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Maslowich
3 years ago
12

Litmus paper can be used to determine the _______ of a substance. A. relative volume B. relative pH C. temperature D. chemical f

ormula
Chemistry
1 answer:
fgiga [73]3 years ago
5 0

Answer:

B

Explanation:

Litmus paper determines the relative pH

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Is the answer B? Help
Masja [62]

Answer:

A

Explanation:

Hmm, so we have the following in the diagram

Pt(s)

Cl2(g)

Ag(s)

NaCl(aq)

AgNO3(aq)

Pt 2+, 4+, 6+  Though it states Pt is inert

Cl 2-

Ag 1+

Na 1+

NO3-

Anode definition: the positively charged electrode by which the electrons leave an electrical device.

Electrode definition: a conductor through which electricity enters or leaves an object, substance, or region.

Cations attracted to cathode pick up electrons

Anions attracted to anode release electrodes+

Reduction at Cathode (red cat gain of e)

Oxidation at Anode (ox anode loss of e)

So from the diagram we can see that the charge is being generated through the 2 metal plates.

So the answer is A, the anode material is Pt and the half reaction is 2Cl- = Cl2 + 2e-

7 0
3 years ago
What is chemistry? I NEED HELP ASAP!
Thepotemich [5.8K]

the branch of science that deals with the identification of the substances of which matter is composed; the investigation of their properties and the ways in which they interact, combine, and change; and the use of these processes to form new substances.the branch of science that deals with the identification of the substances of which matter is composed; the investigation of their properties and the ways in which they interact, combine, and change; and the use of these processes to form new substances.

3 0
3 years ago
A particular radioactive nuclide has a half-life of 1000 years. What percentage of an initial population of this nuclide has dec
Delicious77 [7]

Answer:

91.16% has decayed & 8.84% remains

Explanation:

A = A₀e⁻ᵏᵗ => ln(A/A₀) = ln(e⁻ᵏᵗ) => lnA - lnA₀ = -kt => lnA = lnA₀ - kt

Rate Constant (k) = 0.693/half-life = 0.693/10³yrs = 6.93 x 10ˉ⁴yrsˉ¹

Time (t) = 1000yrs  

A = fraction of nuclide remaining after 1000yrs

A₀ = original amount of nuclide = 1.00 (= 100%)  

lnA = lnA₀ - kt

lnA = ln(1) – (6.93 x 10ˉ⁴yrsˉ¹)(3500yrs) = -2.426

A = eˉ²∙⁴²⁶ = 0.0884 = fraction of nuclide remaining after 3500 years

Amount of nuclide decayed = 1 – 0.0884 = 0.9116 or 91.16% has decayed.

3 0
3 years ago
Why must the spectrophotometer cell be removed from the instrument between readings?
Natali [406]

Spectrophotometric cell or a cuvette is made of quartz for UV spectrophotometers. These cuvettes are used as sample holders for the spectrophotometric determination of the analytes. The material that makes up the cuvette and the condition of the cuvette is to be taken care of in order to avoid erroneous absorbance readings. The sample holder or the cuvette must be removed from the spectrophotometer in between two successive readings. This is to ensure that the light sensing detector of the instrument is not affected.

5 0
3 years ago
Which number best describes the number of stars you would find in the galaxy?
IceJOKER [234]
D. The answer would be D.
4 0
3 years ago
Read 2 more answers
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