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Harlamova29_29 [7]
3 years ago
9

Calculate the percentage mass of calcium chloride, CaCi^2​

Chemistry
1 answer:
Firdavs [7]3 years ago
7 0

Answer: m-%(Ca) = 40.08 / 110.98

Explanation: molar mass of CaCl2 is 40.08+ 2·35.45 = 110.98

Think you have one mole substance. It contains 40.08 g Ca

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Storing sugar as long chains for later use is an example of a(n) ____________ chemical reaction.
DerKrebs [107]

Answer:

Endothermic

Explanation:

Storing sugar for later use is an example of an endothermic reaction because that energy is being absorbed.

8 0
3 years ago
Lunar Eclipe's do not happen every month because the moon, earth and sun must be lined up in a certain way.
Andreyy89

Answer:

A. True

Explanation:

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4 0
3 years ago
Read 2 more answers
How many hours will it take for the concentration of methyl isonitrile to drop to 15.0 % of its initial value?
SpyIntel [72]

Answer:

The concentration  of methyl isonitrile will become 15% of the initial value after 10.31 hrs.

Explanation:

As the data the rate constant is not given in this description, However from observing the complete question  the rate constant is given as a rate constant of 5.11x10-5s-1 at 472k .

Now the ratio of two concentrations is given as

ln (\frac{C}{C_0})=-kt

Here C/C_0 is the ratio of concentration which is given as 15% or 0.15.

k is the rate constant which is given as 5.11 \times 10^{-5} \, s^{-1}

So time t is given as

ln (\frac{C}{C_0})=-kt\\ln(0.15)=-5.11 \times 10^{-5} \times t\\t=\frac{ln(0.15)}{-5.11 \times 10^{-5} }\\t=37125.6 s\\t=37125.6/3600 \\t= 10.31 \, hrs

So the concentration will become 15% of the initial value after 10.31 hrs.

7 0
3 years ago
Read 2 more answers
A buffer solution is made that is 0.347 M in H2C2O4 and 0.347 M KHC2O4.
irga5000 [103]

Answer:

1. pH = 1.23.

2. H_2C_2O_4(aq) +OH^-(aq)\rightarrow HC_2O_4^-(aq)+H_2O(l)

Explanation:

Hello!

1. In this case, for the ionization of H2C2O4, we can write:

H_2C_2O_4\rightleftharpoons HC_2O_4^-+H^+

It means, that if it is forming a buffer solution with its conjugate base in the form of KHC2O4, we can compute the pH based on the Henderson-Hasselbach equation:

pH=pKa+log(\frac{[base]}{[acid]} )

Whereas the pKa is:

pKa=-log(Ka)=-log(5.90x10^{-2})=1.23

The concentration of the base is 0.347 M and the concentration of the acid is 0.347 M as well, as seen on the statement; thus, the pH is:

pH=1.23+log(\frac{0.347M}{0.347M} )\\\\pH=1.23+0\\\\pH=1.23

2. Now, since the addition of KOH directly consumes 0.070 moles of acid, we can compute the remaining moles as follows:

n_{acid}=0.347mol/L*1.00L=0.347mol\\\\n_{acid}^{remaining}=0.347mol-0.070mol=0.277mol

It means that the acid remains in excess yet more base is yielded due to the effect of the OH ions provided by the KOH; therefore, the undergone chemical reaction is:

H_2C_2O_4(aq) +OH^-(aq)\rightarrow HC_2O_4^-(aq)+H_2O(l)

Which is also shown in net ionic notation.

Best regards!

4 0
3 years ago
Which part of a feedback mechanism causes change to make up for the departure from the set point? A. sensor B. effector C. respo
koban [17]
I would have to say C. Response.
5 0
4 years ago
Read 2 more answers
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