Ba^{2+}+2Cl^{-}(aq)

Mg^{2+}(aq)+SO_4^{2-}(aq)+Ba^{2+}+2Cl^{-}(aq)(aq)\rightarrow BaSO_4(s)+Mg^{2+}(aq)+2Cl^{-}(aq)

Removing common ions from both sides, we get the net ionic equation:

Ba^{2+}(aq)+SO_4^{2-}(aq)\rightarrow BaSO_4(s)

5 0
3 years ago
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