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3 years ago

M(8,7) is the midpoint of rs. The coordinates of s are (9,5) what is the coordinates of r

Chemistry

1 answer:

Darya [45]3 years ago

3 0

Answer:

Coordinate or r = (7,9).

Explanation:

Data obtained from the question include the following:

Mid point = (8,7)

Coordinate of S = (9,5)

Coordinate of r =...?

We shall determine the coordinate of r as follow:

Let the coordinate of r be (x2, y2)

Mid point = (x1 + x2)/2 , (y1 + y2)/2

Mid point = (8,7)

Coordinate of S = (9,5)

x1 = 9

y1 = 5

x2 =?

y2 =?

The value of x2 can be obtained as follow:

8 = (x1 + x2)/2

8 = (9 + x2)/2

Cross multiply

9 + x2 = 2 × 8

9 + x2 = 16

Collect like terms

x2 = 16 – 9

x2 = 7

The value of y2 can be obtained as follow:

5 = (y1 + y2)/2

7 = (5 + y2)/2

Cross multiply

5 + y2 = 2 × 7

5 + y2 = 14

Collect like terms

y2 = 14 – 5

y2 = 9

Coordinate of r = (x2, y2)

Coordinate or r = (7,9)

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= -1096.8Kj = -301Kj = 0Kj = 0Kj = -4110.6Kj = -3930.5Kj

ΔHₙ°(rxn) = ∑ (ΔH˚(f)products) - ∑(ΔH˚(f)reactants)

= [3(0)Kj + 17(-241.8)Kj + (-393.5)Kj] - [(-(1096.8)Kj + (-301)Kj + (0)Kj]

= [-(8041.1) - (-1397.8)]Kj

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∴ Standard Heat of Rxn = -6643.3Kj/3moles = -214.8Kj/mole NH₄NO₃(s)

ΔH°(rxn for 14.11g NH₄NO₃(s)) = (11.4g/80.04g·mol⁻¹)(-214.8Kj/mol) = 30.5937Kj ≅ 30.59Kj (4 sig. figs. ~mass of NH₄NO₃(s) given)

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3 years ago

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3 years ago

A 125g metal block at a temperature of 93.2 degrees Celsius was immersed in 100g of water at 18.3 degrees Celsius. Given the spe

nikitadnepr [17]

Answer:

$\large \boxed{34.2\, ^{\circ}\text{C}}$

Explanation:

There are two heat transfers involved: the heat lost by the metal block and the heat gained by the water.

According to the Law of Conservation of Energy, energy can neither be destroyed nor created, so the sum of these terms must be zero.

Let the metal be Component 1 and the water be Component 2.

Data:

For the metal:

$m_{1} =\text{125 g; }T_{i} = 93.2 ^{\circ}\text{C; }\\C_{1} = 0.900 \text{ J$^{\circ}$C$^{-1}$g$^{-1}$}$

For the water:

$m_{2} =\text{100 g; }T_{i} = 18.3 ^{\circ}\text{C; }\\C_{2} = 4.184 \text{ J$^{\circ}$C$^{-1}$g$^{-1}$}$

$\begin{array}{rcl}\text{Heat lost by metal + heat gained by water} & = & 0\\q_{1} + q_{2} & = & 0\\m_{1}C_{1}\Delta T_{1} + m_{2}C_{2}\Delta T_{2} & = & 0\\\text{125 g}\times 0.900 \text{ J$^{\circ}$C$^{-1}$g$^{-1}$} \times\Delta T_{1} + \text{100 g} \times 4.184 \text{ J$^{\circ}$C$^{-1}$g$^{-1}$}\Delta \times T_{2} & = & 0\\112.5\Delta T_{1} + 418.4\Delta T_{2} & = & 0\\112.5\Delta T_{1} & = & -418.4\Delta T_{2}\\\Delta T_{1} & = & -3.719\Delta T_{2}\\\end{array}$

$\Delta T_{1} = T_{\text{f}} - 93.2 ^{\circ}\text{C}\\\Delta T_{2} = T_{\text{f}} - 18.3 ^{\circ}\text{C}$

$\begin{array}{rcl}\Delta T_{1} & = & -3.719\Delta T_{2}\\T_{\text{f}} - 93.2 ^{\circ}\text{C} & = & -3.719 (T_{\text{f}} - 18.3 ^{\circ}\text{C})\\T_{\text{f}} - 93.2 ^{\circ}\text{C} & = & -3.719T_{\text{f}} + 68.06 ^{\circ}\text{C}\\4.719T_{\text{f}} & = & 161.3 ^{\circ}\text{C}\\T_{\text{f}} & = & \mathbf{34.2 ^{\circ}}\textbf{C}\\\end{array}\\\text{The final temperature of the block and the water is $\large \boxed{\mathbf{34.2\, ^{\circ}}\textbf{C}}$}$

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3 years ago

If the average rate of sediment accumulation is 3 cm/1000years, how long would be required for sediments to completely bury seam

stira [4]

Answer:

500,050000 years

Explanation:

Given that :

Rate of Accumulation = 3cm/ 1000 years

Height of seamounts = 1.5km

Representing height of seamounts in cm;

(1.5 *1000 * 100) = 1500150 cm

Time taken = height / rate

Time taken = (1500150 cm / 3cm) * 1000

Time taken = 500,050000 years

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3 years ago