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3 years ago

6

Phosphorus (P) is a nonmetal, which is a poor conductor of heat and electricity. Which of the elements listed below shares these

properties?
Group of answer choices

strontium (Sr)

sodium (Na)

iodine (I)

calcium (Ca)

1 answer:

tankabanditka [31]3 years ago

8 0

Answer:

Sodium is the answer

Explanation:

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TRIPLE POINTS!!! I NEED HELP!!! PLEASE EXPLAIN TOO!!! What is the energy of a quantum of light with a frequency of 4.31 x 1014 1

Lelechka [254]

Answer:

The answer is

$2.86 \times {10}^{ - 19} \: J$

Explanation:

The energy of a quantum of light can be found by using the formula

<h3>E = hf</h3>

where

E is the energy

f is the frequency

h is the Planck's constant which is

6.626 × 10-³⁴ Js

From the question

f = 4.31 × 10¹⁴ Hz

We have

E = 4.31 × 10¹⁴ × 6.626 × 10-³⁴

We have the final answer as

$2.86 \times {10}^{ - 19} \: J$

Hope this helps you

3 0

3 years ago

At equilibrium, no chemical reactions take place. true or false

arsen [322]

False. At equilibrium, the rate of forward reaction is equal to the rate of backward reaction. The net concentration of both products and reactants won't change, but the reactions still take place.

6 0

3 years ago

If a penny (with the this volume .39 cm^3) was made out of pure copper, what should its mass be?

blondinia [14]

Answer:

3.49 g

Explanation:

The mass is the product of volume and density:

(8.96 g/cm³)(0.39 cm³) ≈ 3.49 g

The mass of a pure-copper penny would be 3.49 g.

5 0

3 years ago

How many moles of iron (iii) oxide are in a 120 gram sample of fe2o3?

AnnyKZ [126]

First find the molar mass of Fe2O3, after you would want to do this
120 g Fe2O3 x 1 mol = # mol
Molar Mass

3 0

3 years ago

Start with 100.00 mL of 0.10 M acetic acid, CH3COOH. The solution has a pH of 2.87 at 25 oC. a) Calculate the Ka of acetic acid

jasenka [17]

Answer: a) The $K_a$ of acetic acid at $25^0C$ is $1.82\times 10^{-5}$

b) The percent dissociation for the solution is $4.27\times 10^{-3}$

Explanation:

$CH_3COOH\rightarrow CH_3COO^-H^+$

cM 0 0

$c-c\alpha$ $c\alpha$ $c\alpha$

So dissociation constant will be:

$K_a=\frac{(c\alpha)^{2}}{c-c\alpha}$

Give c= 0.10 M and $\alpha$ = ?

Also $pH=-log[H^+]$

$2.87=-log[H^+]$

$[H^+]=1.35\times 10^{-3}M$

$[CH_3COO^-]=1.35\times 10^{-3}M$

$[CH_3COOH]=(0.10M-1.35\times 10^{-3}=0.09806M$

Putting in the values we get:

$K_a=\frac{(1.35\times 10^{-3})^2}{(0.09806)}$

$K_a=1.82\times 10^{-5}$

b) $\alpha=\sqrt\frac{K_a}{c}$

$\alpha=\sqrt\frac{1.82\times 10^{-5}}{0.10}$

$\alpha=4.27\times 10^{-5}$

$\% \alpha=4.27\times 10^{-5}\times 100=4.27\times 10^{-3}$

5 0

3 years ago