Lily replicates an experiment that found that the number of calories in a particular food is 50 kcal. She obtained data from

Tpy6a [65]

Answer:

the praying mantis is eating a bee to stay alive with the calories inside it's body. Therefore the praying mantis is the predator and the bee is the prey. This clearly dictates a predatory relationship.

Explanation:

the answer is self explainatory, good luck!

7 0

3 years ago

Read 2 more answers

The total volume required to reach the endpoint of a titration required more than the 50 mL total volume of the buret. An initia

Margaret [11]

Answer:

The endpoint volume is 50.52 ± 0.14 mL

Explanation:

In a titration always is necessary to subtract the blank volume to the titrant volume to obtain the real volume of the titrant. Thus in this case, the total endpoint volume is the sum of the initial volume delivered and the second volume delivered, minus the blank volume:

V = (49.16±0.06 mL) + (1.69±0.04 mL) - (0.33±0.04 mL)

V = (49.16 + 1.69 - 0.33) ± (0.06+0.04+0.04) mL

V = 50.52 ± 0.14 mL

It is necessary to consider the sum of the errors too.

7 0

4 years ago

!!!!! HELP DUE IN 2 MIN !!!

olchik [2.2K]

Answer: 3, 2

Explanation: right on edge 2020

6 0

3 years ago

Read 2 more answers

Compare a low-dose rate internal radiation treatment to a high dose rate internal treatment. What do you think are the positives

Assoli18 [71]

Internal radiation is also called brachytherapy. A radioactive implant is put inside the body in or near the tumor. Getting the implant placed is usually a painless procedure. Depending on your type of cancer and treatment plan, you might get a temporary or a permanent implant. Internal radiation therapy (brachytherapy) allows a higher dose of radiation in a smaller area than might be possible with external radiation treatment. It uses a radiation source that’s usually sealed in a small holder called an implant. Different types of implants may be called pellets, seeds, ribbons, wires, needles, capsules, balloons, or tubes. No matter which type of implant is used, it is placed in your body, very close to or inside the tumor. This way the radiation harms as few normal cells as possible.

During intracavitary radiation, the radioactive source is placed in a body cavity (space) , such as the rectum or uterus.
With interstitial radiation, the implants are placed in or near the tumor, but not in a body cavity. The implant procedure is usually done in a hospital operating room designed to keep the radiation inside the room. You’ll get anesthesia, which may be either general (where drugs are used to put you into a deep sleep so that you don’t feel pain) or local (where part of your body is numbed).

One or more implants is put into the body cavity or tissue with an applicator, usually a metal tube or a plastic tube called a catheter. Imaging tests (an x-ray, ultrasound, MRI, or CT scan) are usually used during the procedure to find the exact place the implant needs to go.

Before being placed, implants are kept in containers that hold the radiation inside so it can’t affect others. The health professionals handling the implants may wear special gear that protects them from exposure once the implants are taken out of the container. High-dose-rate (HDR) brachytherapy allows a person to be treated for several minutes at a time with a powerful radioactive source that’s put in the applicator. The source is removed after 10 to 20 minutes. This may be repeated twice a day over a few days, or once a day over the course of a few weeks. The radioactive material is not left in your body. The applicator might be left in place between treatments, or it might be put in before each treatment.

People getting HDR sometimes stay in the hospital if it involves multiple day treatments and if the applicator is left in place. There may be special precautions to take after the treatment, so be sure to talk to the cancer care team about this. In this approach, the implant gives off lower doses of radiation over a longer period.

Some implants are left in from 1 to a few days and then removed. You’ll probably have to stay in the hospital, sometimes in a special room, during treatment. For larger implants, you might have to stay in bed and lie still to keep it from moving.

Some smaller implants (such as the seeds or pellets) are left in place and never taken out. Over the course of several weeks they stop giving off radiation. The seeds or pellets are about the size of rice grains and rarely cause problems. If your implants are to be left in, you may be able to go home the same day they’re put in. There may be special precautions to take, so be sure to talk to the cancer care team about this.

4 0

3 years ago

Consider the titration of a 73.9 mL sample of 0.13 M HC2H3O2 with 6.978 M NaOH. Ka(HC2H3O2) = 1.8x10-5 Determine the initial pH

Alexeev081 [22]

Answer:

1. pH = 2,82

2. 3,20mL of 1,135M NaOH

3. pH = 3,25

Explanation:

The buffer of acetic acid (HC₂H₃O₂) is:

HC₂H₃O₂ ⇄ H⁺ + C₂H₃O₂⁻

The reaction of HC₂H₃O₂ with NaOH produce:

HC₂H₃O₂ + NaOH → C₂H₃O₂⁻ + Na⁺ + H₂O

And ka is defined as:

ka = [H⁺] [C₂H₃O₂⁻] / [HC₂H₃O₂] = 1,8x10⁻⁵ (1)

1. When in the solution you have just 0,13M HC₂H₃O₂ the concentrations in equilibrium will be:

[H⁺] = x

[C₂H₃O₂⁻] = x

[HC₂H₃O₂] = 0,13 - x

Replacing in (1)

[x] [x] / [0,13-x] = 1,8x10⁻⁵

x² = 2,34x10⁻⁶ - 1,8x10⁻⁵x

x² - 2,34x10⁻⁶ + 1,8x10⁻⁵x = 0

Solving for x:

x = - 0,0015 (Wrong answer, there is no negative concentrations)

x = 0,0015

As [H⁺] = x = 0,0015 and pH is -log [H⁺], pH of the solution is 2,82

2. The equivalence point is reached when moles of HC₂H₃O₂ are equal to moles of NaOH. Moles of HC₂H₃O₂ are:

0,0466L × (0,078mol / L) = 3,63x10⁻³ moles of HC₂H₃O₂

In a 1,135M NaOH, these moles are reached with the addition of:

3,63x10⁻³ moles × (L / 1,135mol) = 3,20x10⁻³L = 3,20mL of 1,135M NaOH

3. The initial moles of HC₂H₃O₂ are:

0,0172L × (0,128mol / L) = 2,20x10⁻³ moles of HC₂H₃O₂

As the addition of NaOH spent HC₂H₃O₂ producing C₂H₃O₂⁻. Moles of C₂H₃O₂⁻ are equal to moles of NaOH and moles of HC₂H₃O₂ are initial moles - moles of NaOH. That means:

0,46x10⁻³L NaOH × (0,155mol / L) = 7,13x10⁻⁵ moles of NaOH ≡ moles of C₂H₃O₂⁻

Final moles of HC₂H₃O₂ are:

2,20x10⁻³ - 7,13x10⁻⁵ = 2,2187x10⁻³ moles of HC₂H₃O₂

Using Henderson-Hasselbalch formula:

pH = pka + log₁₀ [C₂H₃O₂⁻] / [HC₂H₃O₂]

Where pka is -log ka = 4,74. Replacing:

pH = 4,74 + log₁₀ [7,13x10⁻⁵] / [2,2187x10⁻³ ]

pH = 3,25

I hope it helps!

4 0

4 years ago