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Whitepunk [10]
3 years ago
13

Type the correct answer in the box.

Chemistry
1 answer:
kompoz [17]3 years ago
5 0

Answer:

3.9g/cm3

Explanation:

Density ( d)=?

Mass(m)=27.3g

Volume (v)=7.0cm3

D=m÷v

D=27.3g÷7.0cm3

D=3.9g/cm3

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If we know this is a second-order reaction, what is the rate law?
Pavel [41]

Answer:

The rate is a mathematical relationship obtained by comparing reaction rate with reactant concentrations.

6 0
3 years ago
How many valence electrons are in atom of radon?
snow_lady [41]

Answer: 8

Explanation: Radon has 8 valence electrons. Radon is considered stable with a complete octet of electrons, filling the s and p orbitals.

4 0
3 years ago
P4+ __02_ P4010 <br> I need help ASAP
klasskru [66]

Answer:

P_4+5O_2\rightarrow P_4O_{10}

Explanation:

Hello!

In this case, when we want to balance chemical reactions such as in this case, the idea is to equal to number of atoms of each element at each side of the equation according to the lay of conservation of mass, just as shown below:

P_4+5O_2\rightarrow P_4O_{10}

Because we have four phosphorous and ten oxygen atoms at each side.

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8 0
3 years ago
What is the percent yield of O2 if 10.2 g of O2 is produced from the decomposition of 17.0 g of H2O?
yawa3891 [41]

The balanced chemical reaction will be:

2H2O = 2H2 + O2

<span>We are given the amount of water used in the decomposition reaction. This will be our starting point.</span>

<span>17.0 g H2O</span> (1 mol  H2O/ 18.02 g H2O) (1 mol O2/2 mol <span>H2O</span>) ( 32.00 g O2/1mol O2) = 15.09 g O2

Percent yield = actual yield / theoretical yield x 100

<span>Percent yield =10.2 g / 15.09  g x 100</span>

Percent yield = 67.58%

5 0
3 years ago
Read 2 more answers
If you assume this reaction is driven to completion because of the large excess of one ion, what is the concentration of [Fe(SCN
viktelen [127]

Answer : The concentration of [Fe(SCN)]^{2+} is, 4.32\times 10^{-4}M

Explanation :

When we assume this reaction is driven to completion because of the large excess of one ion then we are assuming limiting reagent is SCN^- and Fe^{3+} is excess reagent.

First we have to calculate the moles of KSCN.

\text{Moles of }KSCN=\text{Concentration of }KSCN\times \text{Volume of solution}

\text{Moles of }KSCN=0.00180M\times 0.006L=1.08\times 10^{-5}mol

Moles of KSCN = Moles of K^+ = Moles of SCN^- = 1.08\times 10^{-5}mol

Now we have to calculate the concentration of [Fe(SCN)]^{2+}

\text{Concentration of }[Fe(SCN)]^{2+}=\frac{\text{Moles of }[Fe(SCN)]^{2+}}{\text{Volume of solution}}

Total volume of solution = (6.00 + 5.00 + 14.00) = 25.00 mL = 0.025 L

\text{Concentration of }[Fe(SCN)]^{2+}=\frac{1.08\times 10^{-5}mol}{0.025L}=4.32\times 10^{-4}M

Thus, the concentration of [Fe(SCN)]^{2+} is, 4.32\times 10^{-4}M

7 0
3 years ago
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