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2 years ago

On a coordinate plane, which single transformation could map point A (3,-4) onto point A' (1.5,-2)? *

Chemistry

1 answer:

Ksenya-84 [330]2 years ago

3 0

Answer: A dilation with rule: $(x,y)\to(\dfrac12x,\dfrac12y)$

Explanation:

A dilation is a non-rigid transformation that creates an image that is the same shape as the original but has a different size.

It uses a scale factor k such that

$(x,y)\to(kx,ky)$

(x,y)= coordinates of original figure

(kx,ky) = corresponding coordinate in the image.

To transform: A (3,-4) onto point A' (1.5,-2).

Using scale factor k= $\dfrac12$ , we have

$A(3,-4)\to A'(\dfrac12\times3, \dfrac12\times-4)=A'(1.5,-2)$

Required rule: $(x,y)\to(\dfrac12x,\dfrac12y)$

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2 years ago

The rate constant k for a certain reaction is measured at two different temperatures temperature 376.0 °c 4.8 x 108 280.0 °C 2.3

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Answer:

The activation energy for this reaction = 23 kJ/mol.

Explanation:

Using the expression,

$\ln \dfrac{k_{1}}{k_{2}} =-\dfrac{E_{a}}{R} \left (\dfrac{1}{T_1}-\dfrac{1}{T_2} \right )$

Where,

$k_1\ is\ the\ rate\ constant\ at\ T_1$

$k_2\ is\ the\ rate\ constant\ at\ T_2$

$E_a$ is the activation energy

R is Gas constant having value = 8.314×10⁻³ kJ / K mol

$k_2=2.3\times 10^8$

$T_2=280\ ^0C$

The conversion of T( °C) to T(K) is shown below:

T(K) = T( °C) + 273.15

So,

T = (280 + 273.15) K = 553.15 K

$T_2=553.15\ K$

$k_1=4.8\times 10^8$

$T_1=376\ ^0C$

The conversion of T( °C) to T(K) is shown below:

T(K) = T( °C) + 273.15

So,

T = (376 + 273.15) K = 649.15 K

$T_1=649.15\ K$

So,

$\left(\ln \left(\:\frac{4.8\times \:\:\:10^8}{2.3\times \:\:\:10^8}\right)\right)\:=-\frac{E_a}{8.314\times \:10^{-3}\ kJ/mol.K}\times \:\left(\frac{1}{649.15\ K}-\frac{1}{553.15\ K}\right)$

$E_a=-\frac{10^{-3}\times \:8.314\ln \left(\frac{10^8\times \:4.8}{10^8\times \:2.3}\right)}{-\frac{96}{359077.3225}}\ kJ/mol$

$E_a=-\frac{\frac{8.314\ln \left(\frac{4.8}{2.3}\right)}{1000}}{-\frac{96}{359077.3225}}\ kJ/mol$

$E_a=22.87\ kJ/mol$

<u>The activation energy for this reaction = 23 kJ/mol.</u>

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2 years ago

An unknown compound contains 75.69% carbon, 8.80% hydrogen, and 15.51% oxygen by mass. A mass spectrometry analysis reveals that

tiny-mole [99]

Answer:

Molecular formula is C₂₆H₃₆O₄

Explanation:

The compound is 75.69 % C, 8.80 % H and 15.51 % O. This data means, that in 100 g of compound we have 75.69 g, 15.51 g and 8.80 g of, C, O and H, respectively. We know the molar mass of the compound, so we can work to solve the moles of each element.

In 100 g of compound we have 75.69 g C, 15.51 g O and 8.80 g H

In 412 g of compound we would have:

(412 . 75.69) / 100 = 311.8 of C

(412 . 15.51) / 100 = 63.9 g of O

(412 . 8.80) / 100 = 36.2 g of H

Now, we can determine the moles of each, that are contained in 1 mol of compound.

312 g / 12 g/mol 26 C

64 g / 16 g/mol = 4 O

36 g / 1 g/mol = 36 H

Molecular formula is C₂₆H₃₆O₄

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3 years ago