英文字典中文字典


英文字典中文字典51ZiDian.com



中文字典辞典   英文字典 a   b   c   d   e   f   g   h   i   j   k   l   m   n   o   p   q   r   s   t   u   v   w   x   y   z       







请输入英文单字，中文词皆可：


请选择你想看的字典辞典：
单词字典翻译
hopen查看　hopen　在百度字典中的解释百度英翻中〔查看〕
hopen查看　hopen　在Google字典中的解释Google英翻中〔查看〕
hopen查看　hopen　在Yahoo字典中的解释Yahoo英翻中〔查看〕

安装中文字典英文字典查询工具!


中文字典英文字典工具:
选择颜色:
输入中英文单字

































































英文字典中文字典相关资料:


  • 0. 6 mL of acetic acid (CH3COOH) having density 1. 06 g mL is . . .
    0 6 mL of acetic acid (CH 3 COOH) having density 1 06 g mL is dissolved in 1 litre of water The depression in freezing point observed for this is 0 0205°C Calculate van’t Hoff factor and the dissociation constant for the acid
  • 0. 6 mL of acetic acid (CH(3)COOH) having density 1. 06 g mL . . .
    0 6mL of acetic acid (CH 3COOH) having density 1 06gmL−1 is dissolved in 1L of water The depression in freezing point observed for this strength of acid was 0 0205∘C Calculate the Van't Hoff factor and dissociation constant of the acid Kf for H 2O = 1 86Kkg−1mol−1) Step 1 Determining van't hoff factor
  • [Solved] 0. 6 mL of acetic acid (CH3 COOH), having density 1. . . .
    0 6mL of acetic acid (CH3COOH), having density 1 06gmL−1, is dissolved in 1 litre of water The depression in freezing point observed for this strength of acid was 0 0205∘C Calculate the van't Hoff factor and the dissociation constant of acid Not the question you're searching for?
  • 0. 6 mL of acetic acid $\\left( {{\\rm{C}}{{\\rm{H}}_{\\rm{3 . . .
    0 6 mL of acetic acid (C H 3 C O O H), having density 1 06 g m L − 1, is dissolved in 1 litre of water The depression in freezing point observed for this strength of acid was 0 0205 ∘ C Calculate the van’t Hoff factor and the dissociation constant of acid Hint: Here, first we have to calculate the molality of the solution
  • 0. 6 ml of acetic acid having density 1. 06 - Brainly. in
    To find the mass of acetic acid, we'll use its density and volume Then, with the molar mass of acetic acid, we can calculate its moles Given: - Volume of acetic acid (V) = 0 6 mL - Density of acetic acid = 1 06 g mL - Molar mass of acetic acid (CH3COOH) = 60 05 g mol First, let's find the mass of acetic acid: Mass = Volume * Density
  • 0. 6 mL of acetic acid (CH3COOH), having density 1. 06 g mL-1 . . .
    The depression in freezing point observed for this strength of acid was 0 0205°C The van’t Hoff factor is Q 0 6 m L of acetic acid ( C H 3 COO H ) , having density 1 06 g m L − 1 , is dissolved in 1 litre of water
  • 0. 6 mL of acetic acid (CH3COOH), having density 1. 06 g mL–1 . . .
    0 6 mL of acetic acid (CH3COOH), having density 1 06 g mL–1, is dissolved in 1 litre of water The depression in freezing point observed for this strength of acid was 0 0205°C The value of van’t Hoff factor and the dissociation constant of acid are respectively: 1 1 041; 1 86 × 10–52 2 073; 1 86 × 10–53 0 091; 1 76 × 10–54 1 041; 1 76 × 10–5





中文字典-英文字典  2005-2009