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
NNADVOKAT [17]
2 years ago
14

The metal tantalum becomes superconducting at temperatures below 4.483 K. Calculate the temperature at which tantalum becomes su

perconducting in degrees Celsius.
Chemistry
1 answer:
masha68 [24]2 years ago
3 0

Answer:

The correct answer is "-268.667°C".

Explanation:

Given:

Temperature,

= 4.483 K (below)

Now,

The formula of temperature conversion will be:

⇒ T(^{\circ} C)=T(K)-273.15

By putting the values, we get

⇒            =4.483-273.15

⇒            =-268.667^{\circ} C

Thus the above is the correct answer.

You might be interested in
A hydrocarbon has the molar mass 26.04 g/mol. what is a possible formula
jonny [76]
C2H2 is the right answer I believe
6 0
2 years ago
A solution containing 60 grams of nano3 completely dissolved in 50. Grams of water at 50°c is classified as being
Igoryamba

Answer:

<em>A solution containing 60 grams of nano3 completely dissolved in 50. Grams of water at 50°c is classified as being</em> <u>supersaturaded</u>

Explanation:

This question is about solubility.

Regarding solubility, the solutions may be classified as:

  • Unsaturated: the concentration is below the maximum concentration permited at the given temperature.

  • Saturated: the concentration is the maximum permitted at the given temperature, under normal conditions.

  • Supersaturated: the concentration has overcome the maximum permitted at the given temperature. This is possible only under special conditions and is a very unstable state.

Each substance has its own, unique solubility properties. So, in order to tell the state of the solution you need to compare with either solubility tables, or solubility curves; or run you own experiments.

  • In internet you can find the solubility curve of NaNO₃ showing the solubility for a wide range of temperatures.

  • In such curve the solubility of NaNO₃ at 50°C is about 115 g of NaNO₃ per 100 g  of water.

  • Hence, do the proportion to determine the amount of solute that can be dissolved in 50 grams of water at 50°CÑ

       115 g NaNO₃ / 100 g H₂O = x / 50 g H₂O  ⇒ x =  57.5 g NaNO₃

  • <u>Conclusion</u>: 50 grams of water can contain 57.5 g of NaNO₃ dissolved; so, <em>a solution containing 60 g of NaNO₃ completely dissolved in 50 grams of water is supersaturated.</em>

<em />

3 0
3 years ago
Gold is alloyed (mixed) with other metals to increase its hardness in making jewelry.
KiRa [710]

Answer:

A) 54.04%

B) 13-karat

Explanation:

A) From the problem we have

<em>1)</em> Mg + Ms = 9.40 g

<em>2)</em> Vg + Vs = 0.675 cm³

Where M stands for mass, V stands for volume, and g and s stand for gold and silver respectively.

We can rewrite the first equation using the density values:

<em>3)</em> Vg * 19.3 g/cm³ + Vs * 10.5 g/cm³ = 9.40

So now we have<em> a system of two equations</em> (2 and 3) <em>with two unknowns</em>:

We <u>express Vg in terms of Vs</u>:

  • Vg + Vs = 0.675 cm³
  • Vg = 0.675 - Vs

We <u>replace the value of Vg in equation 3</u>:

  • Vg * 19.3 + Vs * 10.5 = 9.40
  • (0.675-Vs) * 19.3 + Vs * 10.5 = 9.40
  • 13.0275 - 19.3Vs + 10.5Vs = 9.40
  • -8.8 Vs + 13.0275 = 9.40
  • <u>Vs = 0.412 cm³</u>

Now we <u>calculate Vg</u>:

  • Vg + Vs = 0.675 cm³
  • Vg + 0.412 cm³ = 0.675 cm³
  • Vg = 0.263 cm³

We <u>calculate Mg from Vg</u>:

  • 0.263 cm³ * 19.3 g/cm³ = 5.08 g

We calculate the mass percentage of gold:

  • 5.08 / 9.40 * 100% = 54.04%

B)

We multiply 24 by the percentage fraction:

  • 24 * 54.04/100 = 12.97-karat ≅ 13-karat
4 0
3 years ago
Describe the appearance of the material. Include its color and state (solid or liquid). Pour ¼ cup of water into a small, clear
Alex777 [14]

<u>Answer:</u>

<em>Here the given material is taken and mixed with water.</em>

<u>Explanation:</u>

The amount of material and water taken are same. Hence if it is not soluble in water it should make a dense and flowy paste like material and if it is soluble in water it should this and thicker density of water should remain.

If the amount of water that we are taking is more than the material will float in water if it is not soluble and lighter than water or would sink if it is heavier than water.

5 0
3 years ago
Read 2 more answers
What is the mass in grams of 1.00 x 10 24 atoms of Mn?
Alex17521 [72]
<h3>Answer:</h3>

91.2 g Mn

<h3>General Formulas and Concepts:</h3>

<u>Math</u>

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

  1. Brackets
  2. Parenthesis
  3. Exponents
  4. Multiplication
  5. Division
  6. Addition
  7. Subtraction
  • Left to Right<u> </u>

<u>Chemistry</u>

<u>Atomic Structure</u>

  • Reading a Periodic Table
  • Avogadro's Number - 6.022 × 10²³ atoms, molecules, formula units, etc.

<u>Stoichiometry</u>

  • Using Dimensional Analysis
<h3>Explanation:</h3>

<u>Step 1: Define</u>

[Given] 1.00 × 10²⁴ atoms Mn

<u>Step 2: Identify Conversions</u>

Avogadro's Numer

[PT] Molar Mass of Mn - 54.94 g/mol

<u>Step 3: Convert</u>

  1. [DA] Set up:                                                                                                      \displaystyle 1.00 \cdot 10^{24} \ atoms \ Mn(\frac{1 \ mol \ Mn}{6.022 \cdot 10^{23} \ atoms \ Mn})(\frac{54.94 \ g \ Mn}{1 \ mol \ Mn})
  2. [DA] Multiply/Divide [Cancel out units]:                                                           \displaystyle 91.2321 \ g \ Mn

<u>Step 4: Check</u>

<em>Follow sig fig rules and round. We are given 3 sig figs.</em>

91.2321 g Mn ≈ 91.2 g Mn

7 0
3 years ago
Read 2 more answers
Other questions:
  • Not all solutions are solids dissolved in liquids. Provide 2 examples of other types of solutions.
    7·1 answer
  • Calculate the vapor pressure (in torr) at 298 k in a solution prepared by dissolving 15.3 g of the non-volatile non-electrolye u
    10·1 answer
  • The complete oxidation of glucose (c6h12o6) to carbon dioxide and water in aerobic respiration consumes how many molecules of ox
    10·1 answer
  • Show the calculation of the final temperature for a 20.8 gram piece of iron heated to 100oC which has been added to a 55.3 gram
    14·1 answer
  • HCI + NaOH →?<br> O NaCl + H20<br> O NaCl + CO2 + H20<br> O NaCl + H2
    13·1 answer
  • Aspartame is a sweetner that is 160 times sweeter than sucrose (table sugar) when dissolved in water. The molecular formula of a
    9·1 answer
  • I need to know the element in group 15 in the periodic table .
    10·1 answer
  • Which of the following correctly identifies a bryophyte
    6·1 answer
  • Where is carbon dioxide found in nature​
    10·2 answers
  • If I have an unknown quantity of gas at a pressure of 1.2 atm, a volume of 77 liters, and a temperature of
    9·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!