Answer:
Starch.
Explanation:
When the triiodide combine with starch, it forms dark blue colour. Amylose in starch is responsible for the occurrence of a deep blue color when the iodine is combine with the starch. The iodine molecule goes inside of the amylose coil which makes a linear triiodide ion complex that goes into the coil of the starch that leads to an intense blue-black color in the end so we can say that starch turns the colour into blue.
20ml of water is mixed with 40gm of fine powder. Calculate the concentration of the solution obtained.
Answer:
[tex]\%m=66.7\%[/tex]
Explanation:
Hello there!
In this case, according to the given information, it turns out possible for us to calculate the concentration of the solution obtained, by knowing 20 mL of water are the same to 20 g and therefore the mass of the solution is 40g+20g=60g.
Next, we apply the following equation to obtain the required concentration:
[tex]\%m=\frac{40g}{60g} *100\%\\\\\%m=66.7\%[/tex]
Regards!
Chrysanthenone is an unsaturated ketone. If Chrysanthenone has M+ = 150 and contains 2 double bond(s) and 2 ring(s); what is its molecular formula? Enter the formula in the form CH first, then all other atoms in alphabetical order; do not use subscripts. The formula is case-sensitive.
Answer:
the Molecular formula will be; C10H14O
Explanation:
Given the data in the question;
Chrysanthenone is an unsaturated ketone,
it has M+ = 150 and contains 2 double bond(s) and 2 ring(s).
molecular formula = ?
we know that ketone contain 1 oxygen and mass of oxygen is 16
so mass of the C and H remaining will be;
⇒ 150 - 16 = 134
Now we determine the number of C atoms;
⇒ 134 / 13 = 10
hydrocarbon with 10 hydrogen atom have CnH2n+2 means
⇒ ( 10 × 2 ) +2 = 22 hydrogens
But then we have 3 unsaturation meaning 6 hydrogens less and also we have ring meaning 2 more hydrogens
⇒ 22 - 6 - 2 = 14
Hence the Molecular formula will be; C10H14O
cuáles son las características de la luz y en qué consisten
Answer:
Cuáles son las características de la luz y en qué consisten?
Explanation:
La luz es una radiación que se propaga en forma de ondas. Las ondas que se pueden propagar en el vacío se llaman ONDAS ELECTROMAGNÉTICAS. La luz es una radiación electromagnética
a) If we have a 4.5 L container of CH 10 gas at a temperature of 178 K and a pressure of 0.50 atm, then how many moles of CaHio do
we have?
b) How many grams of C4H1o do we have?
Answer:
a) 0.15 mol.
b) 8.95 g.
Explanation:
Hello there!
In this case, according to the given information, it is possible for us to infer this problem is solved by using the ideal gas equation:
[tex]PV=nRT[/tex]
And proceed as follows:
a) Here, we solve for the moles, n, as follows:
[tex]n=\frac{PV}{RT} \\\\n=\frac{0.50atm*4.5L}{0.08206\frac{atm*L}{mol*K}*178K} \\\\n=0.15mol[/tex]
b) for the calculation of the mass, we recall the molar mass of butane, 58.12 g/mol, to obtain:
[tex]0.15mol*\frac{58.12g}{1mol} =8.95g[/tex]
Regards!
If 4.00 moles of O2 occupies a volume of 5.0 L at a particular temperature and pressure, what volume will 3.00 moles of oxygen gas occupy under the same condition?
Answer: Volume occupied by 3.00 moles of oxygen gas under the same condition is 3.75 L.
Explanation:
Given: [tex]n_{1}[/tex] = 4.00 moles, [tex]V_{1}[/tex] = 5.0 L
[tex]n_{2}[/tex] = 3.00 moles, [tex]V_{2}[/tex] = ?
Formula used is as follows.
[tex]\frac{V_{1}}{n_{1}} = \frac{V_{2}}{n_{2}}[/tex]
Substitute the values into above formula as follows.
[tex]\frac{V_{1}}{n_{1}} = \frac{V_{2}}{n_{2}}\\\frac{5.0 L}{4.00 mol} = \frac{V_{2}}{3.00 mol}\\V_{2} = 3.75 L[/tex]
Thus, we can conclude that volume occupied by 3.00 moles of oxygen gas under the same condition is 3.75 L.
