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lisov135 [29]
4 years ago
7

Please help me with these asap!

Chemistry
1 answer:
Oksanka [162]4 years ago
8 0

Answer: okay this should get you started

Explanation:

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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
Does Catalysts increase the rate of chemical reactions. true or false
anastassius [24]
True.

A catalyst is a substancr that increases the rate of a chemical reaction without itself undergoing any permanent chemical change.

6 0
3 years ago
When you need to produce a variety of diluted solutions of a solute, you can dilute a series of stock solutions. A stock solutio
8090 [49]

Answer:

Volume of stock solution needed = 6.0299 mL

Explanation:

<u> </u>Dilution consists of lowering the amount of solute per unit volume of solution. It is achieved by adding more diluent to the same amount of solute.

This is deduced when thinking that both the dissolution at the beginning and at the end will have the same amount of moles.

<u>Data:</u>

M1 = 6.01 M stock solution concentration

M2 = 0.3624 M diluted solution concentration

V2 =100 mL diluted solution volume

V1 = ? stock solution volume

M1 * V1 = M2 * V2

V1=\frac{M2*V2}{M1} =\frac{0.3624M*100mL}{6.01M} =6.0299 mL

4 0
3 years ago
If a solution has virtually no hydrogen ions what type of ph does it have?
lora16 [44]
D. It has almost no hydrogen ions, therefore it is very basic.
7 0
3 years ago
Read 2 more answers
Calculate the mass of calcium chloride that contains 3.20 x 1024 atoms of chlorine.
frez [133]

Answer:

294.87 gm  CaCl_2

Explanation:

The computation of the  mass of calcium chloride is shown below:

But before that following calculations need to be done

Number of moles of chlorine atom is

= 3.20 × 10^24 ÷ 6.022 × 10^23

= 5.314 moles

As we know that

1 mole CaCl_2 have the 2 moles of chlorine atoms

Now 5.341 mole chloride atoms would be

= 1 ÷ 2 × 5.314

= 2.657 moles

Now

Mass of CaCl_2 = Number of moles × molar mass of  CaCl_2

= 2.657 moles × 110.98 g/mol

= 294.87 gm  CaCl_2

7 0
3 years ago
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