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Lelu [443]
3 years ago
7

The table below shows the content in mg of some chemical elements found in 100g of milk:

Chemistry
1 answer:
garri49 [273]3 years ago
5 0

Answer:

... chloride, calcium, potassium, and zinc was signifi- ... of cow and goat milk pasteurization on element retention ... certified American Chemical Society (ACS);. Whatman ... goat milk. Table 2 gives the content of 17 elements of ... found .0026 rag/100 g in raw and .0024 mg/100 ... mg/100 g chloride content (27) and another.

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A student sets up the following equation to convert a measurement.
dybincka [34]

Answer:

\frac{1 m}{100 cm}

Explanation:

The final answer has a different set of units. In particular, meters (m) changes to centimeters (cm). To make this change, you need to multiply the first value by proportions.

When writing these proportions, it is important that they are arranged in a way that allows for the cancellation of units. For instance, since m is located in the denominator, it must be located in the numerator of the conversion.

<u>Proportion:</u>

1 m = 100 cm

The full expression:

<h3>-1.7*10^5\frac{V}{m}  ·  \frac{1 m}{100 cm}  =  ? \frac{V}{cm}</h3><h2>                 ^</h2>

As you can see, the old unit (m) cancels out and you are left with cm in the denominator.

7 0
2 years ago
Determine the final temperature of sample with a specific heat of 1.1 J/g°C and a mass of 385 g if it starts out at a temperatur
Assoli18 [71]

Answer:

T2 =21.52°C

Explanation:

Given data:

Specific heat capacity of sample = 1.1 J/g.°C

Mass of sample = 385 g

Initial temperature = 19.5°C

Heat absorbed = 885 J

Solution:

Formula:

Q = m.c. ΔT

Q = amount of heat absorbed or released

m = mass of given substance

c = specific heat capacity of substance

ΔT = change in temperature

ΔT = Final temperature - initial temperature

885J = 385 g× 1.1 J/g.°C×(T2 - 19.5°C )

885 J = 423.5 J/°C× (T2 - 19.5°C )

885 J / 423.5 J/°C = (T2 - 19.5°C )

2.02°C = (T2 - 19.5°C )

T2 = 2.02°C + 19.5°C

T2 =21.52°C

8 0
3 years ago
Suppose you need of Grade 70 tow chain, which has a diameter of and weighs , to tow a car. How would you calculate the mass of t
BabaBlast [244]

Answer: check explanation

Explanation:

In this question we are to find mass. In order to calculate the Mass, We need the values of two parameters, that is, the values given for the grade tow chain, and the value given for the mass per length.

Assuming the mass per length is 3 Kilogram per metre(kg/m) and the grade 70 tow chain length is 5 metre(m).

Therefore, the formula for calculating mass of the chain is given below;

Mass of the chain= mass per unit length(kg/m) × length ---------------------------------------------------------------------------------------------------------------------(1).

Mass of the chain= 3 kg/m × 5 m.

Mass of the chain= 15 kg.

7 0
3 years ago
Sophia was fascinated while studying the role of oceans in the hydrosphere. She decided to illustrate the features of the ocean
ANEK [815]

Answer:

The Major features of the ocean floor are:

  1. Continental Shelf
  2. Continental Slope
  3. Continental Rise
  4. Abyssal Plain
  5. Oceanic Trench
  6. Mid-Ocean Ridge

Explanation:

1. Continental Shelf: This refers to the part of the land on every continent that is covered with water that is not too deep. The types of animals that can be found on the continental shelf are:

Crab, Tuna, Lobster, Dungeness cod, etc. Within the Continental shelf, there are permanent rocks that house other organisms such as sponges, anemones, clams, sponges, oysters. The continental shelf also contains the route of migration for bigger animals such as sea turtles dolphins and even whales.

2. Continental Slope: This spans from the shelf break to the continental rise.  It can slope up to 4 degrees. Slopes can be created by faulting, slumping of huge boulders of sediments, rifting, etc.

Some of the aquatic animals that can be found in this region include but are not limited to:

Sablefish, Dover sole rockfish, etc.

3. Continental rise

This part of the ocean floor usually has a very steep gradient or angle slope. It slopes very steeply into the abyssal plain of the ocean.

The following can help form continental rise:

  • Mass wasting;
  • deposition from contour currents and
  • the longitudinal settling of biogenic and clastic particles

4. Abyssal Plain.

This is the real bottom of the ocean. There is a very high probability that one would find animals such as nematodes, polychaetes, etc which are all types of worms down there. The Abyssal plain is also home to molluscs,  and echinoderms.

5. Oceanic Trench

Sometimes there is a long and narrow indenture or depression along the seafloor. These are called Trenches. Trenches are sometimes formed by the boundaries between one lithospheric plate and another. The deepest trench on earth is found in the Pacific Ocean. It has been nick-named the Challenger Deep and said to be the deepest point known on earth reaching almost 11 kilometers.

6. Mid-Ocean Ridge

This is a mountain range underneath the ocean. It is formed when there is an upward push by convection currents of the mantle beneath the oceanic crust. When this happens and molten magma is ejected or created at the boundary between the plates, the result is a Mid-Oceanic Ridge.

Cheers

7 0
2 years ago
Hydrofluoric acid and Water react to form fluoride anion and hydronium cation, like this HF(aq) + H_2O(l) rightarrow F(aq) + H_3
maksim [4K]

Answer:

Kc = 1.09x10⁻⁴

Explanation:

<em>HF = 1.62g</em>

<em>H₂O = 516g</em>

<em>F⁻ = 0.163g</em>

<em>H₃O⁺ = 0.110g</em>

<em />

To solve this question we need to find the moles of each reactant in order to solve the molar concentration of each reactan and replacing in the Kc expression. For the reaction, the Kc is:

Kc = [H₃O⁺] [F⁻] / [HF]

<em>Because Kc is defined as the ratio between concentrations of products over reactants powered to its reaction coefficient. Pure liquids as water are not taken into account in Kc expression:</em>

<em />

[H₃O⁺] = 0.110g * (1mol /19.01g) = 0.00579moles / 5.6L = 1.03x10⁻³M

[F⁻] = 0.163g * (1mol /19.0g) = 0.00858moles / 5.6L = 1.53x10⁻³M

[HF] = 1.62g * (1mol /20g) = 0.081moles / 5.6L = 0.0145M

Kc = [1.03x10⁻³M] [1.53x10⁻³M] / [0.0145M]

<h3>Kc = 1.09x10⁻⁴</h3>
7 0
3 years ago
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