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vlabodo [156]
3 years ago
14

Can someone help me out with this?

Chemistry
1 answer:
ycow [4]3 years ago
4 0

Answer:

2.8 * 10^(-6) / 1.4 * 10^(-2)=

2* 10^(-8)

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Predict whether litmus paper will turn red or blue when dipped in:
Temka [501]

Answer:

a. blue, b. red, c. red, d. basic - blue

Explanation:

a.washing powder ( containing sodium carbonate ) basic -blue

b. orange juice ( containing citric acid )- acidic -red

c. lemonade ( containing H2CO₃ )-acidic - red

d. cleaner ( containg NH₃ )- basic - blue

7 0
3 years ago
When 7.085 grams of a hydrocarbon, CxHy, were burned in a combustion analysis apparatus, 21.71 grams of CO2 and 10.37 grams of H
Taya2010 [7]

Answer:

- Empirical:

C_3H_7

- Molecular:

C_6H_{14}

Explanation:

Hello,

In this case, based on the information regarding the combustion, the moles of carbon turn out:

n_C=21.71gCO_2*\frac{1molCO_2}{44gCO_2}*\frac{1molC}{1molCO_2}=0.493molC

Moreover, the moles of hydrogen:

n_H=10.37gH_2O*\frac{1molH_2O}{18gH_2O}*\frac{2molH}{1molH_2O}=1.152molH

Thus, the subscripts of carbon and hydrogen in the hydrocarbon turn out:

C=\frac{0.4934}{0.4934}=1\\H=\frac{1.15222}{0.4934}=2.335\\CH_{2.335}

Now, looking for a suitable whole number we obtain the following empirical formula as 2.335 times 3 is 7 for hydrogen:

C_3H_7

In such a way, that compound has a molar mass of 43 g/mol, thus, the whole compound's molar mass is 86.18 g/mol for which the molecular formula is twice the empirical one, therefore:

C_6H_{14}

Which is hexane.

Best regards.

6 0
3 years ago
3.30 moles of pyrophyllite, Al2Si4O10(OH)2<br><br>How many moles of Al are in this compound?
Rom4ik [11]

33.33 moles in AI. Because it is just to use a diffrent site

3 0
2 years ago
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4 0
3 years ago
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What is the value of the rate constant k for this reaction?When entering compound units, indicate multiplication of units explic
vekshin1

The table with the data is in the picture attached.

Answer:

  • k=0.0033M^{-2}.s^{-1}

Explanation:

The reaction equation suggests that the law could have this form:

  • rate=k[A]^a[B]^b[C]^c

Then, the work is to find the values of the exponents that satisfy the initial rate data.

A first glance shows that for the third and fourth trials the initial rates are the same. Since for these two trials only the initial concentration of substance B changed (A and C were kept equal), you conclude that the reaction rate does not depend on B, and ist exponent (lower b) is 0.

Then, so far you can say:

  • rate=k[A]^a[C]^c

When you use trials 1 and 2, you get:

\frac{r_2}{r_1}=\frac{27M/s}{9M/s}=\frac{(0.3M)^a(0.3M)^b(0.9M)^c}{(0.3M)^a(0.3M)^b(0.3M)^c}=3^c\\\\ 3=3^c\\ \\ 1=c

Now, you can use trials 1 and 3 to determine the other exponent:

\frac{r_3}{r_1}=\frac{36M/s}{9M/s}=\frac{(0.6M)^a(0.3M)^b(0.3M)^c}{(0.3M)^a(0.3M)^b(0.3M)^c}=2^a\\\\ 4=2^a\\ \\ 2^2=2^a\\ \\ 2=a

Thus, you have the rate law:

  • r=k[A]^2[C]

Now, you just use any trial to obtain k. Using trail 1:

  • k(0.3M)^2(0.3M)=9.10^{-5}M/s

Which yields:

  • k=0.0033M^{-2}.s^{-1}

4 0
3 years ago
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