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Assoli18 [71]
3 years ago
15

A forward reaction in which adding heat decreases product formation is _________, while a forward reaction in which adding heat

increases product formation is_______.
a.) exothermic, endothermic
b.) irreversible, reversible
c.) endothermic, exothermic
d.) none of the above
Chemistry
1 answer:
ELEN [110]3 years ago
3 0
I believe the correct answer from the choices listed above is option A. <span>A forward reaction in which adding heat decreases product formation is exothermic, while a forward reaction in which adding heat increases product formation is endothermic. Exothermic would mean that heat is being released by the process while the opposite is called endothermic in which it absorbs heat.</span>
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What are the genotypes and phenotypes of the offspring in this image?
soldier1979 [14.2K]

Answer:

Genotypes: Hh, Hh, hh, and hh

Phenotypes: Short Hair, Short Hair, Long Hair, Long Hair

Have a nice day! ^-^

6 0
2 years ago
Read 2 more answers
9. The times of first sprinkler activation (in seconds) for a series of tests of fire-prevention sprinkler systems that use aque
Nezavi [6.7K]

This question is incomplete, the complete question is;

The times of first sprinkler activation (in seconds) for a series of tests of fire-prevention sprinkler systems that use aqueous film-forming foam is as follows; 27 41 22 27 23 35 30 33 24 27 28 22 24

( see " use of AFFF in sprinkler systems," Fire technology, 1975: 5)

The system has been designed so that the true average activation time is supposed to be at most 25 seconds.

Does the data indicate the design specifications have not been met?

Test the relevant hypothesis at significance level 0.05 using the P-value approach  

     

Answer:

since p-value (0.042299) is lesser than the level of significance ( 0.05)

We Reject Null Hypothesis

Hence, There is no sufficient evidence that the true average activation time is at most 25 seconds

Explanation:

Given the data in the question;

lets consider Null and Alternative hypothesis;

Null hypothesis H₀ : There is sufficient evidence that the true average activation time is at most 25 seconds

Alternative hypothesis H₁ : There is no sufficient evidence that the true average activation time is at most 25 seconds

i.e

Null hypothesis H₀ : μ ≤ 25

Alternative hypothesis H₁ :  μ > 25

level of significance σ = 0.05

first we determine the sample mean;

x^{bar} = \frac{1}{n}∑x_{i}

where n is sample size and ∑x_{i} is summation of all the sample;

=  \frac{1}{13}( 27 + 41 + 22 + 27 + 23 + 35 + 30 + 33 + 24 + 27 + 28 + 22 + 24 )

=   \frac{1}{13}( 363

sample mean x^{bar} = 27.9231

next we find the standard deviation

s = √( \frac{1}{n-1}∑(x_{i}-x^{bar})²

x                    (x_{i}-x^{bar})                       (x_{i}-x^{bar})²

27                   -0.9231                          0.8521

41                    13.0769                        171.0053

22                  -5.9231                          35.0831  

27                  -0.9231                          0.8521

23                  -4.9231                          24.2369

35                  7.0769                          50.0825

30                  2.0769                          4.3135

33                  5.0769                          25.7749

24                  -3.9231                          15.3907

27                  -0.9231                          0.8521

28                  0.0769                          0.0059

22                 -5.9231                          35.0831  

24                 -3.9231                          15.3907

sum                                                    378.9229

so ∑(x_{i}-x^{bar})² = 378.9229

∴

s = √( \frac{1}{13-1} ×378.9229 )

s = √31.5769

standard deviation s = 5.6193

now, the Test statistics

t = ( x^{bar} - μ ) / \frac{s}{\sqrt{n} }

we substitute

t = ( 27.9231 - 25 ) / \frac{5.6193}{\sqrt{13} }

t = 2.9231 / 1.5585

t = 1.88

now degree of freedom df = n - 1 = 13 - 1 = 12

next we calculate p-value

p-value = 0.042299 ( using Execl's ( = TDIST(1.88,12,1)))

Here x=1.88, df=12, one tail

now we compare the p-value with the level of significance

since p-value (0.042299) is lesser than the level of significance ( 0.05)

We Reject Null Hypothesis

Hence, There is no sufficient evidence that the true average activation time is at most 25 seconds

3 0
3 years ago
What is the mass, in grams of a pure iron cube that has a volume of 4.20cm^3
Dmitriy789 [7]
d_{Fe}=7874\frac{kg}{m^{3}}=7,874\frac{g}{cm^{3}}\\&#10;V=4,2cm^{3}\\\\&#10;m=dV=7,784\frac{g}{cm^{3}}*4,2cm^{3}\approx 32,7g
6 0
3 years ago
PLEASE HELP
swat32

Answer:

1461.7 g of AgI

Explanation:

We'll begin by writing the balanced equation for the reaction. This is given below:

CaI₂ + 2AgNO₃ —> 2AgI + Ca(NO₃)₂

From the balanced equation above,

1 mole of CaI₂ reacted to produce 2 moles of AgI.

Next, we shall determine the number of mole AgI produced by the reaction of 3.11 moles of CaI₂. This can be obtained as follow:

From the balanced equation above,

1 mole of CaI₂ reacted to produce 2 moles of AgI.

Therefore, 3.11 moles of CaI₂ will react to produce = 3.11 × 2 = 6.22 moles of AgI

Finally, we shall determine the mass of 6.22 moles of AgI. This can be obtained as follow:

Mole of AgI = 6.22 moles

Molar mass of AgI = 108 + 127

= 235 g/mol

Mass of AgI =?

Mass = mole × molar mass

Mass of AgI = 6.22 × 235

Mass of AgI = 1461.7 g

Therefore, 1461.7 g of AgI were obtained from the reaction.

5 0
3 years ago
A piece of lithium is placed on the surface of some water in a beaker. hydrogen is given off. Lithium hydroxide is formed. What
Gala2k [10]
<span>A piece of lithium is placed on the surface of some water in a beaker. Hydrogen is given off and Lithium hydroxide is formed. The word equation would be as follows:
</span>

lithium<span> + </span>water<span>→ </span>lithium hydroxide<span> + </span>hydrogen

Hope this answers the question. Have a nice day.
8 0
2 years ago
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