Chemistry Middle School

Need help please: How many mL of 0.150 Molarity CaBr2 solution must be used to react with 50 mL of 0.115 Molarity AgNO3?

CaBr2 + 2AgNO3 --> Ca(No3)2 + 2AgBr

Answers

Answer 1

Answer:

0.019 L or 19.1 mL

Explanation:

Given the equation of the reaction; we have:

CaBr2 + 2AgNO3 --> Ca(No3)2 + 2AgBr

We must first obtain the number of moles of AgNO3 reacted. This can be obtained from;

Concentration of AgNO3 reacted= 0.115 M

Volume of AgNO3 reacted= 50 ml

Since

Number of moles of AgNO3= concentration of AgNO3 × volume of AgNO3

Number of moles = 0.115 M × 50/1000 = 5.75 ×10^-3 moles

From the reaction equation;

2 moles of AgNO3 reacts with 1 mole of CaBr2

5.75 ×10^-3 moles of AgNO3 reacts with 5.75 ×10^-3 moles × 1 /2 = 2.875×10^-3 moles of CaBr2

Now;

Number of moles of CaBr2= concentration of CaBr2 × volume of CaBr2

Number of moles of CaBr2= 2.875×10^-3 moles

Volume of CaBr2= ????

Concentration of CaBr2= 0.150 M

Hence;

Volume of CaBr2= number of moles of CaBr2/ concentration of CaBr2

Volume of CaBr2= 2.875×10^-3 moles / 0.150

Volume of CaBr2= 0.019 L or 19.1 mL

Answer 2

Answer:

19 mL of CaBr₂ solution should be used to react with AgNO₃

Explanation:

By stoichiometry of the reaction (that is, the relationship between the amount of reagents and products in a chemical reaction) react:

AgNO₃: 2 molesCaBr₂: 1 mole

50 ml of 0.115 M AgNO₃ react. Since molarity is the number of moles of solute per liter of solution, then you can do the following rule of three: if there are 0.115 moles of AgNO₃ in 1 L, in 0.05 L (being 1 L = 1000 mL, then 50 mL = 0.05 L How many moles are there?

$moles of AgNO_{3}=\frac{0.05L*0.115 moles}{1 L}$

moles of AgNO₃=5.75*10⁻³

Now we can make a new rule of three to calculate the amount of moles necessary for CaBr₂ to react with moles of AgNO₂: if stoichiometry 2 moles of AgNO₃ react with 1 mole of CaBr₂, 5.75*10⁻³ moles of AgNO₃ with how many moles of CaBr₂ react ?

$moles of CaBr_{2} =\frac{5.75*10^{-3}moles of AgNO_{3}*1 mole of CaBr_{2} }{2moles of AgNO_{3}}$

moles of CaBr₂=2.875*10⁻³ moles

Being 0.150 the molarity of the CaBr₂ compound, you can apply the following rule of three: if by definition of molarity 0.150 moles are in 1 L, 2.875 * 10⁻³ moles in how much volume is it present?

$volume=\frac{2.875*10^{-3} moles*1L}{0.150 moles}$

volume=0.019 L

Being 1 L = 1000 mL, then 0.019 L = 19 mL

19 mL of CaBr₂ solution should be used to react with AgNO₃

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When the speed of the bottle is 2 m/s, the KE is kg m2/s2. When the speed of the bottle is 3 m/s, the KE is kg m2/s2. When the speed of the bottle is 4 m/s, the KE is kg m2/s2. When the speed of the bottle is 5 m/s, the KE is kg m2/s2. When the speed of the bottle is 6 m/s, the KE is kg m2/s2.
instructions
In this part of the experiment, you will be changing the speed of the bottle by dropping it from different heights. You will use the same mass, 0.250 kg, for each trial, so record this mass in Table B for each velocity. Then, calculate the expected kinetic energy (KE) at each velocity. Use the formula KE = mv2,
where m is the mass and v is the speed. Record your calculations in Table B of your Student Guide.

Answers

Answer:

Based on the given instructions, you need to calculate the expected kinetic energy (KE) at each velocity using the formula KE = mv^2, where m is the mass and v is the speed. The mass is given as 0.250 kg.

Table B: Calculation of Kinetic Energy (KE)

Speed (m/s)Mass (kg)KE (kg m^2/s^2)20.2500.50030.2500.75040.2501.00050.2501.25060.2501.500

To calculate the kinetic energy at each velocity, multiply the mass (0.250 kg) by the square of the velocity.

For example, when the speed of the bottle is 2 m/s: KE = (0.250 kg) * (2 m/s)^2 = 0.250 kg * 4 m^2/s^2 = 1.000 kg m^2/s^2

Perform similar calculations for the other velocities and record the results in Table B.

Answer:

Explanation:mass of the bottle in each case is M = 0.250 kg

now as per given speeds we can use the formula of kinetic energy to find it

1) when speed is 2 m/s