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zzz [600]
3 years ago
11

I have 2 samples of solid chalk (aka calcium carbonate). Sample A has a total mass of 4.12 g and Sample B has a total mass of 19

.37 g. What is the difference between the samples?
A) Sample B has more calcium carbonate molecules
B) Sample B has a larger ratio of carbon, oxygen, and calcium atoms
C) Sample B has more calcium ion than carbonate ions
D) Sample B must have some impurity
Chemistry
1 answer:
IRINA_888 [86]3 years ago
7 0

Answer:

A) Sample B has more calcium carbonate molecules

Explanation:

M = Molar mass of calcium carbonate = 100.0869 g/mol

N_A = Avogadro's number = 6.022\times 10^{23}\ \text{mol}^{-1}

For the 4.12 g sample

Moles of a substance is given by

n=\dfrac{m}{M}\\\Rightarrow n=\dfrac{4.12}{100.0869}\\\Rightarrow n=0.0411\ \text{mol}

Number of molecules is given by

nN_A=0.0411\times 6.022\times 10^{23}=2.48\times 10^{22}\ \text{molecules}

For the 19.37 g sample

n=\dfrac{19.37}{100.0869}\\\Rightarrow n=0.193\ \text{mol}

Number of molecules is given by

nN_A=0.193\times 6.022\times 10^{23}=1.16\times 10^{23}\ \text{molecules}

1.16\times 10^{23}\ \text{molecules}>2.48\times 10^{22}\ \text{molecules}

So, sample B has more calcium carbonate molecules.

The ratio of the elements of carbon, oxygen, calcium atoms, ions, has to be same in both the samples otherwise the samples cannot be considered as calcium carbonate. Same is applicable for impurities. If there are impurites then the sample cannot be considered as calcium carbonate.

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5 0
3 years ago
A mixture 21.7 g NaCl 3.74 g kcl and 9.76 g licl how many moles of nacl are in this mixture
slava [35]
Moles (mol) = mass (g) / molar mass (g/mol)

Mass of NaCl = 21.7 g
Molar mass of NaCl = <span>58.4 g/mol
Hence, moles of NaCl = </span>21.7 g / 58.4 g/mol = 0.372 mol

Hence moles of NaCl in the mixture is 0.372 mol.

Let's assume that mixture has only given compounds and free of impurities.
Then, we can present this as a mole percentage.

mole % = (moles of desired substance / Total moles of the mixture) x 100%

Hence,
mole % of NaCl = (moles of NaCl / Total moles of the mixture) x 100%

Total moles of mixture = moles of NaCl + KCl + LiCl

Mass of KCl = 3.74 g 
Molar mass of NaCl = 74.6 g/mol
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Mass of NaCl = <span>9.76 g
</span>Molar mass of NaCl = 42.4 g/mol
Hence, moles of NaCl = 9.76 g / 42.4 g/mol = 0.230 mol

Total moles =  0.372 mol + 0.050 mol + 0.230 mol = 0.652 mol

mole % of NaCl = (moles of NaCl / Total moles of the mixture) x 100%
                           = (0.372 mol / 0.652 mol) x 100%
                           = 57.06% 

Hence, mixture has 57.06% of NaCl as the mole percentage.
8 0
3 years ago
2.5 piece of lithium is dropped into a 500 g sample of water at 22.1 C.ten temperature of the water increases to 23.5 C. How muc
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Answer:

2.6 kJ  

Explanation:

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q = mCΔT

1. Calculate ΔT

ΔT = 23.5 °C - 22.1 °C = 1.4 °C

2. Calculate q

q₂ = mCΔT = 500 g × 4.184 J·°C⁻¹g⁻¹ × 1.4 °C = 2900 J = 2.9 kJ

3 0
4 years ago
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skelet666 [1.2K]
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6 0
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77julia77 [94]
You have to use the equation q=mcΔT and solve for T(final).  
T(final)=(q/mc)+T(initial)
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m=the mass of the sample (in this case 15.6g)
c= the specific heat capacity of the substance (in this case 2.41 J/g°C)
T(initial)=the initial temperature of the sample (in this case 21.5°C)

When you plug everything in, you should get 44.6°C.
Therefore the final temperature of ethanol is 44.6°C

I hope this helps.  Let me know if anything is unclear.
3 0
3 years ago
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