All categories

3 years ago

How many moles are in 15 grams of Cr2(CO3)3 ?

Chemistry

1 answer:

Rainbow [258]3 years ago

7 0

Answer:

0.052813386714757

Explanation:

That's what I got, not quite sure if it's right tho

You might be interested in

How can the actions of people affect runoff and absorption in a positive way? How do they affect them in a negative way?

lora16 [44]

Answer:

pollution and waste every where like in are ouceans

Explanation:

4 0

2 years ago

What is the formula for 2-methylnonane

Vera_Pavlovna [14]

Answer:

C10H22

sorry if im wrong

5 0

3 years ago

A helium balloon has a volume of 2.5 L. When heated to 343.5K the pressure of the gas is 66.7 kPa. How many moles of He gas woul

Lelu [443]

Answer:

0.06mole

Explanation:

Given parameters:

Volume of balloon = 2.5L

T = 343.5K

P = 66.7kPa; in atm gives = 0.66atm

Unknown:

Number of moles of He = ?

Solution:

To solve this problem, we use the ideal gas equation.

It is mathematically expressed as;

Pv = nRT

where p is the pressure

v is the volume

n is the number of moles

R is the gas constant

T is the temperature

0.66 x 2.5 = n x 0.082 x 343.5

n = 0.06mole

8 0

3 years ago

How many electrons are their in an atom of chlorine?

astraxan [27]

There are 7 valence electrons in the atom :)

4 0

3 years ago

6) (a) Calculate the absorbance of the solution if its concentration is 0.0278 M and its molar extinction coefficient is 35.9 L/

Anvisha [2.4K]

Answer:

6) (a) 0.499; (b) 31.7 %

7) 0.15

Explanation:

6) (a) Absorbance

Beer's Law is

$A = \epsilon cl\\A = \text{35.9 L&\cdot$mol$^{-1}$cm$^{-1}$} $\times$ 0.0278 mol$\cdot$L$^{-1} \times $ 0.5 cm = \mathbf{0.499}$

(b) Percent transmission

$A = \log {\left (\dfrac{1}{T}}\right)}\\\\\%T = 100T\\\\T = \dfrac{\%T}{100}\\\\\dfrac{1}{T} = \dfrac{100 }{\%T}\\\\A = \log \left(\dfrac{100 }{\%T} \right ) = 2 - \log \%T\\\\0.499 = 2 - \log \%T\\\\\log \%T = 2 - 0.499 = 1.501\\\\\%T = 10^{1.501} = \mathbf{31.7}$

7) Absorbance

$A = \log \left (\dfrac{I_{0}}{I} \right ) = \log \left (\dfrac{I_{0}}{0.70I_{0}} \right ) = \log \left (\dfrac{1}{0.70} \right ) = -\log(0.70) = \mathbf{0.15}}$

8 0

3 years ago