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3 years ago

12

Consider the function f(x) =

1 answer:

Butoxors [25]3 years ago

8 0

I think it’s b so yea

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PLEASE HELP URGENT WILL GOVE BRANLIEST WITH EXPLANATION

Ymorist [56]

Answer:

no because it already had it in there and it exploded by a fuse

Explanation:

Hope this helps?? :))

6 0

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

6 0

2 years ago

What is the lewis structure for al?

bogdanovich [222]

The abbreviation Al, with one dot on top of the abbreviation, one on the left, one on the right.

8 0

3 years ago

QveST [7]

Metals only and ionic

3 0

3 years ago

Read 2 more answers

What is the percent composition of dinotrogen pentoxide? With work shown please

iris [78.8K]

Answer:

%N = 25.94%

%O = 74.06%

Explanation:

Step 1: Calculate the mass of nitrogen in 1 mole of N₂O₅

We will multiply the molar mass of N by the number of N atoms in the formula of N₂O₅.

m(N): 2 × 14.01 g = 28.02 g

Step 2: Calculate the mass of oxygen in 1 mole of N₂O₅

We will multiply the molar mass of O by the number of O atoms in the formula of N₂O₅.

m(O): 5 × 16.00 g = 80.00 g

Step 3: Calculate the mass of 1 mole of N₂O₅

We will sum the masses of N and O.

m(N₂O₅) = m(N) + m(O) = 28.02 g + 80.00 g = 108.02 g

Step 4: Calculate the percent composition of N₂O₅

We will use the following expression.

%Element = m(Element)/m(Compound) × 100%

%N = m(N)/m(N₂O₅) × 100% = 28.02 g/108.02 g × 100% = 25.94%

%O = m(O)/m(N₂O₅) × 100% = 80.00 g/108.02 g × 100% = 74.06%

7 0

3 years ago