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Jobisdone [24]
2 years ago
15

Iupac names of the following ​

Chemistry
1 answer:
Nadusha1986 [10]2 years ago
6 0
It’s blurry.... try again
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A 25 kg rock is placed in a graduated cylinder with water.the volume of the fluid is 18.3ml.calculate the density of the rock in
ipn [44]

Answer:

=> 1366.120 g/mL.

Explanation:

To determine the formula to use in solving such a problem, you have to consider what you have been given.

We have;

mass (m)     = 25 Kg

Volume (v) = 18.3 mL.

From our question, we are to determine the density (rho) of the rock.

The formula:

p = \frac{m}{v}

First let's convert 25 Kg to g;

1 Kg    = 1000 g

25 Kg = ?

= \frac{25 × 1000}{1}

= 25000 g

Substitute the values into the formula:

p =  \frac{25000 g}{18.3 ml}

= 1366.120 g/mL.

Therefore, the density (rho) of the rock is  1366.120 g/mL.

8 0
2 years ago
What force types change the direction of motion
lora16 [44]

Forces affect how objects move. They may cause motion; they may also slow, stop, or change the direction of motion of an object that is already moving. Since force cause changes in the speed or direction of an object, we can say that forces cause changes in velocity. Remember that acceleration is a change in velocity. Let’s say an object is moving along a table on earth, suddenly the finite table ends, resulting in the object being present in the air, which means there is no normal contact force N to combat the force by gravity mg, which is why there is an acceleration downwards. This proves as a projectile motion since the direction of motion start changing from horizontal to vertical. Another example is one throwing an object up. It moves up and slows down, reaching its maximum point, leading to it starting to move downwards. This too is a change in motion.

7 0
2 years ago
Read 2 more answers
How many grams of aluminum are required to react with 35 mL of 2.0 M hydrochloric acid, HCl?__ HCl + __ Al --> __ AlCl3 + __
kiruha [24]

Answer:

0.6258 g

Explanation:

To determine the number grams of aluminum in the above reaction;

  • determine the number of moles of HCl
  • determine the mole ratio,
  • use the mole ratio to calculate the number of moles of aluminum.
  • use RFM of Aluminum to determine the grams required.

<u>Moles </u><u>of </u><u>HCl</u>

35 mL of 2.0 M HCl

2 moles of HCl is contained in 1000 mL

x moles of HCl is contained in 35 mL

x \: mol \:  =  \:  \frac{2 \:  \times  \: 35}{1000}  \\  = 0.07 \: moles \:

We have 0.07 moles of HCl.

<u>Mole </u><u>ratio</u>

  • Balanced equation

6HCl(aq) + 2Al(s) --> 2AlCl3(aq) + 3H2(g)

Hence mole ratio = 6 : 2 (HCl : Al

  • but moles of HCl is 0.07, therefore the moles of Al;

=  \frac{2}{6}  \times 0.07 \\  \:  = 0.0233333 \: moles

Therefore we have 0.0233333 moles of aluminum.

<u>Grams of </u><u>Aluminum</u>

We use the formula;

grams \:  = moles \:  \times  \: rfm

The RFM (Relative formula mass) of aluminum is 26.982g/mol.

Substitute values into the formula;

= 0.0233333 \: moles  \:  \times  \: 26.982 \:  \frac{g}{mol}  \\  = 0.625799 \: grams

The number of grams of aluminum required to react with HCl is 0.6258 g.

3 0
2 years ago
A system gains 687 kJ of heat, resulting in a change in internal energy of the system equal to 156 kJ. How much work is done?
Maslowich

Answer:

w = -531 kJ

1. Work was done by the system.

Explanation:

Step 1: Given data

  • Heat gained by the system (q): 687 kJ (By convention, when the system absorbs heat, q > 0).
  • Change in the internal energy of the system (ΔU°): 156 kJ

Step 2: Calculate the work done (w)

We will use the following expression.

ΔU° = q + w

w = ΔU° - q

w = 156 kJ - 687 kJ

w = -531 kJ

By convention, when w < 0, work is done by the system on the surroundings.

4 0
2 years ago
A force of 25 newtons is exerted over an area of 50 square meters. Calculate
snow_lady [41]

Answer:

the pressure exerted in pascals is 0.5 Pa

Explanation:

The computation of the pressure exerted in pascals is shown below:

As we know that

Pressure = force ÷ area

= 25 ÷ 50

= 0.5 Pa

Hence, the pressure exerted in pascals is 0.5 Pa

We simply applied the above formula so that the correct pressure could come

4 0
2 years ago
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