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blondinia [14]
4 years ago
13

The energy released during ________ reactions can be used to drive ________ reactions, and this is called ________ reactions.

Chemistry
2 answers:
aivan3 [116]4 years ago
7 0

Answer:

should be D

Explanation:

SpyIntel [72]4 years ago
6 0

Answer:

D. exergonic; endergonic; coupled

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What is the wavelength of a 2500 kg truck traveling at a rate of 75km/h?
Pepsi [2]

m = mass of the truck traveling = 2500 kg

v = speed of the truck traveling = 75 km/h = 75 (km/h) (1000 m/ 1 km) (1 h /3600 sec) = 20.83 m/s

h = plank's constant = 6.63 x 10⁻³⁴

λ = wavelength of truck = ?

according to de broglie's principle, wavelength of truck is given as

λ = h/(mv)

inserting the values in the above equation

λ = (6.63 x 10⁻³⁴)/((2500) (20.83))

λ = 1.3 x 10⁻³⁸ m

5 0
3 years ago
Read 2 more answers
Which of the following is true for the endothermic reaction Ba(OH)2(aq) + NH4NO3(aq) --> Ba(NO3)2(aq) + 2NH3(aq) + 2H2O(l)?
Alchen [17]
If it's an endothermic reaction, then that means heat is being added to the system therefore H>0. Entropy is disorder, and since there are more moles on the products side, entropy is increasing therefore S>0 as well. 
6 0
3 years ago
Choose the system to which this item belongs: stomach
hram777 [196]
Digestive. This also includes the intestines.
7 0
3 years ago
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A glucose solution is administered intravenously into the bloodstream at a constant rate r. As the glucose is added, it is conve
Helen [10]

Answer:

C(t)=\frac{r}{k}-(\frac{r-kC_o}{k})e^{-kt}

Explanation:

The differential equation is given as:

\frac{dC}{dt} = r- kC

\frac{dC}{r- kC} = dt

Taking integral of both sides; we have:

\int\limits  \frac{dC}{r- kC} = \int\limits dt

-\frac{1}{k} In(r-kC) = t+D\\In(r-kC)=-kt-kD

r-kC=e^{-kt-kD}

r-kC=e^{-kt}e^{-kD}

r-kC=Ae^{-kt}

kC=r-Ae^{-kt}

C=\frac{r}{k}-\frac{A}{k}e^{-kt}

C(t)=\frac{r}{k}-\frac{A}{k}e^{-kt}     ------- equation (1)

If C(0)= C_o ; we have:

C(0)= \frac{r}{k}-\frac{A}{k}e^0         (where; A is an integration constant)

C_o = \frac{r}{k}- \frac{A}{k}

C_o=\frac{r-A}{k}

kC_o=r-A

A=r-kC_o

Substituting A=r-kC_o into equation (1); we have;

C(t)=\frac{r}{k}-(\frac{r-kC_o}{k})e^{-kt}

3 0
3 years ago
Now if your lab partner determined that there were 1.552 x 1022 atoms of H in an unknown sample of C2H4Cl2, how many milligrams
STALIN [3.7K]
To determine the mass of the sample in milligrams in this problem, we use the avogadro's number to convert from atoms to moles, relate the moles of element in the sample to the mole present and the molar mass of the sample. We do as follows:

1.552 x 10^22 atoms H ( 1 mol H / 6.022x10^23 atoms H ) ( 1 mol C2H4Cl2 / 4 mol H ) ( 98.96 g C2H4Cl2 / 1 mol C2H4Cl2 ) = 0.625 g C2H4Cl2 = 625 mg <span>C2H4Cl2</span>
3 0
4 years ago
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