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Mademuasel [1]
2 years ago
15

Pleaaaaseee …… I’m begging, helpppp

Chemistry
1 answer:
Goshia [24]2 years ago
7 0

Answer:

13 b

14 c

..... I am pretty sure its right I got the same questions and I got it right

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Mmncotiuy unscrambled would be Community
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3 years ago
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A student will measure and record the growth of two flowering plants every day for 28 days. Plant A will be watered and fertiliz
Mekhanik [1.2K]

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If Plant A grows taller than Plant B

Explanation:

because plant A is being fertilized whereas plant B isn't

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3 years ago
A 10.00 mL sample of a solution containing formic acid (a weak acid) was placed in a 25 mL volumetric flask and diluted to the m
Eddi Din [679]

Answer:

Molarity: 0.522M

Percentage by mass: 2.36 (w/w) %

Explanation:

Formic acid, HCOOH reacts with NaOH as follows:

HCOOH + NaOH → NaCOOH + H₂O

To solve this question we must find the moles of NaOH added = Moles formic acid. Taken into account the dilution that was made we can find the moles -And molarity of formic acid and its percentage by mass as follows:

<em>Moles NaOH = Moles HCOOH:</em>

0.01580L * (0.1322mol / L) =0.002089 moles HCOOH

<em>Moles in the original solution:</em>

0.002089 moles HCOOH * (25mL / 10mL) = 0.005222 moles HCOOH

<em>Molarity of the solution:</em>

0.005222 moles HCOOH / 0.01000L =

<h3>0.522M</h3>

<em>Mass HCOOH in 1L -Molar mass: 46.03g/mol-</em>

0.522moles * (46.03g / mol) = 24.04g HCOOH

<em>Mass solution:</em>

1L = 1000mL * (1.02g / mL) = 1020g solution

<em>Mass percent:</em>

24.04g HCOOH / 1020g solution * 100

2.36 (w/w) %

7 0
3 years ago
A balloon at 32 °C is filled with 21 L of air. What would its volume be at a temperature of 52 °C, assuming pressure remains con
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Answer:C). 24L

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4 0
3 years ago
The ph of 0.015 m hno2 (nitrous acid) aqueous solution was measured to be 2.63. what is the value of pka of nitrous acid?
Licemer1 [7]
Nitrous acid<span> dissociates as follows:
</span>
HNO₂(s) ⇄ H⁺(aq) + NO₂⁻(aq) 
           
According to the equation, an acid constant has the following form:

Ka = [H⁺] × [NO₂⁻ ] / [HNO₂] 

From pH, we can calculate the concentration of H⁺ and NO₂⁻:

[H⁺] = 10^-pH = 10^-2.63 = 0.00234 M = [NO₂⁻]

Now, the acid constant can be calculated:

Ka = 0.00234 x 0.00234 / 0.015  = 3.66 x 10⁻⁴

And finally,

pKa = -log Ka = 3.44 


7 0
3 years ago
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