\(D=\dfrac{m}{V}=\dfrac{4800}{\left(10.20.30\right).10^{-3}}=800\left(\dfrac{kg}{m^3}\right)\)

1.D

2.A 

3.D

4.A

5.C

6.D

7.D

vẫn ra 58 mà bạn

36.24 bằng 864 mà chị

Chiều dài : 

\(\left(120:2\right):2,5\times1,5=36\left(m\right)\)

Chiều rộng :

\(60-36=24\left(m\right)\)

Diện tích  :

\(36\times24=864\left(m^2\right)\)

Diện tích trồng hoa :

\(864\times\dfrac{1}{5}=172,8\left(m^2\right)\)

Chiều rộng HCN là :

\(\dfrac{10}{21}:\dfrac{5}{7}=\dfrac{2}{3}\left(m\right)\)

\(n_{Fe}=\dfrac{5,6}{56}=0,1\left(mol\right)\)

\(n_{HCl}=0,5.1=0,5\left(mol\right)\)

PTHH :

\(Fr+2HCl\rightarrow FeCl_2+H_2\uparrow\)

0,1     0,2           0,1        0,1 

\(\dfrac{0,1}{1}< \dfrac{0,5}{2}\)

--> Tính theo Fe 

HCl dư 

\(a,m_{HCldư}=\left[0,5.\left(0,1.2\right)\right].36,5=10,95\left(g\right)\)

\(b,m_{FeCl_2}=0,1.127=12,7\left(g\right)\)

\(c,V_{H_2}=0,1.22,4=2,24\left(l\right)\)

\(d,m_{HCl}=0,5.36,5=18,25\left(g\right)\)

\(C\%_{HCl}=\dfrac{18,25}{200}.100\%=9,125\%\)

\(B=-\left(1+\dfrac{1}{3}+\dfrac{1}{6}+\dfrac{1}{10}+\dfrac{1}{15}+...+\dfrac{1}{1225}\right)\)

\(\dfrac{1}{2}B=-\left(\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}+...+\dfrac{1}{2450}\right)\)

\(\dfrac{1}{2}B=-\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{2.4}+\dfrac{1}{4.5}+...+\dfrac{1}{49.50}\right)\)

\(\dfrac{1}{2}B=-\left(1-\dfrac{1}{50}\right)\)

\(\dfrac{1}{2}B=-1+\dfrac{1}{50}\)

\(\dfrac{1}{2}B=\dfrac{-49}{50}\)

\(B=\dfrac{-49}{25}\)

\(n_{HCl}=\dfrac{200.7,3\%}{36,5}=0,4\left(mol\right)\)

PTHH :

\(Zn+2HCl\rightarrow ZnCl_2+H_2\uparrow\)

0,2      0,4          0,2        0,2

\(a,m_1=0,2.65=13\left(g\right)\)

\(m_2=0,2.136=27,2\left(g\right)\)

\(V_{H_2}=0,2.22,4=4,48\left(l\right)\)

\(b,m_{ddZnCl_2}=13+200-\left(0,2.2\right)=212,6\left(g\right)\)

\(C\%_{ZnCl_2}=\dfrac{27,2}{212,6}.100\%\approx12,79\%\)