All categories

3 years ago

How many moles are in 83.1g of CaBr2

Chemistry

1 answer:

9966 [12]3 years ago

8 0

Answer:

0.416 mol CaBr₂

General Formulas and Concepts:

Chemistry - Atomic Structure

Reading a Periodic Table
Using Dimensional Analysis

Explanation:

Step 1: Define

83.1 g CaBr₂

Step 2: Identify Conversions

Molar Mass of Ca - 40.08 g/mol

Molar mass of Br - 79.90 g/mol

Molar Mass of CaBr₂ - 40.08 + 2(79.90) = 199.88 g/mol

Step 3: Convert

$83.1 \ g \ CaBr_2(\frac{1 \ mol \ CaBr_2}{199.88 \ g \ CaBr_2} )$ = 0.415749 mol CaBr₂

Step 4: Check

We are given 3 sig figs. Follow sig fig rules and round.

0.415749 mol CaBr₂ ≈ 0.416 mol CaBr₂

You might be interested in

Cuantos mililitros de HC10.1M contiene 0.025 mol de HC1

jonny [76]

Yes I agree thank you

4 0

3 years ago

natural rubidium has the average mass of 85.4678 and is composed of isotopes 85 Rb(mass = 84.9117) and 87 Rb. The ratio of atoms

mote1985 [20]

Answer:

Mass of Rb-87 is 86.913 amu.

Explanation:

Given data:

Average mass of rubidium = 85.4678 amu

Mass of Rb-85 = 84.9117

Ratio of 85Rb/87Rb in natural rubidium = 2.591

Mass of Rb = ?

Solution:

The ration of both isotope is 2.591 to 1. Which means that for 2.591 atoms of Rb-85 there is one Rb-87.

For 100% naturally occurring Rb = 2.591 + 1 = 3.591

% abundance of Rb-85 = 2.591/ 3.591 = 0.722

% abundance of Rb-87 = 1 - 0.722= 0.278

84.9117 × 0.722 + X × 0.278 = 85.4678

61.306 + X × 0.278 = 85.4678

X × 0.278 = 85.4678 - 61.306

X × 0.278 = 24.1618

X = 24.1618 / 0.278

X = 86.913 amu

8 0

2 years ago

Why do ionic borids form between metal and no metals

mamaluj [8]

Nonmetals don’t have enough electronegativity to transfer electrons so the electrons are shared between them

4 0

3 years ago

A histidine is involved in an interaction with a glutamic acid that stabilizes the charged form of the histidine, such that the

Greeley [361]

Answer:

pKa of the histidine = 9.67

Explanation:

The relation between standard Gibbs energy and equilibrium constant is shown below as:

$\Delta{G^0} =-RT \ln \frac{[His]}{[His+]}$

R is Gas constant having value = 0.008314 kJ / K mol

Given temperature, T = 293 K

Given, $\Delta{G^0}=15\ kJ/mol$

So, Applying in the equation as:-

$15\ kJ/mol=-0.008314\ kJ/Kmol\times 293\ K\times \ln \frac{[His]}{[His+]}$

Thus,

$15\ kJ/mol=-0.008314\ kJ/Kmol\times 293\ K\times \ln \frac{[His]}{[His+]}$

$\frac{[His]}{[His+]}=e^{\frac{15}{-0.008314\times 293}$

$\frac{[His]}{[His+]}=0.00211$

Also, considering:-

$pH=pKa+log\frac{[His]}{[His+]}$

Given that:- pH = 7.0

So, $7.0=pKa+log0.00211$

pKa of the histidine = 9.67

8 0

2 years ago

which industrial city would have fewer air pollution incidents related to temperature inversions one on the great plains or one

egoroff_w [7]

I'm not positive, but I believe it would be one near the Rocky Mountains.

3 0

3 years ago

How many moles are in 83.1g of CaBr2​

How many moles are in 83.1g of CaBr2