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3 years ago

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Is glucose an pure substance or mixture?

2 answers:

mel-nik [20]3 years ago

7 0

Yes glucose is a pure substance

Nesterboy [21]3 years ago

6 0

A pure substance hope this helped

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1. When iron is oxidized, what is reduced?

lys-0071 [83]

Answer:Oxygen

Explanation:

Oxygen gets reduced when iron is oxidized.

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3 years ago

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How does evidence of chemical

Sliva [168]

Answer:

Changes in Properties Changes in properties result when new substances form. For instance, gas production, formation of a precipitate, and a color change are all possible evidence that a chemical reaction has taken place. ... Change in Color A color change can signal that a new substance has formed.

Explanation:

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2 years ago

From the value Kf=1.2×109 for Ni(NH3)62+, calculate the concentration of NH3 required to just dissolve 0.016 mol of NiC2O4 (Ksp

Nina [5.8K]

<u>Answer:</u> The concentration of $NH_3$ required will be 0.285 M.

<u>Explanation:</u>

To calculate the molarity of $NiC_2O_4$ , we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}$

Moles of $NiC_2O_4$ = 0.016 moles

Volume of solution = 1 L

Putting values in above equation, we get:

$\text{Molarity of }NiC_2O_4=\frac{0.016mol}{1L}=0.016M$

For the given chemical equations:

$NiC_2O_4(s)\rightleftharpoons Ni^{2+}(aq.)+C_2O_4^{2-}(aq.);K_{sp}=4.0\times 10^{-10}$

$Ni^{2+}(aq.)+6NH_3(aq.)\rightleftharpoons [Ni(NH_3)_6]^{2+}+C_2O_4^{2-}(aq.);K_f=1.2\times 10^9$

Net equation: $NiC_2O_4(s)+6NH_3(aq.)\rightleftharpoons [Ni(NH_3)_6]^{2+}+C_2O_4^{2-}(aq.);K=?$

To calculate the equilibrium constant, K for above equation, we get:

$K=K_{sp}\times K_f\\K=(4.0\times 10^{-10})\times (1.2\times 10^9)=0.48$

The expression for equilibrium constant of above equation is:

$K=\frac{[C_2O_4^{2-}][[Ni(NH_3)_6]^{2+}]}{[NiC_2O_4][NH_3]^6}$

As, $NiC_2O_4$ is a solid, so its activity is taken as 1 and so for $C_2O_4^{2-}$

We are given:

$[[Ni(NH_3)_6]^{2+}]=0.016M$

Putting values in above equations, we get:

$0.48=\frac{0.016}{[NH_3]^6}}$

$[NH_3]=0.285M$

Hence, the concentration of $NH_3$ required will be 0.285 M.

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3 years ago

A type of reaction that occurs when oxygen combines with iron to form rust

fiasKO [112]

Oxidation
It is oxidation because two O's 1 x 1 d 2 i's 1 n and 1 a

5 0

3 years ago

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A drop of water weighing 0.48 g is vaporized at 100 ?c and condenses on the surface of a 55-g block of aluminum that is initiall

irakobra [83]

<h3><u>Answer;</u></h3>

<em>-49 °C</em>

<h3><u>Explanation and solution;</u></h3>

Considering the fact that, the specific heat capacity of aluminum is 0.903 J/g x C, and the heat of vaporization of water at 25 C is 44.0 KJ/mol.

Moles water = 0.48 g / 18.02 g/mol

=0.0266 moles

<em>Heat lost by water</em> = 0.0266 mol x 44.0 kJ/mol

=1.17 kJ => 1170 J

<em>But heat lost =heat gained</em>

<em>Therefore;</em> Heat gained by aluminium = 1170 J

1170 = 55 x 0.903 ( T - 25) = 49.7 T - 1242

1170 + 1242 = 49.7 T

T = 48.5 °C ( 49 °C <em>at two significant figures)</em>

<em>Hence</em>, final temperature = 49 °C

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3 years ago

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