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mestny [16]
2 years ago
12

So when dealing with molar mass when do you know to round up

Chemistry
1 answer:
aniked [119]2 years ago
8 0

The first rule is to use four digits in atomic weights and consider the calculated formula or molar mass as given with four significant digits. This rule is appropriate for beginners but has some disadvantages.

The second rule is to use all digits in the recommended atomic weights and to estimate the accuracy of the result from the number of decimal places or significant digits in these.

The third rule is to use all digits in the recommended atomic weights and to calculate the absolute maximum uncertainty of the result from the recommended uncertainties of the atomic weights. The uncertainty is rounded up to one (two) significant digit(s), and the formula or molar mass is rounded to the same absolute accuracy as the rounded uncertainty.

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Which of the following represents a chemical change? A.Souring of milk B.Melting of chocolate C.Condensation of water D.Breaking
leonid [27]
Chemical changes<span> occur when a substance combines with another to form a new substance, called </span>chemical<span> synthesis or, alternatively, </span>chemical <span>decomposition into two or more different substances. 

So which one do you think is the answer?</span>
8 0
3 years ago
Read 2 more answers
How many liters of water can be made from 55 grams of oxygen gas and an excess of hydrogen at a pressure of 12.4 atm and a tempe
polet [3.4K]
First, we need the no.of moles of O2 = mass/molar mass of O2
                                                             = 55 g / 32 g/mol
                                                             = 1.72 mol
from the balanced equation of the reaction:
2H2 (g) + O2(g) → 2H2O(g)
we can see that the molar ratio between O2: H2O = 1: 2 
So we can get the no.of moles of H2O = 2 * moles of O2
                                                                  = 2 * 1.72 mol
                                                                  = 3.44 mol
So by substitution by this value in ideal gas formula:
PV = nRT

when P = 12.4 atm  & n H2O = 3.44 mol & R= 0.0821 & T = 85 + 273=358K

12.4 atm *V = 3.44 * 0.0821 * 358 = 8.15 L
 ∴ V ≈ 8.2 L 
4 0
3 years ago
A buffer contains 0.18 mol of propionic acid (C2H5COOH) and 0.26 mol of sodium propionate (C2H5COONa) in 1.20 L. What is the pH
Yuki888 [10]

Answer:

1) pH = 5.05

2) pH = 5.13

3) pH = 4.97

Explanation:

Step 1: Data given

Number of moles of propionic acid = 0.18 moles

Number of moles sodium propionate = 0.26 moles

Volume = 1.20 L

Ka = 1.3 * 10^-5    → pKa = 4.989

Step 2: Calculate concentrations

Concentration = moles / volume

[acid]= 0.18/ 1.2 =0.150 M

[salt]= 0.26/ 1.3 = 0.217 M

pH = 4.89 + log(0.217/0.150)=<u>5.05</u>

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What is the pH of the buffer after the addition of 0.02 mol of NaOH?

moles acid = 0.18 - 0.02 = 0.16

[acid]= 0.16/ 1.2=0.133 M

moles salt = 0.26 + 0.02 = 0.28

[salt]= 0.28/ 12=0.233

pH = 4.89 + log 0.233/ 0.133 = 5.13

What is the pH of the buffer after the addition of 0.02 mol of HI?

moles acid = 0.18+ 0.02 = 0.20 moles

[acid]= 0.20/ 1.2 = 0.167 M

[salt]= 0.26 - 0.02= 0.24 moles

[salt]= 0.24/ 1.2 = 0.20 M

pH = 4.89 + log 0.20/ 0.167= 4.97

8 0
2 years ago
How acidic buffer resist change in pH on addition of acid and base.
andrew-mc [135]

Answer:

Buffers are solutions that resist changes in pH, upon addition of small amounts of acid or base. The can do this because they contain an acidic component, HA, to neutralize OH- ions, and a basic component, A-, to neutralize H+ ions. Since Ka is a constant, the [H+] will depend directly on the ratio of [HA]/[A-].

hope it's help

<h3>#carryONlearning </h3>
5 0
3 years ago
A scientist wants to make a solution of tribasic solution phosphate, na3po4, for a laboratory experiment. How many grams of na3p
timurjin [86]

Answer:

55.75g

Explanation:

From

m/M = CV

Where

m= required mass of solute

M= molar mass of solute

C= concentration of solution

V= volume of solution=675ml

Molar mass of solute= 3(23) + 31 + 4(16)= 69+31+64=164gmol-1

Number of moles of sodium ions present= 1.5× 675/1000= 1.01 moles

Since 1 mole of Na3PO4 contains 3 moles of Na+

It implies that 1.01/3 moles of Na3PO4 are present in solution= 0.34moles

mass of Na3PO4= number of moles × molar mass= 0.34 × 164 =55.75g

3 0
3 years ago
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