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Law Incorporation [45]
2 years ago
6

A monosaccharide that consists of 5 carbon atoms, one of which is in a ketone group, is classified as a(n) _______

Chemistry
1 answer:
Elenna [48]2 years ago
5 0

A monosaccharide that consists of 5 carbon atoms, one of which is in a ketone group, is classified as a(n) ketopentose.

Ketone family

The suffix "-one" is added after locating the carbonyl group and, if necessary, designating it with a location number. Ketones are given their common names by first identifying the alkyl groups linked to the carbonyl (in alphabetical order), followed by the prefix "ketone." Several Regular Ketones. The three most significant ketones in terms of scale are acetone, methyl ethyl ketone, and cyclohexanone. Though less frequently than in general organic chemistry, they are still widespread in biochemistry. The simplest ketone is dimethyl ketone, or CH3COCH3, often known as acetone.

To learn more about the ketone group refer here:

brainly.com/question/14002520

#SPJ4

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How many moles of ions are released when 0.27 mole of cobalt chloride is dissolved in water?
egoroff_w [7]

0.81 moles

Explanation:

cobalt chloride when dissolved in water dissociate into cobalt and chloride ions as shown;

CoCl₂ (s) → [Co²⁺](aq) + 2[Cl⁻] (aq)

The mole ratio between CoCl₂ and [Co²⁺] is 1 ; 1 while the mole ratio between CoCl₂ and [Cl⁻] is 1 : 2

Therefore; 0.27 moles of CoCl₂ will form;

(1 * 0.27) = 0.27 moles of [Co²⁺] ions

(2 * 0.27) = 0.54 moles of [Cl⁻] ions

Total moles of dissociated ions are;

0.27 + 0.54

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If you combine 230.0 mL 230.0 mL of water at 25.00 ∘ C 25.00 ∘C and 120.0 mL 120.0 mL of water at 95.00 ∘ C, 95.00 ∘C, what is t
Thepotemich [5.8K]

<u>Answer:</u> The final temperature of the mixture is  49°C

<u>Explanation:</u>

To calculate the mass of water, we use the equation:

\text{Density of substance}=\frac{\text{Mass of substance}}{\text{Volume of substance}}

  • <u>For cold water:</u>

Density of cold water = 1 g/mL

Volume of cold water = 230.0 mL

Putting values in above equation, we get:

1g/mL=\frac{\text{Mass of water}}{230.0mL}\\\\\text{Mass of water}=(1g/mL\times 230.0mL)=230g

  • <u>For hot water:</u>

Density of hot water = 1 g/mL

Volume of hot water = 120.0 mL

Putting values in above equation, we get:

1g/mL=\frac{\text{Mass of water}}{120.0mL}\\\\\text{Mass of water}=(1g/mL\times 120.0mL)=120g

When hot water is mixed with cold water, the amount of heat released by hot water will be equal to the amount of heat absorbed by cold water.

Heat_{\text{absorbed}}=Heat_{\text{released}}

The equation used to calculate heat released or absorbed follows:

Q=m\times c\times \Delta T=m\times c\times (T_{final}-T_{initial})

m_1\times c\times (T_{final}-T_1)=-[m_2\times c\times (T_{final}-T_2)]      ......(1)

where,

q = heat absorbed or released

m_1 = mass of hot water = 120 g

m_2 = mass of cold water = 230 g

T_{final} = final temperature = ?°C

T_1 = initial temperature of hot water = 95°C

T_2 = initial temperature of cold water = 25°C

c = specific heat of water = 4.186 J/g°C

Putting values in equation 1, we get:

120\times 4.186\times (T_{final}-95)=-[230\times 4.186\times (T_{final}-25)]

T_{final}=49^oC

Hence, the final temperature of the mixture is  49°C

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