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frutty [35]
3 years ago
13

How many atoms of hydrogen and oxygen are present in 5 gm of HNO3

Chemistry
2 answers:
Ghella [55]3 years ago
7 0

ANSWER: note the amounts of atoms of all the component in HNO3, which are 1 <em><u>atom</u></em><em><u> </u></em><em><u>of</u></em><em><u> </u></em><em><u>hydrogen,</u></em><em><u> </u></em><em><u>1</u></em><em><u> </u></em><em><u>atom</u></em><em><u> </u></em><em><u>of</u></em><em><u> </u></em><em><u>nitrogen </u></em><em><u>and </u></em><em><u>3</u></em><em><u> </u></em><em><u>atom</u></em><em><u> </u></em><em><u>of</u></em><em><u> </u></em><em><u>oxygen</u></em><em><u>.</u></em>

Rufina [12.5K]3 years ago
6 0
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What is a polar liquid?
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Calculate the theoretical yield for the bromination of both stilbenes
podryga [215]

Answer:

cinnamic acid - 150 mg

cis-stilbene - 100 μL

trans- stilbene - 100 mg

pyridinium tribromide - 200-385 mg

For this data:

moles of cinnamic acid = 0.150 g/148.16 g/mol = 0.001 mols

Theoretical mass of dibromoproduct formed = 0.001 mol x 307.97 g/mol = 0.312 g

cis-stilbene (100 ul = 0.1 ml)

moles of cis-stilbene = 0.1 ml x 1.01 g/mol/180.25 g/mol = 0.00056 mols

Theoretical mass of dibromoproduct formed = 0.00056 mol x 340.05 g/mol = 0.19 g

trans-stilbene

moles of tran-stilbene = 0.1 g/180.25 g/mol = 0.00055 mols

Theoretical mass of dibromoproduct formed = 0.00055 mol x 340.05 g/mol = 0.19 g

Explanation:

4 0
2 years ago
A pure substance is found to contain 53.7% fluorine and 46.3% xenon by mass. What is the empirical formula of this substance?
evablogger [386]

Answer: XF8

Explanation:

Empirical Formular shows the simplest ratio of elements in a compound.

 Xe = 46.3%             F  = 53.7%

Divide the percentage composition of each element by the atomic  mass.

Xe = 46.3/ 131.3                      F= 53.7/ 19

      = 0.353( approx)               =  2.826 (approx)

Divide through with the smallest of the answers gotten in previous step.

 Xe = 0.353 / 0.353                F = 2.826/ 0.353

       =  1                                       = 8.0

Empirical formular = XF8

3 0
3 years ago
A sample of carbon dioxide gas occuples a volume of 2.5L at standard temperature and pressure (STP). What will be the volume of
MArishka [77]

Answer:

The correct answer is 5.0 L

Explanation:

STP are defined as T=273 K and P= 1 atm

By using the ideal gas equation, we can calculate the number of moles (n) of the gas at a volume V=2.5 L:

PV= nRT

⇒n= (PV)/(RT) =(1 atm x 2,5 L)/(0.082 L.atm/K.mol x 273 K)= 0.112 mol

For a sample of argon gas, with the same number of moles (0.112 mol) but twice the temperature (T = 273 K x 2= 546 K):

V= (nRT)/P = (0.112 mol x 0.082 L.atm/K.mol x 546 K)/1 atm = 5.0 L

That is consistent with the fact that when a gas is heated, it expanses. So, if the temperature increases twice, the volume also increases twice.

5 0
3 years ago
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