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schepotkina [342]

3 years ago

7

Whose personal story did you read (if you read several, pick your favorite)? What was the most interesting or eye-opening part o

f that person's story?

1 answer:

tatuchka [14]3 years ago

7 0

I feel like it would probably be all the things that my mom said about things and I didn’t believe until I saw

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A compound of a transition metal and iodine is 56.7% metal by mass.How many grams of the metal can be obtained from 630 g of thi

erma4kov [3.2K]

Answer:

357 g of the transition metal are present in 630 grams of the compound of the transition metal and iodine

Explanation:

In any sample of the compound, the percentage by mass of the transition metal is 56.7%. This means that for a 100 g sample of the compound, 56.7 g is the metal while the remaining mass, 43.3 g is iodine.

Given mass of sample compound = 630 g

Calculating the mass of iodine present involves multiplying the percentage by mass composition of the metal by the mass of the given sample;

56.7 % = 56.7/100 = 0.567

Mass of transition metal = 0.567 * 630 = 357.21 g

Therefore, the mass of the transition metal present in 630 g of the compound is approximately 357 g

4 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

What type of bonds are shown in this diagram?

qaws [65]

Answer:

What type of bonds are shown in this diagram?

A: covalent bonds

B: ionic bonds

C: hydrogen bonds

D: metallic bonds

(answer) metallic bonds

In what type of bonds do atoms join together because their opposite charges attract each other?

A: metallic bonds and covalent bonds

B: metallic bonds and ionic bonds

C: ionic bonds and covalent bonds

D: ionic bonds and hydrogen bonds

(answer) ionic bonds and hydrogen bonds

What types of bonds are shown in this diagram?

A: covalent bonds

B: ionic bonds

C: hydrogen bonds

D: metallic bonds

(answer) hydrogen bonds

Which statement best describes the types of bonds shown in the diagram?

A: an ionic bond; the hydrogen chloride molecule has an electrical charge

B: an ionic bond; a hydrogen ion is bonding with a chlorine atom

C: a covalent bond; the hydrogen atom’s two electrons are being shared with the chlorine atom

D: a covalent bond; the hydrogen atom’s single electron is being shared with the chlorine atom

(answer) a covalent bond; the hydrogen atom’s single electron is being shared with the chlorine atom

Which of the following bonds is the strongest?

A: hydrogen bonds

B: metallic bonds

C: valence bonds

D: covalent bonds

(answer)

Explanation:

UwU

3 0

8 months ago

Read 2 more answers

Given the two reactions PbCl2(aq)⇌Pb2+(aq)+2Cl−(aq), K3 = 1.89×10−10, and AgCl(aq)⇌Ag+(aq)+Cl−(aq), K4 = 1.25×10−4, what is the

Sergio [31]

Answer:

$1.2\times 10^{-2}$

Explanation:

The given reactions are:

PbCl2(aq)⇌Pb2+(aq)+2Cl−(aq) $K_3 = 1.89\times 10^{-10}$

AgCl(aq)⇌Ag+(aq)+Cl−(aq) $K_4 = 1.25\times 10^{-4}$

Required reaction is:

PbCl2(aq)+2Ag+(aq)⇌2AgCl(aq)+Pb2+(aq)

$K_f = \frac{K_3}{K_4^2}\\=\frac{1.89\times 10^{-10}}{1.25\times 10^{-4}}^2\\=1.2\times 10^{-2}$

4 0

3 years ago

If the atmospheric pressure is 0.975 atm what is the pressure of the enclosed gas

Nuetrik [128]

Missing in your question:
Picture (1)
when its an open- tube manometer and the h = 52 cm.
when the pressure of the atmosphere is equal the pressure of the gas plus the pressure from the mercury column 52 Cm so, we can get the pressure of the gas from this formula:
P(atm) = P(gas) + height (Hg)
∴P(gas) = P(atm) - height (Hg)
= 0.975 - (520/760)
= 0.29 atm
Note: I have divided 520 mm Hg by 760 to convert it to atm
Picture (2)
The pressure of the gas is the pressure experts by the column of mercury and when we have the Height (Hg)= 67mm
So the pressure of the gas =P(atm) + Height (Hg)
= 0.975 + (67/ 760) = 1.06 atm
Picture (3)
As the tube is closed SO here the pressure of the gas is equal the height of the mercury column, and when we have the height (Hg) = 103 mm. so, we can get the P(gas) from this formula:
P(gas) = Height(Hg)
= (103/760) = 0.136 atm

6 0

2 years ago