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dmitriy555 [2]
3 years ago
13

PLEASE ANSWER FAST!

Chemistry
2 answers:
Paraphin [41]3 years ago
3 0

Answer:

0.36 g/cm3 I hope this helps

const2013 [10]3 years ago
3 0

Answer:

0.36 hope it helped

Explanation:

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If someone trys to lur u into a car with candy...say NO. and run away
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Answer:

as u should. candy ain't even that tempting >^<

8 0
3 years ago
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Which atom’s ionization energy is greater than that of phosphorus (P)?
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The first ionisation energy of silicon is greater than that of phosphorus.
7 0
3 years ago
In the following reaction, how many grams of nitroglycerin C3H5(NO3)3 will decompose to give 25 grams of CO2?
gogolik [260]

4C₃H₅(NO₃)₃_{(l)} ------> 12CO₂_{(g)} + 6N₂_{(g)} +  10H₂O_{(g)}  +  O₂_{(g)}

mol of CO₂  =  \frac{mass}{molar mass}

                    =  \frac{25g}{44.01 g/mol}

mol ratio of  CO₂ :  C₃H₅(NO₃)₃

                    12    :     4

∴  if  mole of CO₂  =  0.568 mol

then   "       "   C₃H₅(NO₃)₃  =  \frac{0.568 mol}{12}  *  4

                                           = 0.189 mol


∴ mass of nitroglycerin  =  mole  *  Mr

                                       =  0.189 mol  *  227.0995 g / mol

                                       =  43.00 g

8 0
3 years ago
0.20 L of HNO3 is titrated to equivalence using 0.12 L of 0.2 M NaOH. What is the concentration of the HNO3?
AleksandrR [38]

M_{A}V_{A}=M_{B}V_{B}\\(0.20)(M_{A})=(0.12)(0.2)\\M_{A}=\frac{(0.12)(0.2)}{(0.20)}=\boxed{0.12 \text{ M}}

4 0
2 years ago
A solution is made containing 14.6g of CH3OH in 185g H2O.1. Calculate the mole fraction of CH3OH.2. Calculate the mass percent o
Andre45 [30]

Answer:

* x_{CH_3OH}=0.0425

* \%m/m_{CH_3OH}=7.31\%

* m=2.46m

Explanation:

Hello,

In this case, for the mole fraction of methanol we use the formula:

x_{CH_3OH}=\frac{n_{CH_3OH}}{n_{CH_3OH}+n_{water}}

Thus, we compute the moles of both water (molar mass 18 g/mol) and methanol (molar mass 32 g/mol):

n_{CH_3OH}}=14.6g*\frac{1mol}{32g}=0.456molCH_3OH \\\\n_{water}}=185g*\frac{1mol}{18g}=10.3molH_2O

Hence, mole fraction is:

x_{CH_3OH}=\frac{0.456mol}{0.456mol+10.3mol}\\\\x_{CH_3OH}=0.0425

Next, mass percent is:

\%m/m_{CH_3OH}=\frac{m_{CH_3OH}}{m_{CH_3OH}+m_{water}}*100\%\\\\\%m/m_{CH_3OH}=\frac{14.6g}{14.6g+185g}*100\%\\\\\%m/m_{CH_3OH}=7.31\%

And the molality, considering the mass of water in kg (0.185 kg):

m=\frac{n_{CH_3OH}}{m_{water}} =\frac{0.456mol}{0.185kg}\\ \\m=2.46m

Regards.

7 0
3 years ago
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