Answer:
4.8 g H₂O
Explanation:
To find the mass of water, you need to (1) convert grams B₂H₆ to moles B₂H₆ (via molar mass from periodic table), then (2) convert moles B₂H₆ to moles H₂O (via mole-to-mole ratio from reaction coefficients), and then (3) convert moles H₂O to grams H₂O (via molar mass from periodic table).
It is important to arrange the ratios/conversions in a way that allows for the cancellation of units (the desired unit should be in the numerator). The final answer should have 2 sig figs because the given value (3.7 grams) has 2 sig figs.
Molar Mass (B₂H₆): 2(10.811 g/mol) + 6(1.008 g/mol)
Molar Mass (B₂H₆): 27.67 g/mol
1 B₂H₆ + 3 O₂ ---> 2 HBO₂ + 2 H₂O
^ ^
Molar Mass (H₂O): 15.998 g/mol + 2(1.008 g/mol)
Molar Mass (H₂O): 18.014 g/mol
3.7 g B₂H₆ 1 mole 2 moles H₂O 18.014 g
---------------- x --------------- x ----------------------- x ----------------- = 4.8 g H₂O
27.67 g 1 mole B₂H₆ 1 mole
Answer: a. A worm can move so it must be living.
Explanation:
If a worm is moving it is classified as a moving worm also known to be living if moving.
A worm could reproduce therefore it is living.
Worms do pass dirt into and through there bodies to eat of the minerals and discarding things through and let out.
If you do cut a worm in half it may damage it but since it has basically two heads on girl one boy and cut in half it may split them up, and make damage but they will eventually grow back if the injuries are not severe.
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Answer : The chemical formula for the compound is, 
Explanation :
When the element 'M' react with the 
to give .
The balanced chemical reaction is,
In this reaction, 'M' is in mono-atomic form and is in diatomic form.
By the stoichiometry,
2 moles 'M' react with the 1 mole of to give 2 moles of .
Therefore, the chemical formula of the compound is,
this is a dilution equation where 50.0 mL of 1.50 M H₂SO₄ is taken and added to 200 mL of water.
c1v1 = c2v2
where c1 is concentration and v1 is volume of the concentrated solution
and c2 is concentration and v2 is volume of the diluted solution to be prepared
50.0 mL of 1.50 M H₂SO₄ is added to 200 mL of water so the final solution volume is - 200 + 50.0 = 250 mL
substituting these values in the formula
1.50 M x 50.0 mL = C x 250 mL
C = 0.300 M
concentration of the final solution is A) 0.300 M
