TT:m₂= 2kg ; t^o₁= 30^oC ; t^o₂= 100^oC;

m₁=250g =0,25kg ; c₁=880J/kgK ; c₂=4200J/kgK => Q=? J

GIAI

nhiet luong de am nong den 100 do:

Q₁= m₁.c₁(t^o₂- t^o₁)=15400J

nhiet luong de nuoc nong den 100do:

Q₂= m₂.c₂(t^o₂ - t^o₁) = 588000J

nhiet luong de nuoc soi:Q=Q₁+Q₂= 603400J

TT: m = 65kg ; h= 20m ; t = 120s

=> A=? J ; \(\rho=?W\)

GIAI

trong luong cua vat: P=10m= 650N

cong cua anh cong nhan bo qua ma sat: A=P.h= 13000J

cong suat cua anh cong nhan bo qua ma sat:

\(\rho=\dfrac{A}{t}=\dfrac{13000}{120}\approx108,33W\)

\(n_{CO_2}=\dfrac{88}{44}=2\left(mol\right)\)

C + O₂ \(\underrightarrow{t^o}\) CO₂

DE: 2 (Mol)

\(\Rightarrow n_C=2\left(mol\right)\)

\(m_C=2.12=24g\)

\(\%m_C=\dfrac{24}{36}.100\%\approx66,67\%\)

bn oi de co sai khong?

Cau 1: Các chất đc cấu tạo từ các hạt riêng biệt gọi là nguyên tử, phân tử

Cau 2: TT: m₁=300g =0,3kg

V= 1,5l = 0,0015m³ ;t^o₁= 20^oC ;t^o₂= 100^oC

c₁=380J/kgK ;c₂= 4200J/kgK;D=1000kg/m³

=> Q =? J

GIAI

khoi luong cua nuoc:\(D=\dfrac{m}{V}\Rightarrow m=D.V=1,5kg\)

nhiet luong de nuoc nong den 100^oC:

\(Q_2=m.c_2\left(t^o_2-t^o_1\right)=504000J\)

nhiet luong de am nong den 100^oC:

\(Q_1=m_1.c_1\left(t^o_2-t^o_1\right)=9120J\)

nhiet luong de dun soi am nuoc :

\(Q=Q_1+Q_2=504000+9120=513120J\)

Cau D

Oxit axit:

+ SO₃: luu huynh trioxit

+ CO₂: cacbon dioxit

Oxit bazo:

+ Fe₂O₃: sat(III) oxit

+ Al₂O₃: nhom oxit

Axit

+ HCl: axit clohidric

+ H₂S: axit sunfu hidric

+ H₃PO₄: axit photphoric

+ H₂SO_4: axit sunfuric

Muoi:

+ NaCl: natri clorua

+ FeCl₃: sat(III) clorua

+ Al₂(SO₄)₃: nhom sunfat

+ NaHCO₃: natri hidro cacbonat

+ KHCO₃: kali hidro cacbonat

Bazo:

+ Ba(OH)₂: bari hidroxit

+ Fe(OH)₃: sat(III) hidroxit

+ CuOH: dong(I) hidroxit

TT:m₁= 50g= 0,05kg ; m₂= 100g= 0,1kg

t^o₁=100^oC ; t^o₂=20^oC ; t^o= 60^oC

c₁=130 J/kgK;c₂=380 J/kgK;c₃=4200J/kgK

=> m₃=? kg

GIAI

nhiet luong chi toa ra:\(Q_1=m_1.c_1\left(t^o_1-t^o\right)=260J\)

nhiet luong dong toa ra:

\(Q_2=m_2.c_2\left(t^o_1-t^o\right)=1520J\)

nhiet luong nuoc thu vao:

\(Q_3=m_3.c_3\left(t^o-t^o_2\right)=168000m_3\)

nhiet luong mieng dong chi toa ra:

Q=Q₁+Q₂ = 260+1520= 1780J

theo PTCBN: Q = Q₃

\(\Rightarrow1780=168000m_3\)

\(\Rightarrow m_3\approx0,01kg\)

TT:m₁= 4,3kg ; t^o₁= 27^oC ; c₁=880J/kgK

m₂= 1,5kg ; t^o= 32^oC ; c₂= 4200J/kgK

=> a, Q₁=? b, t^o₂=?

GIAI

a, nhiet luong thu vao cua qua cau:

Q₁=m₁.c₁.( t^o- t^o₁)= 18920J

b, nhiet luong toa ra cua nuoc :

Q₂=m₂.c₂.( t^o₂- t^o)= 6300t^o₂- 201600 J

theo PTCBN => Q₁= Q₂

_{\(\Rightarrow6300t^o_2-201600=18920\)}

\(\Rightarrow6300t^o_2=220520\Rightarrow t^o_2\approx35^oC\)

goi a la HT cua X

2X + 2aHCl \(\rightarrow\) 2XCl_a + aH₂

pt: 2X(g) 2X+71a (g)

de: 20g 55,5g

ta co: 111X = 40X + 1420a

\(\Rightarrow\) 71X = 1420a \(\Rightarrow X=20a\)

bien luan:

* a= 1 \(\Rightarrow X=20\left(loai\right)\)

* \(a=2\Rightarrow X=40\left(lay\right)\)

* \(a=3\Rightarrow X=60\left(loai\right)\)

\(\Rightarrow\) X la Ca

\(n_{Ca}=\dfrac{20}{40}=0,5\left(mol\right)\)

Ca + 2HCl \(\rightarrow\) CaCl₂ + H₂

de: 0,5 \(\rightarrow\) 0,5 (mol)

\(V_{H_2}=22,4.0,5=11,2\left(l\right)\)