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2 years ago

What is the formula for the compound?

Chemistry

2 answers:

4vir4ik [10]2 years ago

7 0

Answer:

the correct answer is CO2 (the last one)

svet-max [94.6K]2 years ago

7 0

I think the answer is CO2, I hope that helps

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Im about to take first year chemistry for engineering. Should i review or will this be done in class? If yes what are some conce

sashaice [31]

You should always study. you can be above your class. thats what i do.. search google for ur course. they can tell u some key points for it. u should even study what you know and want to know as well.

6 0

2 years ago

In a titration experiment a student uses 1.4 m hbr solution and the indicator phenolphthalein to determine the concentration of

Likurg_2 [28]

The question is incomplete. Complete question is attached below:
...........................................................................................................................

Answer:
Given: conc. of HBr = 1.4 M
Volume of HBr = 15.4 mL
Volume of KOH = 22.10 mL

We know that, M1V1 = M2V2
(HBr) (KOH)

Therefore, M2 = M1V1/V2
= 1.4 X 15.4/22.10
= 0.9756 M

Concentration of KOH is 0.9756 M.

3 0

2 years ago

Read 2 more answers

Mrs. Harrison mixes some chemicals together in a glass beaker in front of her science class. She asks her students to watch for

Bumek [7]

The correct answer B. increase the flame setting to produce more heat

7 0

2 years ago

Read 2 more answers

Balance the equation below, and answer the following question: What volume of chlorine gas, measured at STP; is needed to comple

spayn [35]

The balanced chemical reaction would be expressed as follows:

2Na + Cl2 = 2NaCl

We are given the amount of sodium metal to be used reaction. This would be the starting value for the calculations. We do as follows:

6.25 g ( 1 mol / 22.99 g ) ( 1 mol Cl2 / 2 mol Na ) ( 22.4 L / 1 mol ) = 3.045 L Cl2 needed

Hope this answers the question.

6 0

3 years ago

Additional Practice Questions

bezimeni [28]

pH of solution = 13.033

<h3>Further explanation</h3>

Given

2.31 g Ba(OH)₂

250 ml water

Required

pH of solution

Solution

Barium hydroxide is fully ionized, means that Ba(OH)₂ is a strong base

So we use a strong base formula to find the pH

[OH ⁻] = b. Mb where

b = number of OH⁻ /base valence

Mb = strong base concentration

Molarity of Ba(OH)₂(MW=171.34 g/mol) :

$\tt M=\dfrac{mol}{L}=\dfrac{2.31~g\div 171.34~g/mol}{0.25~L}\\\\M=0.054$

Ba(OH)₂ ⇒ Ba²⁺ + 2OH⁻(b=valence=2)

[OH⁻]= 2 . 0.054

[OH⁻] = 0.108

pOH= - log 0.108

pOH=0.967

pOH+pH=14

pH=14-0.967

pH=13.033

4 0

2 years ago