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
kompoz [17]
3 years ago
11

What are the reactants in the following equation: hcl(aq) nahco₃(aq)→ co₂(g) h₂o(l) nacl(aq)

Chemistry
1 answer:
Triss [41]3 years ago
4 0
The reactants are all the formulas to the left of the arrow, which are:
hcI(aq) nahco3(aq)
You might be interested in
Carbon-14 has a half-life of 5,730 years. How long will it take for 112. 5 g of a 120. 0-g sample to decay radioactively? 5,730
vichka [17]

The time taken by Carbon-14 to decay radioactively from 120g to 112.5g is 22,920 years.

<h3>How do we calculate the total time of decay?</h3>

Time required for the whole radioactive decay of any substance will be calculated by using the below link:

T = (n)(t), where

  • t = half life time = 5730 years
  • n = number of half life required for the decay

Initial mass of Carbon-14 = 120g

Final mass of Carbon-14 = 112.5g

Left mass = 120 - 112 = 7.5g

Number of required half life for this will be:

  • 1: 120 → 60
  • 2: 60 → 30
  • 3: 30 → 15
  • 4: 15 → 7.5

4 half lives are required, now on putting values we get

T = (4)(5730) = 22,920 years

Hence required time for the decay is 22,920 years.

To know more about radioactive decay, visit the below link:

brainly.com/question/24115447

#SPJ1

3 0
2 years ago
The nonvolatile, nonelectrolyte testosterone, C19H28O2 (288.40 g/mol), is soluble in benzene C6H6. Calculate the osmotic pressur
egoroff_w [7]

Answer: 6.26atm

Explanation:Please see attachment for explanation

7 0
3 years ago
Read 2 more answers
MARKING BRAINLEIST !!
Murrr4er [49]

Answer:

Option C

Explanation:

The theory is useful as in the usual constitution of the theory the professor stated

3 0
2 years ago
Please help me <br> Vote you brainiest but please just help
Goryan [66]
Their average speed during the trip is 53 km/ hr approximately

Explanation: The family traveled 80 km/hr for an hour and then trabeled 40 km/ hr for 2 hours. So they traveled 80 km in one hour then 80 km in the next 2 hours. Total they traveled 160 km in 3 hours.
Average speed = distance traveled/ taken time
= 160/3 = 53.33
4 0
3 years ago
For a voltaic cell consisting of Al(s) in Al(NO3)3(aq) and Cu(s) in Cu(NO3)2(aq), what is Ecell, given [Al3 ]
xz_007 [3.2K]

Answer:

2.0 V

Explanation:

For the oxidation half cell;

Al(s) -------> Al^3+(aq) + 3e.

For reduction half cell;

Cu^2+(aq) +2e ------> Cu(s).

E°cell = E°cathode - E°anode

But;

E°cathode= 0.34 V

E°anode = -1.66 V

E°cell= 0.34 -(-1.66)

E°cell= 2.0 V

5 0
3 years ago
Other questions:
  • For which structure(s) does the electron-bearing carbon atom have a formal charge of zero
    13·1 answer
  • Rank the following from highest to lowest priority according to the Cahn-Ingold-Prelog system: -CH2CH3, -CHCH2, -CCH, -CH3.
    9·1 answer
  • If 0.0714 moles of N2 gas occupies 1.25 L space, how many moles of N2 have a volume of 25.0 L? Assume temperature and pressure s
    6·1 answer
  • Which of the following samples will have the greatest volume at STP?
    5·1 answer
  • The process of a liquid converting to a solid is called
    10·1 answer
  • Consider the reaction: 2HI(g) ⇄ H2(g) + I2(g). It is found that, when equilibrium is reached at a certain temperature, HI is 35.
    12·1 answer
  • Identify the conjugate base in each pairs
    14·1 answer
  • Caluclate the volume of0,25 mol dm -^3 hydrochloric acid required to neutralize 2.0 dm^3 of 0,15 mol dm-^3 barium hydroxide
    11·2 answers
  • ugiugiugiugiugiugiugiugugiiugiugiugiugiugiugiugiugugiiugigiuugiutgiugiugigigigiugiugiugiugiuiugiugigiuiugiuiugiuiuuigiugiugigui
    10·1 answer
  • These steps are followed when using the half-life of carbon-14 to determine
    15·2 answers
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!