1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
Daniel [21]
3 years ago
14

Shawn wants to buy a pair of gloves online. She needs to

Chemistry
1 answer:
Ksivusya [100]3 years ago
5 0
The answer would be millimeters
so answer choice d
You might be interested in
Which example does not involve knowledge of chemistry?
Trava [24]

I would honestly say C. All others have ingredients in them that makes them work, the combination of stuff to put out the fire, film is made from a certain material and drugs are an ovbious mixture. Looking at the stars is the only thing you don't need chemistry for.

6 0
4 years ago
Read 2 more answers
Will mark BRAINLIEST question and choices in photos
ohaa [14]

Answer:

what to do in this question

8 0
3 years ago
A 76.0-gram piece of metal at 96.0 °C is placed in 120.0 g of water in a calorimeter at 24.5 °C. The final temperature in the ca
anygoal [31]

Answer:

S(metal) = 0.66J/g°C

Explanation:

We can find specific heat of a material, S, using the equation:

q = m*S*ΔT

<em>Where q is change in heat, m is the mass of the substance, S specific heat and ΔT change in temperature.</em>

The heat given by the metal is equal to the heat that water absorbs, that is:

m(Metal)*S(metal)*ΔT(Metal) = m(Water)*S(water)*ΔT(water)

<em>Where:</em>

m(Metal) = 76.0g

S(metal) = ?

ΔT(Metal) = 96.0°C-31.0°C = 65.0°C

m(Water) = 120.0g

S(water) = 4.184J/g°C

ΔT(water) = 31.0°C-24.5°C = 6.5°C

Replacing:

76.0g*S(metal)*65.0°C = 120.0g*4.184J/g°C*6.5°C

S(metal) = 0.66J/g°C

<em />

The law of conservation applies because the energy is not been created or destroyed. The energy that the metal gives is absorbed by the water.

3 0
3 years ago
In the following net ionic equation, identify each reactant as either a Bronsted-Lowry acid or a Bronsted-Lowry base. HCN(aq) H2
vfiekz [6]

Answer:

Explanation:

The definition of acids and bases by Arrhenius Theory  was modified and extended by  Bronsted-Lowry.

Bronsted-Lowry defined acid as a molecule or ion which donates a proton while a base is a molecule or ions that accepts the proton. This definition can be extended to include acid -base titrations in non-aqueous solutions.

In this theory, the reaction of an acid with a base constitutes a transfer of a proton from the acid to the base.

From the given information:

\mathsf{HCN _{(aq)} + H_2O_{(l)} \to CN^{-}_{(aq)} + H_3O_{(aq)}}

From above:

We will see that HCN releases an H⁺ ion, thus it is a Bronsted-Lowry acid

H_2O accepts the H⁺ ion ,thus it is a Bronsted-Lowry base.

The formula of the reactant that acts as a proton donor is <u>HCN</u>

The formula of the reactant that acts as a proton acceptor is <u>H2O</u>

8 0
3 years ago
The arsenic in a 1.223 g sample of a pesticide was converted to H3AsO4 by suitable treatment. The acid was then neutralized, and
Vladimir79 [104]

Answer:

5.471% As₂O₃ in the sample.

Explanation:

<em>...the reaction is: Ag+ + SCN- => AgSCN(s) Calculate the percent As2O3 in the sample. (F.W. As2O3 = 197.84 g/mol).</em>

<em />

First, with the amount of KSCN we can find the moles of Ag in the filtrates. As we know the amount of Ag added we can know the precipitate of Ag and the moles of AsO₄ = 1/2 moles of As₂O₃ in the sample:

<em>Moles KSCN = Moles Ag⁺ in the filtrate:</em>

0.01127L * (0.100mol / L)= 0.001127moles Ag⁺

<em>Total moles Ag⁺:</em>

0.0400L * (0.0781mol/L) = 0.0031564 moles Ag⁺

<em>Moles of Ag⁺ in the precipitate:</em>

0.0031564 - 0.001127 = 0.0020294 moles Ag⁺

<em>Moles AsO₄ = Moles As:</em>

0.0020294 moles Ag⁺ * (1mol As / 3 moles Ag⁺) = 6.765x10⁻⁴ moles AsO₄

<em>Moles As₂O₃:</em>

6.765x10⁻⁴ moles AsO₄ * (1 mol As₂O₃ / 2 mol AsO₄) =

3.382x10⁻⁴ moles As₂O₃

<em>Mass As₂O₃:</em>

3.382x10⁻⁴ moles As₂O₃ * (197.84g/mol) = 0.0669g As₂O₃

Percent is:

0.0669g As₂O₃ / 1.223g sample * 100 =

<h3>5.471% As₂O₃ in the sample</h3>

<em />

7 0
3 years ago
Other questions:
  • The amount of heat needed to raise 25 g of a substance by 15°C is 293 J. What is the specific heat of the substance? Use the equ
    6·1 answer
  • A reaction was performed in which 3.7 g of benzoic acid was reacted with excess methanol to make 2.1 g of methyl benzoate. Calcu
    6·1 answer
  • A government laboratory wants to determine whether water in a certain city has any traces of fluoride and whether the concentrat
    5·2 answers
  • If you add water to a 15.00 mL solution of 3.22 M CaCl2 (MM=110.98 g/mol) in order to create a solution that is 15.00% CaCl2 by
    14·1 answer
  • 30) How many grams of CH3OH must be added to water to prepare 210 mL of a solution that is
    15·1 answer
  • Percentage composition
    14·2 answers
  • How many seconds are there in one light-year?
    6·1 answer
  • I NEED APPROPRIATE ANSWERS SO IF ANYONE CAN PLEASE ANSWER THESE TWO QUESTIONS I WILL GIVE YOU BRAINIEST. Can someone please help
    6·1 answer
  • Someone answer this question by me
    6·2 answers
  • How many molecules are present in this formula? 3H2 (SO)4
    14·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!