1. How many grams of NaCl required to prepare each of the following solutions:
1. 1. How many grams of NaCl required to prepare each of the following solutions:
Penjelasan:
Sorry sir, where is the question and the answer choices?
2. 5. The solutions of |6-3x| = 2x + 7 is
Penjelasan dengan langkah-langkah:
good luck with the answer.
3. how many grams of NaC1 required to prepare each of the following solutions2500 ppm (w/v) NaC1 250 mL solutions
Penjelasan:
Formula : m NaCl (in Molar) = n ppm × mL
= 2500 ppm = masses of Salt / mL
= masses of Salt (µg) = 2500 ppm × mL
= massea of salt (µg) = 2500 ppm × 250 mL
= masses of salt (µg) = 625 µg or in mg = 625 µg × 10^-4 = 0,0625 grams
4. Solve the following differential equations by series and also by an elementary method and verify that your solutions agree. note that the goal of these problems is not to get the answer (that's easy by computer or by hand) but to become familiar with the method of series solutions which we will be using later. check your results by computer. 1. xy'=xy+y
[tex]\bold{\underline{\bigstar~Solution:}}\\\\problem:differential\ equations\\\\xy' = xy + y\\xy' = y(x + 1)~~~~~~~[$masing-masing ruas dibagi x, sehingga$]\displaystyle\\\\~~~y' = \frac{y(x+1)}{x}\\\\~~\frac{dy}{dx} = \frac{y(x+1)}{x}\\\\~~\frac{dy}{y} = \frac{(x+1)dx}{x}[/tex]
[tex]\displaystyle\\\\~~ \int\frac{dy}{y} = \int\frac{(x+1)dx}{x}\\\\~~ \int\frac{dy}{y} = \int(1+\frac{1}{x})dx}\\\\~~ \int\frac{dy}{y} = \int(1+x^{-1})dx\\\\~~ \int\frac{dy}{y} = \int\ dx + \int\ (x^{-1})dx\\\\~~~\ln(y) = x + \ln(x) + C\\\\~~~\ln(y) = \ln (e^x)+\ln(x) + \ln(C)\\\\~~~\ln(y) = \ln ($Cxe^x$)[/tex]
[tex]\\\\~~~~\boxed{y = ~$Cxe^x}[/tex]
[tex]========================================\\\bold{Note:}\\remember\ this\ rule:\\e^y=x~~~~and\ than\ base\ e\ logarithm\ of\ x\ is:\\\boxed{\ln(x) = \log_{e}(x) = y}\\\\For\ x>0\ and\ than:\\\boxed{f(f^{-1}(x))=e^{\ln(x)}=x}~~~~or~~~~\boxed{f^{-1}(f(x))=\ln(e^x)=x}[/tex]
5. 20. The number of integer solutions x and y satisfying x + y = 7 is ....
Jawaban:
0+7
1+6
2 +5
3+4
4+3
5+2
7+0
6. solve the following differential equations by series and also by an elementary method and verify that your solutions agree. note that the goal of these problems is not to get the answer (that's easy by computer or by hand) but to become familiar with the method of series solutions which we will be using later. check your results by computer. 1. xy'=xy+y
[tex]\bold{\underline{\bigstar~Solution:}}\\\\problem:differential\ equations\\\\xy' = xy + y\\xy' = y(x + 1)~~~~~~~[$masing-masing ruas dibagi x, sehingga$]\displaystyle\\\\~~~y' = \frac{y(x+1)}{x}\\\\~~\frac{dy}{dx} = \frac{y(x+1)}{x}\\\\~~\frac{dy}{y} = \frac{(x+1)dx}{x}[/tex]
[tex]\displaystyle\\\\~~ \int\frac{dy}{y} = \int\frac{(x+1)dx}{x}\\\\~~ \int\frac{dy}{y} = \int(1+\frac{1}{x})dx}\\\\~~ \int\frac{dy}{y} = \int(1+x^{-1})dx\\\\~~ \int\frac{dy}{y} = \int\ dx + \int\ (x^{-1})dx\\\\~~~\ln(y) = x + \ln(x) + C\\\\~~~\ln(y) = \ln (e^x)+\ln(x) + \ln(C)\\\\~~~\ln(y) = \ln ($Cxe^x$)[/tex]
[tex]\\\\~~~~\boxed{y = ~$Cxe^x}[/tex]
[tex]========================================\\\bold{Note:}\\remember\ this\ rule:\\e^y=x~~~~and\ than\ base\ e\ logarithm\ of\ x\ is:\\\boxed{\ln(x) = \log_{e}(x) = y}\\\\For\ x>0\ and\ than:\\\boxed{f(f^{-1}(x))=e^{\ln(x)}=x}~~~~or~~~~\boxed{f^{-1}(f(x))=\ln(e^x)=x}[/tex]
[tex]x \frac{dy}{dx} = y(x + 1) \\ \frac{dy}{dx} = \frac{y(x + 1)}{x} \\ \frac{dy}{y} = \frac{dx(x+1)}{x} \\ $ integralkan kedua ruas \\ \int\limits {\frac{dy}{y}} = \int\limits {\frac{(x+1)}{x}} \, dx \\ ln (y) = x + ln (x) + c \\ y = e^{x + ln (x) + c}[/tex]
7. (i) If x = 5 is a solution of the equation x^2 - qx + 10 = 0, find the value of q.(ii )Hense, find the other solutions of the equation.(i) q =(ii) x =
Penjelasan dengan langkah-langkah:
x²-qx+10=0
5²-5q+10=0
25+10=5q
35 = 5q
q = 7
x²-7x+10 = (x-5)(x-2)
x=5 atau x=2
8. The strength of electrolyte solutions that do not match the data is..
Jawaban:
•C.C12H22O11 < CH3COOH
Penjelasan:
semoga membantu
9. Let p; q be the solutions of the quadratic equation x^2 - 2mx - 5n = 0 and let m; n be the solutions of x^2 - 2px - 5q = 0. Then the value of p + q + m + n is ....
p + q = -b/a = -(-2m)/1 = 2m
m+n = -b/a = -(-2p)/1 = 2p
p + q + m + n = 2m + 2p = 2(m + p)
10. The number of solutions for the following equation: |x+1|+|x—2| = 4 is .... (A) 6 (B) 5 (C) 4 (D) 3 (E) 2
|x + 1| + |x - 2| = 4
Ix + 1 + x - 2 = 4
2x - 1 = 4
2x = 5
x = 5/2
IIx + 1 + x - 2 = -4
2x - 1 = -4
2x = -3
x = -3/2
11. the number of integer solutions x and y satisfying x^2+y^2=7 is....
tidak ada karena bilangan kuadrat jika dibagi 4 sisa 0 atau 1. jadi jumlah 2 kuadrat selalu sisa 0,1,2 jika dibagi 4. tapi 7 bagi 4 sisa 3
12. 1. The Henderson-Hasselbach formula is commonly used to approximate the pH value of a buffer solution. [H,AⓇ) pH = pK, - log [1,-148-1) Where the constant 8 is the charge of the acid H,A. (i) Derive the Henderson-Hasselbach equation. We have defined the pH value of an optimum acid buffer condition which is equal to the pK, of the acid. Some textbooks and teacher state that to obtain this optimum acid buffer, one can use one to one ratio between the acid and the salt that contain the conjugate anion of that acid. (ii) Justify whether the statement regarding the one to one ratio of acid and conjugate base is true for all case. (iii) Show that the actual ratio between the concentration of acid and the concentration of the conjugate ion in the optimum acid buffer condition is, [H, AⓇ] 2K, +1 [H.-148-1] [H.,-, A8-1] (iv) Hence, or otherwise, show that for a very dilute system Henderson-Hasselbach equation may not work well to approximate the pH value of the buffer solution.
Jawaban:
aku ff asup
akuanak ff kuat asayuku
13. Suppose (x^2 + x − 1)^x−2 = 1 , the number of solutions of the equation is .... (A) 1 (B) 2 (C) 3 (D) 4 (E) 5
(x²+x-1)^(x-2)=1
(x²+x-1)^(x-2)=(x²+x-1)^0
x-2=0
x=2
(B)2
14. afua mixes a solution of potassium iodide with a solution of lead nitrate.birth of the solutions are clear and colourless.the resulting mixture is cloudy and yellow.the yellow precipitate is lead iodide,one of the products on this reaction.what is the name of the other products
Jawaban:
Potassium nitrate KNO[tex]_{3}[/tex]
Penjelasan:
2KI + Pb(NO[tex]_{3}[/tex])[tex]_{2}[/tex] ------> 2KNO[tex]_{3}[/tex] + PbI[tex]_{2}[/tex]
Jawaban:
nowiwohuqlbbhuwongaonvvg
15. Psd pacific digital solutions limited is an international technology company which
Solusi digital pasifik PSD terbatas adalah perusahaan teknologi internasional yang
Penjelasan:
kalau di suruh translate itu ya
•MAAF KALAU ADA YANG SALAH•
Jawaban:
Translate Indonesia : Solusi digital pasifik PSD terbatas adalah perusahaan teknologi internasional yang
answers to questions:
Psd pacific digital solutions limited is an international technology company which provide software, training and support service.
Semoga bermanfaat^^
16. were there any solutions at the end of the story
of course there is a mandate from every story, not just films but lots, like fables, comics, books that have stories, poems, songs and others
this is my answer :)
17. the number of integer solutions x and y satisfying x^2+y^2=7 is .....
i dont under stand what u mean lol
18. Solution which is included in strong electrolyte solutions is
#larutan elektrokit
#X semester 2
yang termasuk dalam larutan elektrolit kuat :
H2SO4
HCl
HNO3
BaCl2
NaCl
NaOH
Ba(OH)2
KOH
(basa kuat, asam kuat atau garam)
@kimiaischemistry
19. 3x + 4y = -22y - x = -19 what is the solutions ( x , y ) to the system of equations above?
Jawab:
(x,y)=(3,-8)
Penjelasan dengan langkah-langkah:
Consider the equations
3x + 4y=-23….(1) and 2y - x =-19….(2)
3*(2)=> 6y - 3x = -57
(1) => 4y + 3x =-23
Adding the above equations we get
(6y - 3x) + (4y + 3x)=-57–23
10y = -80
y=-80/10
y= -8
Put the value of y in equation (2) we get
2(-8)-x =-19
-16 - x =-19
-x =-19+16
-x =-3
x= 3
So the solution is (x,y)=(3,-8)
========================================may be usefulglad to help youkeep the spirit[tex]by:asya291006[/tex]
20. were there any solutions at the end of the story
solutions , u got it when the problems get down, and after tension problem downed
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