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2 years ago

Need help with chemistry

Chemistry

1 answer:

AlekseyPX2 years ago

3 0

Answer:

you need to use the 2 because I already did it

Explanation:

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A newly discovered element, Z, has two naturally occurring isotopes. 90.3 percent of the sample is an isotope with a mass of 267

kirill115 [55]

Ok so, remember that the average atomic mass is what is seen on the periodic table. It is the average mass of all of the isotopes with their frequency taken into account. What you need to do is add the products of the masses and frequencies Just like this:

0.903*267.8 + 0.097*270.9
When you add it the result is what you are looking for

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3 years ago

One atom of silicon can properly be combined in a compound with

Nina [5.8K]

One atom of silicon can properly be combined in a compound with two atoms of oxygen.

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3 years ago

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Most carbon based molecules are classified as

kherson [118]

the answer is organic molecule

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3 years ago

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How many sulfur atoms are present in 25.6 g of Al2(S2O3)3

IgorC [24]

Given the mass of $Al_{2}(S_{2}O_{3})_{3}$ =25.6 g

The molar mass of $Al_{2}(S_{2}O_{3})_{3}$ =390.35g/mol

Converting mass of $Al_{2}(S_{2}O_{3})_{3}$ to moles:

$25.6 g Al_{2}(S_{2}O_{3})_{3}*\frac{1molAl_{2}(S_{2}O_{3}}{390.35 gAl_{2}(S_{2}O_{3}} =0.0656molAl_{2}(S_{2}O_{3}$

Converting mol $Al_{2}(S_{2}O_{3})_{3}$ to mol S:

$0.0656mol Al_{2}(S_{2}O_{3})_{3}*\frac{6molS}{1mol Al_{2}(S_{2}O_{3})_{3}}=0.3936 molS$

Converting mol S to atoms of S using Avogadro's number:

1 mol = $6.022*10^{23}atoms$

$0.3936mol S *\frac{6.022*10^{23}atoms S}{1 mol S}=2.37*10^{23} S atoms$

5 0

3 years ago

After balancing the following reaction under acidic conditions, how many mole equivalents of water are required and on which sid

nordsb [41]

Answer:

d. 8 moles of H2O on the product side

Explanation:

Hello,

In this case, we need to balance the given redox reaction in acidic media as shown below:

$MnO_4^{1-} (aq) + Cl^{1-} (aq) \rightarrow Mn^{2+} (aq) + Cl_2 (g)\\\\(Mn^{7+}O^{2-}_4)^{1-} (aq) + Cl^{1-} (aq) \rightarrow Mn^{2+} (aq) + Cl_2 (g)\\\\\\\\(Mn^{7+}O^{2-}_4)^{1-} (aq)+8H^++5e^- \rightarrow Mn^{2+}+4H_2O\\\\2Cl^{1-}\rightarrow Cl_2^0+2e^-\\\\2*[(Mn^{7+}O^{2-}_4)^{1-} (aq)+8H^++5e^- \rightarrow Mn^{2+}+4H_2O]\\\\5*[2Cl^{1-}\rightarrow Cl_2^0+2e^-]\\\\\\\\2(Mn^{7+}O^{2-}_4)^{1-} (aq)+16H^++10e^- \rightarrow 2Mn^{2+}+8H_2O\\\\10Cl^{1-}\rightarrow 5Cl_2^0+10e^-\\$

Then, we add the half reactions:

$2(Mn^{7+}O^{2-}_4)^{1-} (aq)+16H^++10Cl^{1-} \rightarrow 2Mn^{2+}+8H_2O+5Cl_2^0$

Thereby, we can see d. 8 moles of H2O on the product side.

Best regards.

4 0

3 years ago