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nikklg [1K]
3 years ago
13

Can you identify what object is being represented by the model on the right?

Chemistry
2 answers:
jasenka [17]3 years ago
7 0
Earth...............
Dovator [93]3 years ago
3 0

Answer:

earth

Explanation:

You might be interested in
If 125 g of CaCO3 is mixed and reacted with 125 g of HCl, which reactant is limiting and how many grams of CO2 can be made?
Trava [24]

Answer:

CaCO3 is the limiting reactant

55 g of CO2 is made

Explanation:

First we must put down the reaction equation;

CaCO3(s) + 2HCl(aq) ---------> CaCl2(s) + H2O(l) + CO2(g)

Number of mole of CaCO3 = 125g/100gmol-1 = 1.25 moles

From the reaction equation;

1 mole of CaCO3 yields 1 mole of CO2

Hence 1.25 moles of CaCO3 yields 1.25 moles of CO2

For HCl;

number of moles of HCl = 125g/36.5 g mol-1 = 3.42 moles

From the reaction equation;

2 moles of HCl yields 1 mole of CO2

3.42 moles of HCl yields 3.42 * 1/2 = 1.71 moles of CO2

Hence CaCO3 is the limiting reactant.

Mass of CO2 produced = 1.25g * 44 gmol-1 = 55 g of CO2

6 0
2 years ago
What is the final volume (l) of a 10.0 l system that has the pressure quartered?
mihalych1998 [28]
According to Boyle's Law, P1V1 = P2V2

where P1 and V1 are initial pressure and volume respectively. P2 and V2 are final pressure and volume receptively.

Given: P2 = 4 P1 and V1 = 10.0l

∴ V2 = 2.5 l

Answer: Final volume of system is 2.5 l


5 0
2 years ago
What type of compound is CH3―O―CH2―CH2―CH3?
Lelu [443]
Ether

methoxypropane (methyl propyl ether)
5 0
3 years ago
Find the force of gravity of 1000 kg car.
ANEK [815]

Answer:

The force of gravity acting on the car is <u>9800 N vertically downward.</u>

Explanation:

Given:

Mass of the car given is 1000 kg.

We know that the force of gravity is the force applied by the center of Earth on any body. The force of gravity is also called the weight of the body and always act towards the center of the Earth.

From Newton's second law, we know that the force acting on a body is equal to its mass and acceleration.

Here, the acceleration acting on the car is due to gravity and thus has a constant value of 9.8 m/s² on the surface of Earth.

Therefore, the force of gravity acting on the car is given using the Newton's second law as:

Force of gravity = Mass of car \times Acceleration due to gravity.

Force of gravity = (1000 kg) \times (9.8 m/s²)

Force of gravity = 9800 N          [1 kg.m/s² = 1 N]

Therefore, the force of gravity acting on the car is 9800 N vertically downward.

6 0
3 years ago
Must show units and how they cancelli 1.) 175 km to um 3.) 385 nm to dm 5.) 492 um tom 7.) 52 x 103 dm to mm 9.) 321x 1035 mm to
morpeh [17]

Explanation:

1.) 175 km to μm

1 km=10^9 \mu m

175 km=175\times 10^9\mu m=1.75\times 10^{11} \mu m

3.) 385 nm to dm

1 nm=10^{-8} dm

385 nm=385\times 10^{-8} dm=3.85\times 10^{-6} dm

5.) 492 μm  to m

1 μm =  10^{-6} m

492 \μm=492\times 10^{-6} m=4.92\times 10^{-4} m

7.) 52\times 10^3 dm to mm

1 dm = 100 mm

52\times 10^3 dm=52\times 10^3\times 100 mm=5.2\times 10^{6}dm

9.) 321\times 10^{35} mm to km

1 mm = 10^{-6} km

321\times 10^{35} mm=321\times 10^{35}\times 10^{-6} km=3.21\times 10^{31} km

11.) 456\times 10^3 m to km

m = 0.001 km

456\times 10^3m =456\times 10^3 m\times 0.001 km=456 km

13.) 422\times 10^3 m to nm

1 m = 10^{9} nm

422\times 10^3 m=422\times 10^3\times 10^{9} nm=4.22\times 10^{14} nm

15.) 4.87\times 10^{30} m to pm

1 m = 10^{12} pm

4.87\times 10^{30} m=4.87\times 10^{30}\times 10^{12} pm=4.82\times 10^{42} pm

17.) 5.26\times 10^3 m to um

1 m =  10^{6} \mu m

5.26\times 10^3 m=5.26\times 10^3\times 10^6 \mu m=5.26\times 10^{9} \mu m

19.) 1.25\times 10^{35}m to Mm

1 m =  10^{-6} Mm

1.25\times 10^{35} m=1.25\times 10^{35}\times 10^{-6} Mm=1.25\times 10^{-29} Mm

21.) 4.22\times 10^3 Tm to nm

1 Tm = 10^{21} nm

4.22\times 10^3 Tm=4.22\times 10^3\times 10^{21} nm=4.22\times 10^{24} nm

6 0
3 years ago
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