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maksim [4K]
3 years ago
14

Can someone PLEASE help me with this im struggling bad

Chemistry
1 answer:
Art [367]3 years ago
5 0

Answer:

  1. 4.5+2.34= 6.84
  2. 4.5-5 =-0.5
  3. 6.00+3.411= 9.411
  4. 3.4×2.32 = 7.888
  5. 7.77/2.3= 3.37

7. 1200×23.4=28080

9. = 78.512

10. =341.199

11= 7.45

12 =65.0023

13.=3400210.34

Explanation:

please mark me as branliest

hope it is helpful

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4 0
4 years ago
What pressure is exerted by 932.3 g of CH4 in a 0.560 L steel container at 136.2 K?
Roman55 [17]
Molar mass CH4 = 16.0 g/mol

* number of moles:

932.3 / 16 => 58.26875 moles

T = 136.2 K

V = 0.560 L

P = ?

R = 0.082

Use the clapeyron equation:

P x V = n x R x T

P x 0.560 =  58.26875 x 0.082 x 136.2

P x 0.560 = 650.76

P = 650.76 / 0.560

P = 1162.07 atm
5 0
4 years ago
While 1 gram of fat provides 9 calories, 1 gram of glucose provides 4 calories. Why is that?
seropon [69]

Answer:

fat always has more calories than glucose does

Explanation:

hope hope this helps!

6 0
3 years ago
How many half-lives would something have gone through if you had 75% daughter product and 25% parent?
levacccp [35]

Answer:

Two Half-lives

Explanation:

Let number of Parent nuclei Initially present be X,

Then, finally \frac{X}{4} Parent nuclei Will remain with \frac{3X}{4} daughter nuclei.

In one half- life , parent nuclei becomes half of initial.

So, starting with X parent nuclei,

After one half-life, it will degrade to \frac{X}{2} .

After another half life , Parent nuclei will become half of \frac{X}{2}

Which is equal to \frac{X}{4}.

So, Parent nuclei have to go through Two half-lives.

6 0
4 years ago
4. Find the pH at each of the following points in the titration of 25 mL of 0.3 M HF with 0.3 M NaOH. The Ka value is 6.6x10-4 a
yawa3891 [41]

Explanation:

Since HF is a weak acid, the use of an ICE table is required to find the pH. The question gives us the concentration of the HF.

HF+H2O⇌H3O++F−HF+H2O⇌H3O++F−

Initial0.3 M-0 M0 MChange- X-+ X+XEquilibrium0.3 - X-X MX M

Writing the information from the ICE Table in Equation form yields

6.6×10−4=x20.3−x6.6×10−4=x20.3−x

Manipulating the equation to get everything on one side yields

0=x2+6.6×10−4x−1.98×10−40=x2+6.6×10−4x−1.98×10−4

Now this information is plugged into the quadratic formula to give

x=−6.6×10−4±(6.6×10−4)2−4(1)(−1.98×10−4)−−−−−−−−−−−−−−−−−−−−−−−−−−−−√2x=−6.6×10−4±(6.6×10−4)2−4(1)(−1.98×10−4)2

The quadratic formula yields that x=0.013745 and x=-0.014405

However we can rule out x=-0.014405 because there cannot be negative concentrations. Therefore to get the pH we plug the concentration of H3O+ into the equation pH=-log(0.013745) and get pH=1.86

6 0
3 years ago
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