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3 years ago

Would a block of mass of 32 g the leisure 3 cm in all directions float or sink in water

1 answer:

7 0

A body will float if its density is smaller than the density of water and will sink if its density is bigger than the density of water.

So, you need to calculate the density of the block.

Density = mass / volume

mass = 32 g

volumen = (3cm)^3 = 9 cm^3

=> density = 32 g / 9 cm^3 = 3.55 g/cm^3

Given that the density of water is 1.0 g /cm^3, the density of the block is greater and you conclude it will sink in water.

Answer: the block will sink in water

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A gas occupies 200ml at a temperature of 26 degrees Celsius and 76mmHg pressure. Find the volume at -3degree Celsius with the pr

sergey [27]

Answer:

184.62 ml

Explanation:

Let $p_1, v_1,$ and $T_1$ be the initial and $p_2, v_2,$ and $T_2$ be the final pressure, volume, and temperature of the gas respectively.

Given that the pressure remains constant, so

$p_1=p_2$ ...(i)

$v_1$ = 200 ml

$T_1= 26 ^{\circ}C = 273+26 =299$ K

$T_2= 3 ^{\circ}C = 273+3 =276$ K

From the ideal gas equation, pv=mRT

Where p is the pressure, v is the volume, T is the temperature in Kelvin, m is the mass of air in kg, R is the specific gas constant.

For the initial condition,

$p_1v_1=mRT_1 \\\\mR= \frac{p_1v_1}{T_1}\cdots(ii)$

For the final condition,

$p_2v_2=mRT_2 \\\\mR= \frac{p_2v_2}{T_2}\cdots(iii)$

Equating equation (i), and (ii)

$\frac{p_1v_1}{T_1}=\frac{p_2v_2}{T_2}$

$\frac{v_1}{T_1}=\frac{v_2}{T_2}$ [from equation (i)]

$v_2=\frac{T_2}{T_1} \times v_1$

Putting all the given values, we have

$v_2=\frac{276}{299} \times 200 = 184.62 \; ml$

Hence, the volume of the gas at 3 degrees Celsius is 184.62 ml.

7 0

2 years ago

Scientists have categorized trees based on whether they keep or lose their leaves each year. Another logical way to categorize t

Zolol [24]

Answer:

Explanation:

I think but it is an better attempt than the other guy answer.

5 0

2 years ago

You need to produce a buffer solution that has a pH of 5.50. You already have a solution that contains 10 mmol (millimoles) of a

balandron [24]

Answer:

56.9 mmoles of acetate are required in this buffer

Explanation:

To solve this, we can think in the Henderson Hasselbach equation:

pH = pKa + log ([CH₃COO⁻] / [CH₃COOH])

To make the buffer we know:

CH₃COOH + H₂O ⇄ CH₃COO⁻ + H₃O⁺ Ka

We know that Ka from acetic acid is: 1.8×10⁻⁵

pKa = - log Ka

pKa = 4.74

We replace data:

5.5 = 4.74 + log ([acetate] / 10 mmol)

5.5 - 4.74 = log ([acetate] / 10 mmol)

0.755 = log ([acetate] / 10 mmol)

10⁰'⁷⁵⁵ = ([acetate] / 10 mmol)

5.69 = ([acetate] / 10 mmol)

5.69 . 10 = [acetate] → 56.9 mmoles

6 0

2 years ago

What form of matter is made from only one type of atom? compound molecule material element

Lelechka [254]

<span>Elements are matter made of one kind of atom.</span>

6 0

2 years ago

Read 2 more answers

Find the mass, in grams, of 3.758 mol CH4. ( show work )

vladimir2022 [97]

Answer:

The mass of CH4 is 60, 29 grams.

Explanation:

We use the weight of the atoms C and H for calculate the molar mass:

Weight of CH4= weight C+ 4 x weight H= 12,01 g/mol +4 x 1,008g/mol=

Weight of CH4 =16, 042 g/mol

1molCH4-----16, 042grams

3,758 mol CH4--X= (3,758 mol CH4 x 16, 042 grams)/1 mol CH4=60,285836 grams

5 0

3 years ago