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Hunter-Best [27]

2 years ago

10

The energy stored in the chemical bonds of a carbohydrate molecule is

1 answer:

USPshnik [31]2 years ago

4 0

Answer:

Potential energy is available between chemical bonds. The more stable bond means weak potential energy. Potential energy releases when stronger chemical bonds form, and potential energy amount increases when chemical bond break

Explanation:

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What's the answer to this question?

Lena [83]

The answer is B- 124 degrees

4 0

2 years ago

Read 2 more answers

Wat two common uses for Gallium??

Ann [662]

Answer:

mirrors and metal thermometers

Explanation:

ooga booga

6 0

3 years ago

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Consider the following reaction between mercury(II) chloride and oxalate ion.

Alina [70]

<u>Answer:</u> The rate law of the reaction is $\text{Rate}=k[HgCl_2][C_2O_4^{2-}]^2$

<u>Explanation:</u>

Rate law is defined as the expression which expresses the rate of the reaction in terms of molar concentration of the reactants with each term raised to the power their stoichiometric coefficient of that reactant in the balanced chemical equation.

For the given chemical equation:

$2 HgCl_2(aq.)+C_2O_4^{2-}(aq.)\rightarrow 2Cl^-(aq.)+2CO_2(g)+Hg_2Cl_2(s)$

Rate law expression for the reaction:

$\text{Rate}=k[HgCl_2]^a[C_2O_4^{2-}]^b$

where,

a = order with respect to $HgCl_2$

b = order with respect to $C_2O_4^{2-}$

Expression for rate law for first observation:

$3.2\times 10^{-5}=k(0.164)^a(0.15)^b$ ....(1)

Expression for rate law for second observation:

$2.9\times 10^{-4}=k(0.164)^a(0.45)^b$ ....(2)

Expression for rate law for third observation:

$1.4\times 10^{-4}=k(0.082)^a(0.45)^b$ ....(3)

Expression for rate law for fourth observation:

$4.8\times 10^{-5}=k(0.246)^a(0.15)^b$ ....(4)

Dividing 2 from 1, we get:

$\frac{2.9\times 10^{-4}}{3.2\times 10^{-5}}=\frac{(0.164)^a(0.45)^b}{(0.164)^a(0.15)^b}\\\\9=3^b\\b=2$

Dividing 2 from 3, we get:

$\frac{2.9\times 10^{-4}}{1.4\times 10^{-4}}=\frac{(0.164)^a(0.45)^b}{(0.082)^a(0.45)^b}\\\\2=2^a\\a=1$

Thus, the rate law becomes:

$\text{Rate}=k[HgCl_2]^1[C_2O_4^{2-}]^2$

3 0

2 years ago

What was the velocity over the entire trip? <br><br> A. -2<br> B. -1<br> C. 0<br> D. 1<br> E. 2

zhannawk [14.2K]

You’re right its A -2

5 0

2 years ago

A 215-g sample of copper metal at some temperature is added to 26.6 g of water. The initial water temperature is 22.22 oC, and t

andrezito [222]

The initial temperature of the copper metal was 27.38 degrees.

Explanation:

Data given:

mass of the copper metal sample = 215 gram

mass of water = 26.6 grams

Initial temperature of water = 22.22 Degrees

Final temperature of water = 24.44 degrees

Specific heat capacity of water = 0.385 J/g°C

initial temperature of copper material , Ti=?

specific heat capacity of water = 4.186 joule/gram °C

from the principle of:

heat lost = heat gained

heat gained by water is given by:

q water = mcΔT

Putting the values in the equation:

qwater = 26.6 x 4.186 x (2.22)

qwater = 247.19 J

qcopper = 215 x 0.385 x (Ti-24.4)

= 82.77Ti - 2019.71

Now heat lost by metal = heat gained by water

82.77Ti - 2019.71 = 247.19

Ti = 27.38 degrees

8 0

3 years ago