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2 years ago

Extension: Cedric has been in the hospital for 15 weeks, how many minutes is that? Use

Chemistry

1 answer:

almond37 [142]2 years ago

6 0

Answer:

151200 minutes.

Explanation:

From the question given above, the following data were obtained:

Time (in week) = 15 weeks

Time (in min) =?

Next, we shall convert 15 weeks to days. This can be obtained as follow:

1 week = 7 days

Therefore,

15 weeks = 15 weeks × 7 days / 1 week

15 weeks = 105 days

Next, we shall convert 105 days to hours. This can be obtained as follow:

1 day = 24 h

Therefore,

105 days = 105 days × 24 h / 1 day

105 days = 2520 h

Finally, we shall convert 2520 h to mins. This can be obtained as follow:

1 h = 60 mins

Therefore,

2520 h = 2520 h × 60 mins / 1 h

2520 h = 151200 mins

Thus, 15 weeks is equivalent to 151200 minutes.

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Answer:

1. analyze the samples while they are frozen

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2 years ago

Read 2 more answers

Learn how changes in binding free energy affect binding and the ratio of unbound and bound molecules.

Delvig [45]

C)[D]/[ED] = 5.20

D)[D]/[ED] = 5.20

E)[D']_T = 1.495* 10 ^-7 M

F)[D'] / [ED'] = 0.0579

Explanation:

E = 250 nM =2.5* 10 ^-7 mol/L , T=298.15 K

Dissociation constant of K_D = 1.3 μM (1.3 *10 ^-6 M)

E + D ⇄ ED → K_a = [ED] / [D][E] (association constant)

ED ⇄ E + D → K_D = [E][D] / [ED] (dissociation constant)

[E] =2.5*10^-7 mol/L

K_D = 1.3* 10^-6 M

K_D = [E][D] / [ED] → [D]/[ED] = K_D / [E]

= [D]/[ED] = 1.3* 10 ^-6 / 2.5 *10^-7

= 13/25 * 10

=130/25 = 5.20

[D]/[ED] = 5.20

ΔG =RTln Kd

ΔG_2 for E and D = 1.987 * 298.15 * ln 1.3*10^-6

ΔG_2 592.454 * [ln 1.3 +ln 10^-6]

ΔG_1 = 592.424 [0.2623 - 13.8155]

ΔG_2 = -592.424 * 13.553

ΔG_1 = -8184.633 cal/ mol

ΔG_1 = -8184.633 * 4.18 J/mol = -34244.508 J?mol

ΔG_1 = -34.245 KL/mol

so, ΔG_2 = ΔG_1 - 10.5 KJ/mol

ΔG_2 = -34.245 - 10.5

ΔG_2 = -44.745KJ / mol

ΔG_2 =RT ln K_D

-44.745 *10^3

=8.314 *298.15 lnK_D

lnK_D' = - 44745 / 2478.81 g

ln K_D' = -18.051

K_D' = -18.051

K_D' = e^-18.051

[D]/[ED] = 5.20

[E] = 2.5* 10 ^-7 mol/ L = a

K_D' = [E][D] / [ED'] E +D' → ED'

K_D' = a/2(x-(a/2) / (a/2)

KD' = x - a/2

=2.447 *10^-8 = (2.5/2) * 10^-7

x=2.447 * 10^-8 + 1.25 * 10^-7

x = 2.447 *10^-8 + 1.25 * 10 ^-7

x= 10^-7 [1.25 + 0.2447]

x = 1.4947 * 10^-7

[D']_T = 1.495* 10 ^-7 M

K_D' = [E][D'] / [ED']

[D'] / [ED'] = KD' / [E]

[D'] / [ED'] = 1.447 *10^-8 / 2.5* 10^-7

[D'] / [ED'] = 0.5788 * 10^-1

[D'] / [ED'] = 0.0579

5 0

3 years ago

True or false? the iron recommendation for girls exceeds that of boys during adolescence.

sashaice [31]

It is true that the iron recommendation for girls exceeds that of boys during adolescence.

During an adolescence, girls and boys require iron for a large growth spurt and the gain of adult phenotypes and biologic rhythms.

In this period of the life, iron recommendation increase in both girls and boys, because of the increase in lean body mass, the expansion of the total blood volume, the increase and start of menstruation at girls.

Iron is essential for oxygen transport, red blood cell creation, cognitive performance and immunological function.

The overall iron requirements for girls are up to twice as boys.

More about adolescence: brainly.com/question/13528489

#SPJ4

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1 year ago

For a particular reaction, Δ=−111.4 kJ and Δ=−25.0 J/K.

vichka [17]

Answer:

$\Delta G =-103.95kJ$

Explanation:

Hello there!

In this case, since the thermodynamic definition of the Gibbs free energy for a change process is:

$\Delta G =\Delta H-T\Delta S$

It is possible to plug in the given H, T and S with consistent units, to obtain the correct G as shown below:

$\Delta G =-111.4kJ-(298K)(-25.0\frac{J}{K}*\frac{1kJ}{1000J} )\\\\\Delta G =-103.95kJ$

Best regards!

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3 years ago

Use the molar heat capacity for aluminum from table 1 to calculate the amount of energy needed to raise the temperature of 260.5

Nimfa-mama [501]

Unfortunately, you failed to include the table 1 from which the molar heat capacity of aluminum could have been obtained. However, as a general rule, the heat needed to raise the temperature of a certain substance by certain degrees is calculated through the equation,
H = mcpdT
where H is heat, m is mass, cp is specific heat capacity, and dT is change in temperature. From a reliable source, cp for aluminum is equal to 0.215 cal/g°C. Substituting this to the equation,
H = (260.5 g)(0.215 cal/g°C)(125°C - 0)
H = 7000.94 cal

6 0

3 years ago