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2 years ago

Lisa’s book measures 40 cm x 10 cm x 2 cm. What is the volume of her book

Chemistry

2 answers:

notsponge [240]2 years ago

8 0

The volume would be 800 because volume=lxwxh

White raven [17]2 years ago

3 0

$\text{Hello there! :)}$

$\large\boxed{V = 800cm^{3} }$

$\text{Formula for the volume of a rectangular prism:}\\\\V = l * w * h\\\\\text{In this instance:}\\\\l = 40 cm\\w = 20 cm\\h = 10cm\\\\\text{Use the formula to solve for the volume:}\\V = 40 * 10 * 2\\\\V = 800cm^{3}$

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Arsenic produces a blue flame when heated. Calcium produces an orange-red flame. Which of these best explains why this differenc

MrMuchimi

Flame colors are produced from the movement of the electrons in the metal ions present in the compounds. When you heat it, the electrons gain energy and can jump into any of the empty orbitals at higher levels Each of these jumps involves a specific amount of energy being released as light energy, and each corresponds to a particular color. As a result of all these jumps, a spectrum of colored lines will be produced. The color you see will be a combination of all these individual colors.

6 0

2 years ago

What is the overall charge of the compound frbr

sammy [17]

Answer:

Zero

Explanation:

FrBr is an ionic compound .

Fr is in Group 1. Br is in Group 17.

The charges on the ions are +1 and -1, respectively.

The compound consists of Fr⁺Br⁻ ions.

However, there are equal numbers of + and - charges, so

The overall charge of the compound is zero.

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2 years ago

Arkeisha and Rodney are having a discussion about the correct use of the term "battery acid." Arkeisha insists that an alkaline

mote1985 [20]

Arkeisha is correct because the fluid in an alkaline battery has a ph between 7.1 and 14.0

8 0

2 years ago

A 60.0 mL solution of 0.112 M sulfurous acid (H2SO3) is titrated with 0.112 M NaOH. The pKa values of sulfurous acid are 1.857 (

djverab [1.8K]

Answer:

a)4.51

b) 9.96

Explanation:

Given:

NaOH = 0.112M

H2S03 = 0.112 M

V = 60 ml

H2S03 pKa1= 1.857

pKa2 = 7.172

a) to calculate pH at first equivalence point, we calculate the pH between pKa1 and pKa2 as it is in between.

Therefore, the half points will also be the middle point.

Solving, we have:

pH = (½)* pKa1 + pKa2

pH = (½) * (1.857 + 7.172)

= 4.51

Thus, pH at first equivalence point is 4.51

b) pH at second equivalence point:

We already know there is a presence of SO3-2, and it ionizes to form

SO3-2 + H2O <>HSO3- + OH-

$Kb = \frac{[ HSO3-][0H-]}{SO3-2}$

$Kb = \frac{10^-^1^4}{10^-^7^.^1^7^2} = 1.49*10^-^7$

[HSO3-] = x = [OH-]

mmol of SO3-2 = MV

= 0.112 * 60 = 6.72

We need to find the V of NaOh,

V of NaOh = (2 * mmol)/M

= (2 * 6.72)/0.122

= 120ml

For total V in equivalence point, we have:

60ml + 120ml = 180ml

[S03-2] = 6.72/120

= 0.056 M

Substituting for values gotten in the equation $Kb=\frac{[HSO3-][OH-]}{[SO3-2]}$

We noe have:

$1.485*10^-^7=\frac{x*x}{(0.056-x)}$

$x = [OH-] = 9.11*10^-^5$

$pOH = -log(OH) = -log(9.11*10^-^5)$

=4.04

pH = 14- pOH

= 14 - 4.04

= 9.96

The pH at second equivalence point is 9.96

4 0

2 years ago

One of the emission spectral lines for Be31 has a wavelength of 253.4 nm for an electronic transition that begins in the state w

max2010maxim [7]

Explanation:

The given data is as follows.

$\lambda$ = 253.4 nm = $253.4 \times 10^{-9}m$ (as 1 nm = $10^{-9}$ )

$n_{1}$ = 5, $n_{2}$ = ?

Relation between energy and wavelength is as follows.

E = $\frac{hc}{\lambda}$

= $\frac{6.626 \times 10^{-34} Js \times 3 \times 10^{8} m/s}{253.4 \times 10^{-9}}$

= $0.0784 \times 10^{-17}$ J

= $7.84 \times 10^{-19} J$

Hence, energy released is $7.84 \times 10^{-19} J$ .

Also, we known that change in energy will be as follows.

$\Delta E = -2.178 \times (Z)^{2}[\frac{1}{n^{2}_{2}} - \frac{1}{n^{2}_{1}}$

where, Z = atomic number of the given element

$7.84 \times 10^{-19} J = -2.178 \times (4)^{2}[\frac{1}{n^{2}_{2}} - \frac{1}{(5)^{2}}$

$\frac{7.84 \times 10^{-19} J}{34.848} = \frac{1}{n^{2}_{2}} - \frac{1}{(5)^{2}}$

0.02 + 0.04 = $\frac{1}{n^{2}_{1}}$

$n_{1}$ = $\sqrt{\frac{1}{0.06}}$

= 4

Thus, we can conclude that the principal quantum number of the lower-energy state corresponding to this emission is n = 4.

4 0

2 years ago