a) 4Na + O₂ --t^o--> 2Na₂O

b) \(n_{Na_2O}=\dfrac{24,8}{62}=0,4\left(mol\right)\)

4Na + O₂ --t^o--> 2Na₂O

0,8<--0,2<-------0,4

=> m_Na = 0,8.23 = 18,4 (g)

c) m_O2 = 0,2.32 = 6,4 (g)

V_O2 = 0,2.24,79 = 4,958(l)

Câu 1.

\(M=\left(-\dfrac{2a^3b^2}{3}xy^2z\right)^3.\left(-\dfrac{3}{4}ab^{-3}x^2yz^2\right)^2.\left(-xy^2z^2\right)^2\)

\(=\left(-\dfrac{8}{27}a^9b^6x^3y^6z^3\right).\left(\dfrac{9}{16}a^2b^{-6}x^4y^2z^4\right).\left(x^2y^4z^4\right)\)

\(=-\dfrac{8}{27}.\dfrac{9}{16}.a^{11}x^9y^{12}z^{11}\)

\(=-\dfrac{1}{6}a^{11}x^9y^{12}z^{11}\)

Hệ số: \(-\dfrac{1}{6}\)

Bậc: \(43\)

Câu 2.

a) \(A\left(x\right)=\dfrac{1}{2}x^5+\dfrac{3}{4}x-12x^4-1\dfrac{2}{3}x^3+5+x^2+\dfrac{5}{3}x^3-\dfrac{11}{4}x+1\dfrac{1}{2}x^5+4x\)

\(=\left(\dfrac{1}{2}x^5+\dfrac{3}{2}x^5\right)+\left(-12x^4\right)+\left(-\dfrac{5}{3}x^3+\dfrac{5}{3}x^3\right)+x^2+\left(\dfrac{3}{4}x-\dfrac{11}{4}x+4x\right)+5\)

\(=2x^5-12x^4+x^2+2x+5\)

\(B\left(x\right)=-2x^5+\dfrac{3}{7}x+12x^4-\dfrac{7}{3}x^3-3-6x^2+\dfrac{13}{3}x^3+3\dfrac{4}{7}x\)

\(=\left(-2x^5\right)+12x^4+\left(-\dfrac{7}{3}x^3+\dfrac{13}{3}x^3\right)-6x^2+\left(\dfrac{3}{7}x+\dfrac{25}{7}x\right)-3\)

\(=-2x^5+12x^4+2x^3-6x^2+4x-3\)

b) \(C\left(x\right)=A\left(x\right)+B\left(x\right)=\left(2x^5-12x^4+x^2+2x+5\right)+\left(-2x^5+12x^4+2x^3-6x^2+4x-3\right)\)

\(=\left(2x^5-2x^5\right)+\left(-12x^4+12x^4\right)+2x^3+\left(x^2-6x^2\right)+\left(2x+4x\right)+\left(5-3\right)\)

\(=2x^3-5x^2+6x+2\)

\(D\left(x\right)=A\left(x\right)-B\left(x\right)=\left(2x^5-12x^4+x^2+2x+5\right)-\left(-2x^5+12x^4+2x^3-6x^2+4x-3\right)\)

\(=\left(2x^5+2x^5\right)+\left(-12x^4-12x^4\right)-2x^3+\left(x^2+6x^2\right)+\left(2x-4x\right)+\left(5+3\right)\)

\(=4x^5-24x^4-2x^3+7x^2-2x+8\)

c) \(2x^3-5x^2+6x+2-2x^3+5x^2=-4\)

\(\Rightarrow\left(2x^3-2x^3\right)+\left(-5x^2+5x^2\right)+6x+2\)

\(\Rightarrow6x+2=-4\)

\(\Rightarrow6x=-6\)

\(\Rightarrow x=-1\)

Câu 3.

1) \(M-3xy^2+2xy-x^3+2x^2y=2xy-3x^3+3x^2y-xy^2\)

\(\Rightarrow M=\left(3xy^2+2xy-x^3+2x^2y\right)+\left(2xy-3x^3+3x^2y-xy^2\right)\)

\(=\left(3xy^2-xy^2\right)+\left(2xy+2xy\right)+\left(-x^3-3x^3\right)+\left(2x^2y+3x^2y\right)\)

\(=2xy^2+4xy-4x^3+5x^2y\)

2)

Để cho \(f\left(x\right)\) có nghiệm thì \(6-3x=0\)

\(\Rightarrow3x=6\)

\(\Rightarrow x=2\)

Để cho \(g\left(x\right)\) có nghiệm thì \(x^2-1=0\)

\(\Rightarrow x^2=1\)

\(\Rightarrow\left[{}\begin{matrix}x=1\\x=-1\end{matrix}\right.\)

6.

\(0,3a^3b^2\sqrt{\dfrac{9}{a^4b^8}}=0,3a^3b^2.\dfrac{3}{a^2b^4}=\dfrac{0,9.a}{b^2}\)

Đáp án B

7.

\(-\dfrac{1}{3}ab^3\sqrt{\dfrac{9a^2}{b^6}}=-\dfrac{1}{3}ab^3.\dfrac{3\left|a\right|}{\left|b^3\right|}=-ab^3.\dfrac{-a}{b^3}=a^2\)

Đáp án C

1 nối với C

2 nối với D

3 nối với A

4 nối với B

1 nối với C

2 nối với D

3 nối với A

4 nối với B

III)

36:A

37:B

38:B

39:A(mk ko chắc lắm)

40:B

IV)

1:It is about 100 km from Hanoi to Thanh hoa

2:How long is it from Hai duong to Hue

3:My father used to live in a small village when he was a child

4:Although he is young,he performs excellently

5:There are five rooms in Hoa's house

6:The red car is cheaper than the black car

7:Mr hung has fewer days off than Mr Minh

8:Let's play chess tonight

9:The children enjoy reading comics

100 - y = 53 + 7

100 - y = 60

         y = 100 - 60 

         y = 40

100-y=53+7

100-y=60

       y=100-60

       y=40 

Lần sau gặp bài toán này bạn tìm tổng rùi hẵng tìm số trừ

Của bạn đây 100 - y =53 + 7

                        y + 100 = 60

                         y.          =100 - 60 

                         y.            = 40 

m _Na =\(\dfrac{4,6}{23}\)=0,2 mol

2Na+Cl₂->2NaCl

0,2---0,1-----0,2

=>V_Cl2=0,1.22,4=2,24l

=>mNaCl=0,2.58,5=11,7g