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3 years ago

9

carbohydrates are compounds contains carbon, hydrogen, and oxygen in which the hydrogen to oxygen ratio is 2:1. A certain carboh

ydrate contains 40.0 percent carbon by mass. Calculate the empirical and molecular formula of the compound if the approximate mass is 178grams

1 answer:

iris [78.8K]3 years ago

5 0

Answer:

empircal: c2h2o

molecular: probs c6h6o3

178 grams

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Which explains how burning a magnesium ribbon highlights a toolmark?

Setler [38]

<h2>Answer:</h2>

The magnesium ribbon, <u>D. It forms a material to cast the tool mark</u>.

<h2>Explanation:</h2>

When a magnesium ribbon is burnt in the presence of oxygen it gives out strong light and heat is produced. Apart from it, it leads to the production of substance called as magnesium oxide which is formed as the product due to the reaction of magnesium with the oxygen present in the air.

Tool marks are the mark which is created by tools while using them. In order to identify or locate them castes made up of magnesium oxide is utilized. When this is pasted on the suspected area, the tool mark of the suspected tool gets pasted on it.

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3 years ago

calculate the difference in slope of the chemical potential against temperature on either side of the normal freezing point of w

kipiarov [429]

Answer:

(a) The normal freezing point of water (J·K−1·mol−1) is $-22Jmole^-^1k^-^1$

(b) The normal boiling point of water (J·K−1·mol−1) is $-109Jmole^-^1K^-^1$

(c) the chemical potential of water supercooled to −5.0°C exceed that of ice at that temperature is 109J/mole

Explanation:

Lets calculate

(a) - General equation -

$(\frac{d\mu(\beta )}{dt})p-(\frac{d\mu(\alpha) }{dt})_p$ = $-5_m(\beta )+5_m(\alpha )$ = $-\frac{\Delta H}{T}$

$\alpha ,\beta$ → phases

ΔH → enthalpy of transition

T → temperature transition

$(\frac{d\mu(l)}{dT})_p -(\frac{d\mu(s)}{dT})_p$ = $= -\frac{\Delta_fH}{T_f}$

= $\frac{-6.008kJ/mole}{273.15K}$ ( $\Delta_fH$ is the enthalpy of fusion of water)

= $-22Jmole^-^1k^-^1$

(b) $(\frac{d\mu(g)}{dT})_p-(\frac{d\mu(l)}{dT})_p= -\frac{\Delta_v_a_p_o_u_rH}{T_v_a_p_o_u_r}$

= $\frac{40.656kJ/mole}{373.15K}$ ( $\Delta_v_a_p_o_u_rH$ is the enthalpy of vaporization)

= $-109Jmole^-^1K^-^1$

(c) $\Delta\mu =\Delta\mu(l)-\Delta\mu(s)$ = $-S_m\DeltaT$

$[\mu(l-5$ ° $C)-\mu(l,0$ ° $C)]$ = $[\mu(s-5$ ° $C)-\mu(s,0$ ° $C)]$ $=-S_m$ ΔT

$\mu(l,-5$ ° $C)-\mu(s,-5$ ° $C)=-Sm\DeltaT [\mu(l,0$

$\Delta\mu=(21.995Jmole^-^1K^-^1)\times (-5K)$

= 109J/mole

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2 years ago

Compare the diameter of the moon to the diameter of the Earth

hichkok12 [17]

Answer:

Ok Hold up. I will answer after I think of question

Explanation:

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2 years ago

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Elina [12.6K]

1: viewing any chemical reaction in a laboratory

2: dangerous to look at when it burns & used in photography, fireworks, and flares

3: the product

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3 years ago

Read 2 more answers

At 20°C the vapor pressure of benzene (C6H6) is 75 torr, and that of toluene (C7H8) is 22 torr. Assume that benzene and toluene

valina [46]

Answer: The mole fraction of benzene will be 0.34 and mole fraction of toluene is 0.66

Explanation:

According to Raoult's law, the vapor pressure of a component at a given temperature is equal to the mole fraction of that component multiplied by the vapor pressure of that component in the pure state.

$p_1=x_1p_1^0$ and $p_2=x_2P_2^0$

where, x = mole fraction in solution

$p^0$ = pressure in the pure state

According to Dalton's law, the total pressure is the sum of individual pressures.

$p_{total}=p_1+p_2\\$ $p_{total}=x_{benzene}p_{benzene}^0+x_{toluene}P_{toluene}^0$

$x_{benzene}=x$ ,

$x_{toluene}=1-x_{benzene}=1-x$ ,

$p_{benzene}^0=75torr$

$p_{toluene}^0=22torr$

$p_{total}=40torr$

$40=x\times 75+(1-x)\times 22$

$x=0.34$

Thus (1-x0 = (1-0.34)=0.66

Thus the mole fraction of benzene will be 0.34 and that of toluene is 0.66

8 0

3 years ago