Bài 13:

nO2= 32/32=1(mol)

a) PTHH: 2 Mg + O2 -to-> 2 MgO

nMg=nMgO=nO2.2=2(mol)

=> mMg= 2.24=48(g)

b) mMgO=40.2=80(g)

Bài 14:

nCaCl2= 55,5/111= 0,5(mol)

a) PTHH: Ca +2 HCl ->  CaCl2 + H2

Ta có: nH2=nCa=0,5(mol); nHCl=2.0,5=1(mol)

=> mHCl=1.36,5=36,5(g)

mCa= 40.0,5=20(g)

b) V(H2,đktc)=0,5.22,4=11,2(l)

a, 12- (-4)= 16
b, 12- (-14)= 26
c, (-13)-(-5)=-8
d, (-2)-(-10)= 8

    -5 + (-37 - 45 + 51) - (-37 + 51)

=  -5  - 37 - 45 + 51 + 37 - 51

=  - (5 + 45) - (37 - 37) + (51 - 51)

= - 50 - 0 + 0

= - 50

  - (15 - 47 + 58) + (15 - 47 ) + 3

= - 15 + 47 - 58 + 15 - 47 + 3

= (-15 + 15) + (47 - 47) - (58 - 3)

= 0 + 0 - 55

= - 55

  (53 - 45 - 49) - (53 + 45 - 49)

= 53 -  45 - 49 - 53 - 45 + 49

= (53 - 53) + (-49 + 49) - (45 + 45)

= 0 + 0 - 90

= - 90 

    13 - 15 + 49 - 13 + 15 - 48

=  (13 - 13) + (-15 + 15) + (49 - 48)

= 0 + 0 + 1

=  1 

15+13 -15-13 +49-48

a)

13 x 81 + 13 x 19 

= 13 x ( 81 + 19 )

= 13 x 100 = 1300

b)

246 x 83 + 246 x 17

= 246 x ( 83 + 17 )

= 246 x 100

= 24600

c)

246 x 83 + 247 x 17 

= 246 x 83 + 246 x 17 + 17

= 246 x ( 83 + 17 ) + 17

= 246 x 100 + 17 

= 24600 + 17

= 24617

          Ủng hộ mik nhak !!!

   13x81+13x19

=13x(19+81)

=13x100

=1300

  246x83+246x17

=246x(83+17)

=246x100

=24600

(81+19) x13=

100x13

=1300

cac cau con lai cung vay do

a) x + 45 -[ 90 + (-20) + 5 - (-45)] +3x

= x +45 -( 90 - 20 + 5 + 45) + 3x

= x +45 -120 +3x

= -75 +4x

b) x +(-294 +13 -2x) +(94 -13) +9x

= x -281 -2x +81 +9x

= -200 +8x

Là 3 phần 13 nhân 10 hay 3 phần 13 nhân 16 mới đúng ?

\(\dfrac{\left(3\cdot4\cdot2^{16}\right)^2}{11\cdot2^{13}\cdot4^{11}-16^9}=\dfrac{\left(3\cdot2^2\cdot2^{16}\right)^2}{11\cdot2^2\cdot\left(2^2\right)^{11}-\left(2^4\right)^9}\)

\(=\dfrac{3^2\cdot\left(2^2\right)^2\cdot\left(2^{16}\right)^2}{11\cdot2^2\cdot2^{22}-2^{36}}=\dfrac{3^2\cdot2^4\cdot2^{32}}{11\cdot2^{24}-2^{36}}\)

\(=\dfrac{3^2\cdot2^{34}}{11\cdot2^{24}-2^{36}}=\dfrac{3^2\cdot2^{24}\cdot2^{10}}{11\cdot2^{24}-2^{12}\cdot2^{24}}\)

\(=\dfrac{3^2\cdot2^{24}\cdot2^{10}}{\left(11-2^{12}\right)\cdot2^{24}}=\dfrac{3^2\cdot2^{10}}{11-2^{12}}\)

\(n_{C_2H_2}=\dfrac{13}{26}=0,5\left(mol\right)\\ PTHH:2C_2H_2+5O_2\rightarrow\left(t^o\right)4CO_2+2H_2O\\ CO_2+Ca\left(OH\right)_2\rightarrow CaCO_3\downarrow+H_2O\\ n_{O_2}=\dfrac{5}{2}.0,5=1,25\left(mol\right)\\ V_{O_2\left(đktc\right)}=1,25.22,4=28\left(l\right)\\ a,V_{kk}=\dfrac{100}{20}.V_{O_2\left(đktc\right)}=5.28=140\left(l\right)\\ b,n_{CaCO_3}=n_{CO_2}=\dfrac{4}{2}.0,5=1\left(mol\right)\\ m_{k.tủa}=m_{CaCO_3}=100.1=100\left(g\right)\)

\(a.C_2H_2+\dfrac{5}{2}O_2-t^{^{ }0}->2CO_2+H_2O\\ V_{KK}=\dfrac{\dfrac{13}{26}\cdot\dfrac{5}{2}\cdot22,4}{0,2}=140\left(L\right)\\ CO_2+Ca\left(OH\right)_2->CaCO_3+H_2O\\ m_{KT}=100\cdot0,5\cdot2=100g\)

  13/15 + 4/7 - 101/105

= 91/105 + 60/105 - 101/105

= 50/105 = 10/21

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