Answer:
im gunna save my answer so no one can take it:)
Explanation:
I tried my best
the prefix of the second word indicates that a molecule of carbon dioxide indicates that the compound contains two oxygen atoms
Answer:
3.8 M
Explanation:
Volume of acid used VA= 57.0 - 37.5 = 19.5 ml
Volume of base used VB= 67.8 - 45.0 = 22.8 ml
Equation of the reaction
2HNO3(aq) + Ca(OH)2(aq) --------> Ca(NO3)2(aq) + 2H2O(l)
Number of moles of acid NA= 2
Number of moles of base NB= 1
Concentration of acid CA= ???
Concentration of base CB= 1.63 M
CAVA/CBVB = NA/NB
CAVANB = CBVBNA
CA= CBVBNA/VANB
CA= 1.63 × 22.8 × 2/ 19.5 × 1
CA= 3.8 M
HENCE THE MOLARITY OF THE ACID IS 3.8 M.
Answer:
340 grams Ca₃P₂ (2 sig. figs.)
Explanation:
3Ca + 2P => Ca₃P₂
5.6 mole + excess => ? grams
Convert the 'known' to a coefficient of 1 by dividing all coefficients by 3.
=> Ca + 2/3P => 1/3Ca₃P₂
From the above, 1 mole of Ca => 1/3 mole Ca₃P₂
∴ 5.6 mole Ca in an excess of P => 1/3(5.6 mole) Ca₃P₂
=> 1.8666 mol Ca₃P₂ (calculator answer) ≅ 1.9 mol Ca₃P₂
=> 1.9 mole x 182 g Ca₃P₂/mol Ca₃P₂ = 339.73333 grams Ca₃P₂
≅ 340 grams Ca₃P₂ (2 sig. figs.)
Answer:
169.67Ω
Explanation:
This question is asking for the inductive reactance, which is calculated as follows:
X(L) = 2πfL
Where;
X(L) = inductive reactance (Ω)
f = frequency (Hz)
π = 3.142
L = inductance (Henry)
Given the information provided, f = 60Hz, L = 0.450H
X(L) = 2πfL
X(L) = 2 × 3.142 × 60 × 0.450
X(L) = 6.284 × 60 × 0.450
X(L) = 6.284 × 27
X(L) = 169.668
X(L) = 169.67Ω
