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2 years ago

Please help!!

Chemistry

1 answer:

xxTIMURxx [149]2 years ago

6 0

Answer:

In an experiment, a student transferred 4.50 mL of a liquid into a pre-weighed beaker (the weight of which was determined to be 35.986 g ).

Explanation:

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If 5 moles of P4 reacted with 22 moles Cl2 according to the above reaction, determine:

GaryK [48]

Solution:

$P_4+6Cl_2\rightarrow 4PCl_3$

a) By stoichoimetry If 6 moles of $Cl_2$ gives 4 mole of $PCl_3$ , then 22 moles of $Cl_2$ will give: $\frac{4}{6}\times 22=\frac{44}{3}=14.66$ moles of $PCl_3$

b) If 6 moles of $Cl_2$ reacts with one mole of $P_4$ , then 22 moles of $Cl_2$ will react : $\frac{1}{6}\times 22=\frac{11}{3}=3.66$ moles of $P_4$

Moles of $P_4$ left after the reaction = $5-3.66=\frac{4}{3}=1.34$ moles.

c) 0 Moles of $Cl_2$ , since it is present in less amount it will get completely consumed in the reaction. Hence, it is limiting reagent.

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3 years ago

How many moles of aluminum are needed to make 9 moles of molecular hydrogen? given the reaction: 2 al + 6 hcl → 2 alcl3 + 3h2 6

sergiy2304 [10]

1) Chemical equation

2Al + 6 HCl ---> 2Al Cl3 + 3 H2

2) molar ratios

2 mol Al : 3 moles H2

3) Proportion

2 mol Al / 3mol H2 = x / 9 mol H2

4) Solve for x

x = 9 mol H2 * 2 mol Al / 3 mol H2 = 6 mol Ag

Answer: 6 moles

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The unit of momentum is the product of the units of mass and velocity.

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What volume would be needed to prepare 375 mL of a .45 M CaCl2 using only a solution of 1.0 M CaCl2 and water?

Neporo4naja [7]

Answer:

168.75 ml

Explanation:

M1V1=M2V2

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