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3 years ago

What is the correct IUPAC name for Al(IO₃)₃?

Chemistry

1 answer:

Marysya12 [62]3 years ago

7 0

Answer:

Aluminum Iodate

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Why won't tap water boil at 100 degrees

diamong [38]

Its not pure water, it has other minerals and elements in it whose boiling point may be above 100 degrees

7 0

3 years ago

Read 2 more answers

Complete the calculation to find the mass of 1 mole of water (H2O). Use the atomic weights given in the periodic table. Select t

elena55 [62]

Answer:

2x( 1.008 g/mol) +1 x (15.999 g/mol) = 18.015 g/mol

Explanation:

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3 years ago

Cyclopropane, C3H6, is used as a general anesthetic. If a sample of cyclopropane stored in a 2.36-L container at 10.0 atm and 25

alekssr [168]

Explanation:

The given data is as follows.

$V_{1}$ = 2.36 L, $T_{1}$ = $25^{o}C$ = (25 + 273) K = 298 K,

$P_{1}$ = 10.0 atm, $V_{2}$ = 7.79 L,

$T_{2}$ = ?, $P_{2}$ = 5.56 atm

And, according to ideal gas equation,

$\frac{P_{1}V_{1}}{T_{1}} = \frac{P_{2}V_{2}}{T_{2}}$

Hence, putting the given values into the above formula to calculate the value of final temperature as follows.

$\frac{10.0 atm \times 2.36 L}{298 K} = \frac{5.56 atm \times 7.79 L}{T_{2}}$

$T_{2}$ = $\frac{43.3124}{0.0792}$ K

= 546.87 K

Thus, we can conclude that the final temperature is 546.87 K.

4 0

3 years ago

A package contains 1.33 lb of ground round. If there’s 29% fat, how many grams of fat are in the round?

OlgaM077 [116]

The answer is: 175 grams of fat are in the round.

m(ground round) = 1.33 ib.

First convert ib (pounds) to g (grams):

1 pound (lb) is equal to 453.592 grams (g).

m(ground round) = 1.33 ib · 453.592 g/ib.

m(ground round) = 603.28 g.

ω(fat) = 29% ÷ 100%.

ω(fat) = 0.29; mass percentage of fat.

m(fat) = ω(fat) · m(ground round).

m(fat) = 0.29 · 603.28 g.

m(fat) = 175 g; mass of fat.

3 0

3 years ago

8. The time period of artificial satellite in a circular orbit of radius R is T. The radius of the orbit in which time period is

Elena-2011 [213]

Explanation:

It is given that,

The time period of artificial satellite in a circular orbit of radius R is T. The relation between the time period and the radius is given by :

$T^2\propto R^3$

The radius of the orbit in which time period is 8T is R'. So, the relation is given by :

$(\dfrac{T}{T'})^2=(\dfrac{R}{R'})^3$

$(\dfrac{T}{8T})^2=(\dfrac{R}{R'})^3$

$\dfrac{1}{64}=(\dfrac{R}{R'})^3$

$R'=4\times R$

So, the radius of the orbit in which time period is 8T is 4R. Hence, this is the required solution.

4 0

3 years ago