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2 years ago

5

1 or 2 topics or two lessons should be explained in an illustrated childrens book minimum of 10 pages must have 3 or more senten

ces

1 answer:

marysya [2.9K]2 years ago

8 0

Answer:

Yes because same topic are long

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Find the linear function Cequalsf(F) that gives the reading on the Celsius temperature scale corresponding to a reading on the

padilas [110]

Answer:

C = (5/9) F - (160/9)

They both read equal at Z = - 40

Explanation:

We are looking for a linear function so we can write the following condition

Y = aX + b

Applying it to the exercise we got C = a F + b

Let's use the facts that C = 0 when F = 32 and C = 100 when F = 212

0 = 32 a + b (1)

100 = 212 a + b (2)

From (1) b = - 32 a , when we replace this in (2) we obtain a = (5/9)

and b = - (5/9)32 = - 160/9

Finally the linear function is C = (5/9) F - (160/9)

Both readings are equal at a Z number so

Z = (5/9) Z - 160/9

(4/9) Z = -160/9 and Z = - 40

6 0

3 years ago

Lastly, Snape thinks we should try one more calculation. What is the retention factor if the distance traveled by the solvent fr

Neporo4naja [7]

Answer:

0.1 is the retention factor.

Explanation:

Distance covered by solvent , $d_s= 2.0 cm$

Distance covered by solute or ion, $d = 0.20 cm$

Retention factor $(R_f)$ is defined as ratio of distance traveled by solute to the distance traveled by solvent.

$R_f=\frac{d}{d_s}$

$R_f=\frac{0.20 cm}{2.0 cm}=0.1$

0.1 is the retention factor.

7 0

3 years ago

Zn + O2= ZnO<br><br> How many moles of zinc are needed to make 500. g of zinc oxide?

s2008m [1.1K]

Answer:

5.15 moles

Explanation:

2zn + o2 = 2zno

5.15 2.57 5.15 moles

nzno=500/(16x2+65)= 5.15 moles

-> nzn = 5.15 x 2 ÷ 2 = 5.15 moles

6 0

2 years ago

What pressure, in atm, would be exerted by 0.023 grams of oxygen (O2) if it occupies 31.6 mL at 91

Vikentia [17]

Answer: A pressure of 0.681 atm would be exerted by 0.023 grams of oxygen $(O_2)$ if it occupies 31.6 mL at $91^{o}C$ .

Explanation:

Given : Mass of oxygen = 0.023 g

Volume = 31.6 mL

Convert mL into L as follows.

$1 mL = 0.001 L\\31.6 mL = 31.6 mL \times \frac{0.001 L}{1 mL}\\= 0.0316 L$

Temperature = $91^{o}C = (91 + 273) K = 364 K$

As molar mass of $O_2$ is 32 g/mol. Hence, the number of moles of $O_2$ are calculated as follows.

$No. of moles = \frac{mass}{molar mass}\\= \frac{0.023 g}{32 g/mol}\\= 0.00072 mol$

Using the ideal gas equation calculate the pressure exerted by given gas as follows.

PV = nRT

where,

P = pressure

V = volume

n = number of moles

R = gas constant = 0.0821 L atm/mol K

T = temperature

Substitute the value into above formula as follows.

$PV = nRT\\P \times 0.0316 L = 0.00072 mol \times 0.0821 L atm/mol K \times 364 K\\P = \frac{0.00072 mol \times 0.0821 L atm/mol K \times 364 K}{0.0316 L}\\= 0.681 atm$

Thus, we can conclude that a pressure of 0.681 atm would be exerted by 0.023 grams of oxygen $(O_2)$ if it occupies 31.6 mL at $91^{o}C$ .

4 0

3 years ago

Is a burning log an exothermic or endothermic event if the log is the system?

ziro4ka [17]

Yes it is a exothermic reaction.

7 0

3 years ago

Read 2 more answers