All categories

2 years ago

Force exerted on a lever is applied on the lever's

1 answer:

7 0

A force called the effort force is applied at one point on the lever in order to move an object, known as the resistance force, located at some other point on the lever

You might be interested in

A calorimeter contains 72.0 g of water at 19.2 oC. A 141 g piece of metal is heated to 89.0 oC and dropped into the water. The e

kumpel [21]

Answer:

The specific heat of the metal is 0.212 J/(g°C).

Explanation:

We can calculate the specific heat of the metal by the following equilibrium:

$q_{a} = -q_{b}$

$m_{a}C_{a}\Delta T_{a} = -m_{b}C_{b}\Delta T_{b}$

$m_{a}C_{a}(T_{f_{a}} - T_{i_{a}}) = -m_{b}C_{b}(T_{f_{b}} - T_{i_{b}})$

In the above equation, we have that the heat loses by the metal (b) is gained by the water (a).

$m_{a}$ : is the water's mass = 72.0 g

$C_{a}$ : is the specific heat of water = 4.184 J/(g°C)

$T_{i_{a}}$ : is the initial temperature of the water = 19.2 °C

$T_{f_{a}}$ : is the final temperature of the water = 25.5 °C

$m_{b}$ : is the metal's mass = 141 g

$C_{b}$ : is the specific heat of metal =?

$T_{i_{b}}$ : is the initial temperature of the metal = 89.0 °C

$T_{f_{b}}$ : is the final temperature of the water = 25.5 °C

$m_{a}C_{a}(T_{f_{a}} - T_{i_{a}}) = -m_{b}C_{b}(T_{f_{b}} - T_{i_{b}})$

$72.0 g*4.184 J/(g^{\circ} C)*(25.5 ^{\circ} C - 19.2 ^{\circ} C) = -141 g*C_{b}*(25.5 ^{\circ} C - 89.0 ^{\circ} C)$

$C_{b} = -\frac{72.0 g*4.184 J/(g^{\circ} C)(25.5 ^{\circ} C - 19.2 ^{\circ} C)}{141 g(25.5 ^{\circ} C - 89.0 ^{\circ} C)} = 0.212 J/(g^{\circ} C)$

Therefore, the specific heat of the metal is 0.212 J/(g°C).

I hope it helps you!

7 0

2 years ago

Calculate the mass in grams of 8.35 × 1022 molecules of CBr4.

antiseptic1488 [7]

Answer: 45.983 g CBr₄

Explanation:

To convert from moles to grams, you know that we will need molar mass and Avogadro's number.

Avogadro's number: 6.022×10²³ molecules/mol

Molar mass: 331.627 grams/mol

Now that we have what we need, you can use these to solve for grams. $8.35*10^2^2molecules CBr_{4} *\frac{1mol}{6.022*10^2^3molecules} *\frac{331.627 g}{1 mol} =45.983 g$

Our final answer is 45.983 g CBr₄.

4 0

3 years ago

Give the following for SO2 and BrF5:

frosja888 [35]

Answer:

The given molecules are SO2 and BrF5.

Explanation:

Consider the molecule SO2:

The central atom is S.

The number of domains on S in this molecule is three.

Domain geometry is trigonal planar.

But there is a lone pair on the central atom.

So, according to VSEPR theory,

the molecular geometry becomes bent or V-shape.

Hybridization on the central atom is

$sp^{2}$ .

Consider the molecule BrF5:

The central atom is Br.

The number of domains on the central atom is six.

Domain geometry is octahedral.

But the central atom has a lone pair of electrons.

So, the molecular geometry becomes square pyramidal.

The hybridization of the central atom is $sp^{3} d^{2}$ .

The shapes of SO2 and BrF5 are shown below:

6 0

3 years ago

Find the pH of the equivalence point and the volume (mL) of 0.0372 M NaOH needed to reach the equivalence point in the titration

elixir [45]

Answer:

8.54

Explanation:

At equivalence point :

42.2 X 0.052 = Vol. NaOH X 0.0372

Vol of NaOH = 2.1944/0.0372 = 58.99 ml

So volume of NaOH recquired to reach equivalence point = 58.99 ml

Number of miliimoles of CH3COOH = molarity X volume in ml = 42.2 X 0.052 = 2.1944 millimoles

Number of millimoles of NaOH = 58.99 X 0.0372 = 2.1944

Now CH₃COOH and NaOH reacts to give CH₃COONa according to the reaction :

CH₃COOH + NaOH ------> CH₃COONa + H₂O

1 mole of CH₃COOH reacts with 1 mole of NaOH to give 1 mole of CH₃COONa

So 2.1944 millimoles of CH₃COOH will react with 2.1944 millimoles of NaOH to give 2.1944 millimoles of CH₃COONa

So all the acid (CH₃COOH) and base (NaOH) has been converted into salt (CH₃COONa) so there is no acid or base left.

Now molarity of CH₃COONa = number of millimoles of CH₃COONa/total volume in ml = 2.1944/(58.99 + 42.2) = 2.1944/101.19 = 0.02169 M

So using the hydrolysis equation :

pH = 1/2 [ pKw + pKa + log c ]

Ka for acetic acid = 1.75 X 10⁻⁵

so pKa = -log (1.75 X 10⁻⁵) = 4.74

Kw = 10⁻¹⁴

so pKw = -log 10⁻¹⁴ = 14

c = 0.02169

so log c = log 0.02169 = -1.66

putting the values....

pH = 1/2 [14 + 4.74 - 1.66 ]

pH = 1/2 [ 17.08] = 8.54

6 0

3 years ago

Read 2 more answers

Be(OH)2 is solid why?

pickupchik [31]

Answer:

no it's not solid rather it's an aqueous

Explanation:

B/c Barium hydroxide is used in analytical chemistry for the titration of weak acids, particularly organic acids. Its clear aqueous solution is guaranteed to be free of carbonate, unlike those of sodium hydroxide and potassium hydroxide, as barium carbonate is insoluble in water.

8 0

1 year ago