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s344n2d4d5 [400]
3 years ago
13

What is the conversion of 12.044x10^23 to a gram

Chemistry
1 answer:
postnew [5]3 years ago
6 0

Answer:

8.4g

Explanation:

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7. Write a formula for each of the following:
sergeinik [125]

Answer : The formula for each of the following is:

(a) C_{22}H_{46}

(b) C_{17}H_{34}

(c) C_{13}H_{24}

Explanation :

  • Alkanes are hydrocarbon in which the carbon atoms are connected with single covalent bonds.

The general formula of alkanes is C_n H_{2n+2} where n is the number of the carbon atoms present in a molecule of alkane.

  • Alkenes are hydrocarbon in which the carbon atoms are connected with double covalent bonds.

The general formula of alkenes is C_n H_{2n} where n is the number of the carbon atoms present in a molecule of alkene.

  • Alkynes are hydrocarbon in which the carbon atoms are connected with triple covalent bonds.

The general formula of alkynes is C_n H_{2n-2} where n is the number of the carbon atoms present in a molecule of alkyne.

(a) An alkane with 22 carbon atoms

Putting n = 22 in the general formula of alkane, we get the formula of alkane as, C_{22}H_{2(22)+2} or C_{22}H_{46}

(b) An alkene with 17 carbon atoms

Putting n = 17 in the general formula of alkene, we get the formula of alkene as, C_{17}H_{2(17)} or C_{17}H_{34}

(c) An alkyne with 13 carbon atoms

Putting n = 13 in the general formula of alkyne, we get the formula of alkyne as, C_{13}H_{2(13)-2} or C_{13}H_{24}

3 0
3 years ago
Read 2 more answers
Phosphorus-32 is radioactive and has a half life of 14.3 days. What percentage of a sample would be left after 12.6 days
nadya68 [22]

Answer:

There is 54.29 % sample left after 12.6 days

Explanation:

Step 1: Data given

Half life time = 14.3 days

Time left = 12.6 days

Suppose the original amount is 100.00 grams

Step 2: Calculate the percentage left

X = 100 / 2^n

⇒ with X = The amount of sample after 12.6 days

⇒ with n = (time passed / half-life time) = (12.6/14.3)

X = 100 / 2^(12.6/14.3)

X = 54.29

There is 54.29 % sample left after 12.6 days

8 0
2 years ago
Organic molecules always:
trapecia [35]
Organic chemistry is the study of molecules containing carbon, so A
8 0
3 years ago
In a particular titration experiment a 30.0 ml sample of an unknown hcl solution required 25.0 ml of 0.200 m naoh for the end po
Nikitich [7]
The balanced equation for the acid base reaction is as follows
NaOH + HCl ---> NaCl + H₂O
stoichiometry of NaOH to HCl is 1:1
the number of NaOH moles reacted - 0.200 mol/L x 0.0250 L = 0.005 mol
according to molar ratio
number of NaOH moles reacted = number of HCl moles reacted 
therefore number of HCl moles - 0.005 mol 
volume of 30.0  mL contains 0.005 mol
therefore 1000 mL contains - 0.005 mol / 0.030 L = 0.167 M
concentration of HCl is 0.167 M
5 0
3 years ago
(b) in what way are carbon dioxide and orange juice similar?​
ArbitrLikvidat [17]

Answer:

The effects of supercritical CO2 (SC-CO2) on the microbiological, sensory (taste, odour, and colour), nutritional (vitamin C content), and physical (cloud, total acidity, pH, and °Brix) qualities of orange juice were studied. The CO2 treatment was performed in a 1 litre capacity double-walled reactor equipped with a magnetic stirring system. Freshly extracted orange juice was treated with supercritical CO2, pasteurised at 90°C, or left untreated. There were no significant differences in the sensory attributes and physical qualities between the CO2 treated juice and freshly extracted juice. The CO2 treated juice retained 88% of its vitamin C, while the pasteurised juice was notably different from the fresh juice and preserved only 57% of its vitamin C content. After 8 weeks of storage at 4°C, there was no microbial growth in the CO2 treated juice.

6 0
2 years ago
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