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Marysya12 [62]
3 years ago
8

What is the molecular weight (rmm) of NaOH? 40 20 50 60

Chemistry
1 answer:
Molodets [167]3 years ago
4 0

Answer: A/40 it is actually 39.997 but since that is not an answer they rounded up

Explanation:

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List at least 10 ways in which you can save heat energy
guapka [62]

Answer:

Here are 10 ways to start conserving energy yourself:

Adjust your day-to-day behaviors

Replace your light bulbs

Use smart power strips

Install a programmable thermostat

Use energy efficient appliances

Reduce water heating expenses

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Explanation:

7 0
3 years ago
SO2 can act as both oxidising and reducing agent​
KiRa [710]

Answer:

No, it acts only as a reducing agent

Explanation:

Sulphur dioxide is a strong reducing agent. It is mainly used to prevent dried fruit from being oxidised and discoloured.

3 0
3 years ago
How does the number of reactants in a decomposition reaction compare with
dem82 [27]
Answer is B the number of reactants is the same as the number of products
6 0
3 years ago
Two moles of a monatomic ideal gas are contained at a pressure of 1 atm and a temperature of 300 K; 34,166 J of heat are transfe
anygoal [31]

Answer:

Final temperature is 302 K

Explanation:

You can now initial volume with ideal gas law, thus:

V = \frac{n.R.T}{P}

Where:

n are moles: 2 moles

R is gas constant: 0,082 \frac{atm.L}{mol.K}

T is temperature: 300 K

P is pressure: 1 atm

V is volume, with these values: <em>49,2 L</em>

The work in the expansion of the gas, W, is: 1216 J - 34166 J = <em>-32950 J</em>

This work is:

W = P (Vf- Vi)

Where P is constant pressure, 1 atm

And Vf and Vi are final and initial volume in the expansion

-32950 J = -1 atm (Vf-49,2L) × \frac{101325 J}{1 atm.L}

Solving: <em>Vf = 49,52 L</em>

Thus, final temperature could be obtained from ideal gas law, again:

T = \frac{P.V}{n.R}

Where:

n are moles: 2 moles

R is gas constant: 0,082 \frac{atm.L}{mol.K}

P is pressure: 1 atm

V is volume: 49,52 L

T is final temperature: <em>302 K</em>

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I hope it helps!

4 0
3 years ago
How many grams of sodium formate, NaCHO2, would have to be dissolved in 3.0 L of 0.12 M formic acid (pKa 3.74) to make the solut
kolbaska11 [484]

Answer:

Mass_{sodium\ formate}= 889.57\ g

Explanation:

Considering the Henderson- Hasselbalch equation for the calculation of the pH of the acidic buffer solution as:

pH=pK_a+log\frac{[salt]}{[acid]}

Given that:-

[Acid] = 0.12 M

Volume = 3.0 L

pKa = 3.74

pH = 5.30

So,  

5.30=3.74+log\frac{[sodium\ formate]}{0.12}

Solving, we get that:-

[Sodium formate] = 4.36 M

Considering:

Molarity=\frac{Moles\ of\ solute}{Volume\ of\ the\ solution}

So,

Moles =Molarity \times {Volume\ of\ the\ solution}

So, Moles of sodium formate = 4.36*3.0 moles = 13.08 moles

Molar mass of sodium formate = 68.01 g/mol

The formula for the calculation of moles is shown below:

moles = \frac{Mass\ taken}{Molar\ mass}

Thus,

13.08\ mole= \frac{Mass}{68.01\ g/mol}

Mass_{sodium\ formate}= 889.57\ g

5 0
4 years ago
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