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anzhelika [568]
3 years ago
15

A body of mass 5kg has momentum of 125kgm/s. find the velocity of the body in motion.

Chemistry
1 answer:
borishaifa [10]3 years ago
8 0

Explanation:We have momemtum = mass X velocity

    p = mv

    OR, p/m = v

    v = (125kg m/s)25kg

    v = 125/25 m/s

       v = 5 m/s

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Answer is 2KClO3 3O2 + 2KCl
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A(n)<br> ______ Is a group of organs that work together to perform a com
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3 years ago
Rhodium-101 has a half life of 3.3 years. What percent of the sample will remain after 6.6 years?
babunello [35]

Answer:

25%

Explanation:

Half life means that 50% of the sample is gone at 3.3 years. This means that an additional 3.3 years (total 6.6 years) will reduce the sample a further 50% from the point at 3.3 years. In numbers, this means 50% of 50% (0.50*0.50), which is 25%.

5 0
3 years ago
Given that the nucleophilic substitution reaction used 5.0 mL of t-pentyl alcohol and 12.0 mL of conc. hydrochloric acid to prod
Vladimir [108]

Answer:

4.90 g

Explanation:

Given that:

volume of t-pentyl alcohol = 5 mL

the standard density of t-pentyl alcohol = 0.805 g/mL

Recall that:

density = mass(in wt) /volume

mass = density × volume

mass = 0.805 g/mL × 5 mL

mass = 4.03 g

Volume of HCl used = 12 mL

The reaction for this equation is shown in the image attached below.

From the reaction,

88.15 g of t-pentyl alcohol reacts with concentrated HCl to yield 106.59 g pf t-pentyl chloride.

4.03 g of t-pentyl alcohol forms,

= \dfrac{106.59 \ g \times 4.03 \ g}{88.15 \ g} of t-pentyl chloride.

Therefore,

Theoretical yield of t-pentyl chloride = 4.90 g

8 0
3 years ago
Answer the following for the reaction: 3AgNO3(aq)+Na3PO4(aq)→Ag3PO4(s)+3NaNO3(aq)
Brums [2.3K]

Answer:1) Volume of AgNO_3 required is 55.98 mL.

2) 0.62577 grams of Ag_3PO_4 is produced.

Explanation:

3AgNO_3(aq)+Na_3PO_4(aq)\rightarrow Ag_3PO_4(s)+3NaNO_3(aq)

1) Molarity of AgNO_3,M_1=0.225 M

Volume of AgNO_3.V_1=?

Molarity of Na_3PO_4,M_2=0.135 M

Volume of Na_3PO_4,V_2=31.1 mL=0.0311 L

Molarity=\frac{\text{number of moles}}{\text{volume of solution in liters}}

\text{number of moles }Na_3PO_4=M_2\times V_2=0.135 mol/L\times 0.0311 L=0.0041985 moles

According to reaction, 1 mole of Na_3PO_4 reacts with 3 mole of AgNO_3, then, 0.0041985 moles of Na_3PO_4 will react with:

\frac{3}{1}\times 0.0041985 moles of AgNO_3 that is 0.0125955 moles.

M_1=0.225 M=\frac{\text{number of moles of }AgNO_3}{V_1}

V_1=\frac{0.0125955 moles}{0.225 M}=0.05598 L=55.98 mL

Volume of AgNO_3 required is 55.98 mL.

2)

Molarity=0.195 M=\frac{\text{number of moles}}{\text{volume of solution in liters}}

Number of moles of AgNO_3=0.195\times 0.023 L=0.004485 moles

According to reaction, 3 moles of AgNO_3 gives 1 mole of Ag_3PO_4, then 0.004485 moles of AgNO_3 will give:\frac{1}{3}\times 0.004485 moles of Ag_3PO_4 that is 0.001495 moles.

Mass of Ag_3PO_4 =

Moles of Ag_3PO_4 × Molar Mass of Ag_3PO_4

= 0.001495 moles × 418.58 g/mol = 0.62577 g

0.62577 grams of Ag_3PO_4 is produced.

7 0
3 years ago
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