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NISA [10]
3 years ago
7

How many donuts are in a mole of donuts ?​

Chemistry
2 answers:
zhannawk [14.2K]3 years ago
7 0

6.02×10²³ donuts is how much donuts are in a mole.

pychu [463]3 years ago
5 0

Whats the donar mass

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How many grams of K2Cr207 are in 200 mL 0.1M
charle [14.2K]

Answer:

407grams ithink lang po

4 0
3 years ago
A circle has a radius of 5 feet and a central angle AOB that measures 10°. What is the length of the intercepted arc AB? Use 3.1
algol [13]
We are asked to solve for the arc length of the intercepted arc and the formula is shown below:
Arc length = 2*pi*r(central angle/360°)
r = 5 feet
central angle = 10°

Solving for the arc length, we have:
Arc length = 2*3.14*5 (10/360)
Arc length = 0.872 feet

The arc length is 0.872 feet.
5 0
3 years ago
0 Balance the following equation: CO2 (g) + H2O (1) → C12H24012 (s) + O2 (g) +​
Drupady [299]

Explanation:

Answer: 12CO2(g) +12H2O(l)->C12H24O12(s)+12O2(g)

Once you make balancing equations don't disturb the given numbers because it is fix you need to solve by the side of the chemical name.

5 0
3 years ago
The normal boiling point of a liquid is 282 °C. At what temperature (in
ElenaW [278]

Answer:

The temperature at which the liquid vapor pressure will be 0.2 atm = 167.22 °C

Explanation:

Here we make use of the Clausius-Clapeyron equation;

ln\left (\frac{p_{2}}{p_{1}}  \right )=-\frac{\Delta H_{vap}}{R}\cdot \left (\frac{1}{T_{2}}-\frac{1}{T_{1}}  \right )

Where:

P₁ = 1 atm =The substance vapor pressure at temperature T₁ = 282°C = 555.15 K

P₂ = 0.2 atm = The substance vapor pressure at temperature T₂

\Delta H_{vap} = The heat of vaporization = 28.5 kJ/mol

R = The universal gas constant = 8.314 J/K·mol

Plugging in the above values in the Clausius-Clapeyron equation, we have;

ln\left (\frac{0.2}{1}  \right )=-\frac{28.5 \times 10^3}{8.3145}\cdot \left (\frac{1}{T_{2}}-\frac{1}{555.15}  \right )

\therefore T_2 = \frac{-3427.95}{ln(0.2)-6.175}

T₂ = 440.37 K

To convert to Celsius degree temperature, we subtract 273.15 as follows

T₂ in °C = 440.37 - 273.15 = 167.22 °C

Therefore, the temperature at which the liquid vapor pressure will be 0.2 atm = 167.22 °C.

5 0
3 years ago
Help Pls!
Salsk061 [2.6K]

Answer:

d= 50.23 g/cm³

Explanation:

Given data:

radius = 137.9 pm

mass is = 5.5 × 10−22 g

density = ?

Solution:

volume of sphere= 4/3π r³

First of all we calculate the volume:

v= 4/3π r3

v= 1.33× 3.14× (137.9)³

v= 1.33 × 3.14 × 2622362.939 pm³

v= 1.095 × 10∧7 pm³

v= 1.095 × 10∧-23 cm³

Formula:

Density:

d=m/v

d= 5.5 × 10−22 g/ 1.095 × 10∧-23 cm³

d= 5.023 × 10∧+1 g/cm³

d= 50.23 g/cm³

8 0
3 years ago
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