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
Alina [70]
3 years ago
13

A student performs an experiment to determine the density of a sugar solution. She obtains the following results: 1.79 g/mL, 1.8

1 g/mL, 1.80 g/mL, 1.81 g/mL. If the actual value for the density of the sugar solution is 1.80 g/mL, which statement below best describes her results?
Chemistry
1 answer:
ivann1987 [24]3 years ago
4 0

Answer:

her results has good precision

Explanation:

accuracy and precision

You might be interested in
Heeellllpppp ASAP ......
iren [92.7K]

Answer:

I saw B, he vented SUS

Explanation:

Iodoethane Is the answer.

7 0
2 years ago
A coffee-cup calorimeter initially contains 125 g water at 24.28C. Potassium bromide (10.5 g), also at 24.28C, is added to the w
iren2701 [21]

Answer:

The solution is given below

Explanation:

Heat, q= mc∆T

q= 125g x 4.18 J/g∙°C x (21.18x- 24.28) °C

q=  -1619.75J

NEGATIVE SIGN INDICATES THAT HEAT IS ABSORBED.

Enthalpy Change, ∆H = 1619.75 7/ 10.5 g

                                     = 154.26 J/g

No. of moles of KBr = Mass of KBr/ Molecular Weight of KBr

                                =10.5g/119gmol-1

                                =0.088 mol

∆H= 1619.75 J/ 0.088 mol

      = 18.41 kJ/mol  

6 0
3 years ago
15.00 g of NH4HS(s) is introduced into a 500. mL flask at 25 °C, the flask is sealed, and the system is allowed to reach equilib
Ahat [919]

Answer:

0.328 atm

Explanation:

Kp is the equilibrium constant calculated based on the pressure, and it depends only on the gas substances. It will be the multiplication of partial pressures of the products raised to their coefficients divided by the multiplication of partial pressures of the reactants raised to their coefficients.

For the equation given, the stoichiometry is 1 mol of NH₃ for 1 mol of H₂S, so they will have the same partial pressure in equilibrium, let's call it p. So:

Kp = pxp

0.108 = p²

p = √0.108

p = 0.328 atm, which is the partial pressure of the ammonia.

3 0
3 years ago
What type of compound is Sodium hydroxide?<br> A.acid<br> B.base<br> C.element<br> D.salt
morpeh [17]
The answer is B
b base
3 0
2 years ago
Read 2 more answers
A power plant is driven by the combustion of a complex fossil fuel having the formula C11H7S. Assume the air supply is composed
AlekseyPX

(a) 4C_11 H_7S + 55O_2 → 44CO_2 + 14H_2O + 4SO_2 + 20.68N_2;

(b) 4C_11 H_7S + 66O_2 → 44CO_2 + 14H_2O + 4SO_2 + 248.2N_2 + 11O_2;

(c) 23 900 kg air; (d) air:fuel = 10.2; (e) air:fuel = 12.2:1

(a) <em>Balanced equation including N_2 from air</em>  

The balanced equation <em>ignoring</em> N_2 from air is  

4C_11 H_7S + 55O_2 → 44CO_2 + 14H_2O + 4SO_2  

Moles of N_2 =55 mol O_2 × (3.76 mol N_2/1 mol O_2) = 206.8 mol N_2  

<em>Including</em> N_2 from air, the balanced equation is  

4C_11 H_7S + 55O_2 → 44CO_2 + 14H_2O + 4SO_2 + 206.8N_2  

(b) <em>Balanced equation for 120 % stoichiometric combustion</em>  

Moles of O_2 = 55 mol O_2 × 1.20 = 66.00 mol O_2  

Excess moles O_2 = (66.00 – 55) mol O_2 = 11.00 mol O_2  

Moles of N_2 = 66.00 mol O_2 × (3.76 mol N_2/1 mol O_2) = 248.2 mol N_2  

The balanced equation is

4C_11 H_7S + 66O_2 → 44CO_2 + 14H_2O + 4SO_2 + 248.2N_2 + 11O_2

(c) <em>Minimum mass of air</em>  

Moles of O_2 required = 1700 kg C_11 H_7S

× (1 kmol C_11 H_7S/185.24 kg C_11 H_7S) × (55 kmol O_2/4 kmol C_11 H_7S)

= 126.2 kmol O_2  

Mass of O_2 = 126.2 kmol O_2 × (32.00 kg O_2/1 kmol O_2) = 4038 kg O_2  

Mass of N_2 required = 126.2 kmol O_2 × (3.76 kmol N_2/1 kmol O_2)

× (28.01 kg N_2/1 kmol N_2) = 13 285 kg N_2  

Mass of air = Mass of N_2 + mass of O_2 = (4038 + 13 285) kg = 17 300 kg air  

(d) <em>Air:fuel mass ratio for 100 % combustion</em>  

Air:fuel = 17 300 kg/1700 kg = <em>10.2 :1 </em>

(e) <em>Air:fuel mass ratio for 120 % combustion </em>

Mass of air = 17 300 kg × 1.20 = 20 760 kg air  

Air:fuel = 20 760 kg/1700 kg = 12.2 :1  

6 0
2 years ago
Other questions:
  • Is the flame test a test for the cation (metal listed first) or the anion (nonmetal listed second)?
    5·1 answer
  • The vapor pressure of water at 25.0 oc is 23.8 torr. determine the mass of glucose (molar mass = 180. g/mol) needed to add to 27
    12·1 answer
  • What is the structure of hair?
    7·2 answers
  • An objects starting position is at a point A and it's final point in at B to figure out what the speed is in between what other
    12·1 answer
  • Which most likely would be a new social issue resulting from the installation of a dam?
    7·2 answers
  • For which one of the following molecules is the indicated type of hybridization not appropriate for
    14·1 answer
  • For a principal quantum number n, how many atomic orbitals are possible?
    6·1 answer
  • Why does it make sense that Hydrogen and Helium would be the first elements to form in the Universe?
    15·1 answer
  • Since gasses weigh nothing when they are produced in a reaction they do not have to be accounted for.
    8·1 answer
  • Based on the ksp experiment describe a way to soften water by removing calcium ions
    14·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!