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3 years ago

In each pair of statements, choose the example that is more reliable.

Chemistry

2 answers:

TEA [102]3 years ago

3 0

Answer:

"statement 2" for the first pair and "statement 1" for the second pair

Explanation:

BlessedZay2 years ago

0 0

In each pair of statements, choose the example that is more reliable.

Statement 1
The company Herbal Remedies Inc. announced today that its antiobesity skin treatment is safe and effective.

Statement 2
The US Food and Drug Administration (FDA) announced today that cancer treatment X, manufactured by Pharma Inc., is safe and effective.

Statement 1
An independent chemical analysis paid for by local residents determined that wells are contaminated with dioxin at a concentration of 3 ppb.

Statement 2
The chemical company Chemicals Inc. insists that local groundwater has not been affected adversely by a leak in its waste handling system.

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All the nonmetals have some similar physical properties.<br> True<br> False

OLEGan [10]

TRUE.

Nonmetals have some similar physical properties.

En general:

- They are either gases or brittle solids.
- They are opaque
- They are poor conductors of heat and electricity.

5 0

4 years ago

Read 2 more answers

the spectral lines observed for hydrogen arise from transitions from excited states back to the n=2 principle quantum level. Cal

Sunny_sXe [5.5K]

Rydberg formula is given by:

$\frac{1}{\lambda } = R_{H}\times (\frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}} )$

where, $R_{H}$ = Rydberg constant = $1.0973731568508 \times 10^{7} per metre$

$\lambda$ = wavelength

$n_{1}$ and $n_{2}$ are the level of transitions.

Now, for $n_{1}$ = 2 and $n_{2}$ = 6

$\frac{1}{\lambda} = 1.0973731568508 \times 10^{7} \times (\frac{1}{2^{2}}-\frac{1}{6^{2}} )$

= $1.0973731568508 \times 10^{7} \times (\frac{1}{4}-\frac{1}{36} )$

= $1.0973731568508 \times 10^{7} \times (0.25-0.0278 )$

= $1.0973731568508 \times 10^{7} \times 0.23$

= $0.2523958\times 10^{7}$

$\lambda = \frac{1}{0.2523958\times 10^{7}}$

= $3.9620\times 10^{-7} m$

= $396.20\times 10^{-9} m$

= $396.20 nm$

Now, for $n_{1}$ = 2 and $n_{2}$ = 5

$\frac{1}{\lambda} = 1.0973731568508 \times 10^{7} \times (\frac{1}{2^{2}}-\frac{1}{5^{2}} )$

= $1.0973731568508 \times 10^{7} \times (0.25-0.04 )$

= $1.0973731568508 \times 10^{7} \times (0.21 )$

= $0.230 \times 10^{7}$

$\lambda= \frac{1}{0.230 \times 10^{7}}$

= $4.3478 \times 10^{-7} m$

= $434.78\times 10^{-9} m$

= $434.78 nm$

Now, for $n_{1}$ = 2 and $n_{2}$ = 4

$\frac{1}{\lambda} = 1.0973731568508 \times 10^{7} \times (\frac{1}{2^{2}}-\frac{1}{4^{2}} )$

= $1.0973731568508 \times 10^{7} \times (0.25-0.0625 )$

= $1.0973731568508 \times 10^{7} \times (0.1875 )$

= $0.20575 \times 10^{7}$

$\lambda= \frac{1}{0.20575 \times 10^{7}}$

= $4.8602 \times 10^{-7} m$

= $486.02 \times 10^{-9} m$

= $486.02 nm$

Now, for $n_{1}$ = 2 and $n_{2}$ = 3

$\frac{1}{\lambda} = 1.0973731568508 \times 10^{7} \times (\frac{1}{2^{2}}-\frac{1}{3^{2}} )$

= $1.0973731568508 \times 10^{7} \times (0.25-0.12 )$

= $1.0973731568508 \times 10^{7} \times (0.13 )$

= $0.1426585\times 10^{7}$

$\lambda= \frac{1}{0.1426585\times 10^{7}}$

= $7.0097 \times 10^{-7} m$

= $700.97 \times 10^{-9} m$

= $700.97 nm$

5 0

3 years ago

Read 2 more answers

What functional groups of the emulsifying agent provide the hydrophilic character? Do these groups affect the pH?

podryga [215]

Following this question....

7 0

4 years ago

Read 2 more answers

Convert 7.1x10^25 molecules of water to moles

ruslelena [56]

Answer:

<h2>117.94 moles</h2>

Explanation:

To find the number of moles in a substance given it's number of entities we use the formula

$n = \frac{N}{L} \\$

where n is the number of moles

N is the number of entities

L is the Avogadro's constant which is

6.02 × 10²³ entities

From the question we have

$n = \frac{7.1 \times {10}^{25} }{6.02 \times {10}^{23} } \\ = 117.940199...$

We have the final answer as

<h3>117.94 moles</h3>

Hope this helps you

8 0

3 years ago

What is the ph of a solution of 1.699 l of 1.25 m hcn, ka = 6.2 x 10-10, and 1.37 moles of nacn?

BlackZzzverrR [31]

The pH of a solution is 9.02.

c(HCN) = 1.25 M; concentration of the cyanide acid

n(NaCN) = 1.37 mol; amount of the salt

V = 1.699 l; volume of the solution

c(NaCN) = 1.37 mol ÷ 1.699 l

c(NaCN) = 0.806 M; concentration of the salt

Ka = 6.2 × 10⁻¹⁰; acid constant

pKa = -logKa

pKa = - log (6.2 × 10⁻¹⁰)

pKa = 9.21

Henderson–Hasselbalch equation for the buffer solution:

pH = pKa + log(cs/ck)

pH = 9.21 + log (0.806M/1.25M)

pH = 9.21 - 0.19

pH = 9.02; potential of hydrogen

More about buffer: brainly.com/question/4177791

#SPJ4

8 0

1 year ago