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Gennadij [26K]
3 years ago
8

If 2 objects have the same volume of 10cm3 and object A has a mass of 2 grams and object B has a mass of 4 grams, how do their d

ensities relate to one another?
I need a step by step explanation also. I don’t fully understand the subject. I also need an answer ASAP. This is due tomorrow and my teacher will not give me extra time even though I was excused.
Chemistry
2 answers:
siniylev [52]3 years ago
7 0

Answer:

density of object B is double that density of object A

Explanation:

Density is calculated as follows:

density = mass / volume

then,

density of object A = 2 grams / 10 cm³ = 0.2 g/cm³

density of object B = 4 grams / 10 cm³ = 0.4 g/cm³

The relation between them is:

density of object B / density of object A  =  0.4 / 0.2 = 2

Mrac [35]3 years ago
5 0

Explanation:

density =mass / volume

now, when the mass of an object increases the volume of the object will decrease when density is kept constant.

so if an A object has a higher mass than object Bs mass which has same density, than object B will have higher volume than object A.

in other words mass and volume is indirectly proportional when density is kept constant

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Anna71 [15]

The compound that is most likely present when treated with bromine and the sample absorbs the red bromine color is ; Cycloheptene

<h3>What is cycloheptene </h3>

Cycloheptene is an unsaturated colorless oily liquid which is insoluble in water it absorbs Bromine when used in performing unsaturation test due it unsaturated nature.

The red bromine color will be absorbed when used to treat Cycloheptene but will not be absorbed when used to treat cycloheptane due to its saturated nature.  

Hence we can conclude that the compound that is present is Cycloheptene.

Learn more about Unsaturation : brainly.com/question/561845

5 0
3 years ago
To dilute a HCl solution from 0.400 M to 0.100 M the final volume must be
sp2606 [1]

<u>Answer:</u>

<em>The final volume must be 400 mL.</em>

<em></em>

<u>Explanation:</u>

Let us assume the Initial volume as 100ml

Using dilution factor formula  

\\$M_{1} \times V_{1}=M_{2} \times V_{2}$\\\\$0.400 M \times 100 m l=0.100 M \times V_{2}$

So,

$V_{2}=\frac{0.400 M \times 100 m l}{0.100 M}=400 m l$

Thus, the final volume must be 400 mL.

5 0
4 years ago
Read 2 more answers
A 4.0 L balloon has a pressure of 406 kPa. When the pressure increases to 2,030 kPa, what is the volume? *
elena-14-01-66 [18.8K]

The new volume when pressure increases to 2,030 kPa is 0.8L

BOYLE'S LAW:

The new volume of a gas can be calculated using Boyle's law equation:

P1V1 = P2V2

Where;

  1. P1 = initial pressure (kPa)
  2. P2 = final pressure (kPa)
  3. V1 = initial volume (L)
  4. V2 = final volume (L)

According to this question, a 4.0 L balloon has a pressure of 406 kPa. When the pressure increases to 2,030 kPa, the volume is calculated as:

406 × 4 = 2030 × V2

1624 = 2030V2

V2 = 1624 ÷ 2030

V2 = 0.8L

Therefore, the new volume when pressure increases to 2,030 kPa is 0.8L.

Learn more about Boyle's law calculations at: brainly.com/question/1437490?referrer=searchResults

3 0
3 years ago
Which substance can not be decomposed by a chemical change?<br> (1) AlCl3 (3) HI<br> (2) H2O (4) Cu
user100 [1]
Cooper is the substance that can't be decomposed by a chemical chAnge. So 4)
5 0
3 years ago
Read 2 more answers
To aid in the prevention of tooth decay, it is recommended that drinking water contains 0.900 ppm fluoride (F-). A) How many g o
Romashka-Z-Leto [24]

Answer:

a) <u>1.740 g</u> of F- must be added to a cylindrical water reservoir

b) Grams of sodium fluoride, NaF, that contain this much fluoride:

3.84 g

Explanation:

Step 1. calculate the volume of the tank:

Volume of cylinder =

\pi  r^{2}h ,

Here r = radius of the cylinder = d/2

h = depth = 21.80m

r=\frac{d}{2}

=\frac{3.36x10^{2}}{2}

= 168 m

Volume =

=\frac{22\times 168^{2}\times 21.80}{7}

=1.93\times 10^{6} m^{3}

2.Convert ppm to g/m3 and Solve for mass of F-

1ppm = 1g/m^{3}

0.9ppm = 0.9g/m^{3}

Because both ppm and g/m3 are same quantity .

g/m^{3} =\frac{mass\ of\ F-(g)}{Volume\ m^{3}}\times 10^{6}

0.9 =\frac{mass\ of\ F-}{1.93\times 10^{6} m^{3}}\times 10^{6}

mass\ of\ F- =1.740g

mass of F- required = 1.740 g

3. Apply <u>mole concept </u>to calculate grams of sodium fluoride produced

mass of 1 mole of F2 = 38 g

mass of 1 mole of NaF = 42 g

(from periodic table calculate molar mass)

2Na+F_{2}\rightarrow 2NaF

Here 1 mole of F2 produce = 2 mole of NaF

So,

38 g  of F2 produce = 2 x 42 g of NaF

38 g of F2 produce = 84 g of NaF

1 g of F2 produce = 84/38 g of NaF

1.74 g F2 produce =

\frac {84}{38}\times 1.74

1.74 g F2 produce = 3.84 g of NaF

3.84 g of NaF is produced

3 0
3 years ago
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