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sp2606 [1]
4 years ago
10

What is the term for an object's strength of motion?

Chemistry
1 answer:
Andrei [34K]4 years ago
3 0

I'm pretty sure its momentum

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"how many grams of sucrose (c12h22o11) are in 1.55 l of 0.758 m sucrose solution"
Ierofanga [76]
<span>You would use the equation: Molarity M equals amount of solution over volume of solution. So, 0.758 M equals x L divided by 1.55 L OR 0.758 M = x L / 1.55 L . To solve: 1.55 multiplied by 0.758 equals 1.1479. The answer is 1.1479.</span>
3 0
3 years ago
Which prototype is the most cost- and time-effective solution? The Rockets First company needs to hire an engineer to develop a
MArishka [77]

Answer:

Infinity & Beyond

Explanation:

hope I helped!

4 0
3 years ago
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Calcium has a cubic closest packed structure (fcc) as a solid. Assuming that calcium has an atomic radius of 197 pm, calculate t
Paha777 [63]

Answer:

1.536 g/cm^3 is the density of solid calcium.

Explanation:

Formula used :  

\rho=\frac{Z\times M}{N_{A}\times a^{3}}

where,

\rho = density

Z = number of atom in unit cell

M = atomic mass

(N_{A}) = Avogadro's number  

a = edge length of unit cell

We have:

r = 197 pm = 197\times 10^{-12} m = 197\times 10^{-12}\times 100 cm

r=197\times 10^{-10} m

a=r\times 2\sqrt{2}

a=197\times 10^{-10} m\times 2\sqrt{2}=557.200\times 10^{-10} cm

M = 40 g/mol

Z = 4

On substituting all the given values , we will get the value of 'a'.

\rho =\frac{4\times 40 g/mol}{6.022\times 10^{23} mol^{-1}\times (557.200\times 10^{-10} cm)^{3}}

\rho =1.536 g/cm^3

1.536 g/cm^3 is the density of solid calcium.

5 0
3 years ago
What is the volume of 1.60 grams of O2 gas at STP? 1.72 liters 1.45 liters 1.32 liters 1.12 liters
elixir [45]

Answer:

The required volume is 1.12 L.

Explanation:

Given,

Mass of O2 gas= 1.60 grams

We know,

Number of moles = Mass/Molecular mass

                             = 1.60/32                     [Molecular mass of O2 gas is 32 gram]

                             = 0.05

Again,

We know that at STP,

Number of moles = Volume in L / 22.4 L

or, Volume = Number of moles X 22.4

or, Volume = 0.05 X 22.4

or, Volume = 1.12

∴The required volume is 1.12 L.

5 0
3 years ago
HELP FAST PLEASE!!
Anettt [7]

Answer:

the answer is A

Explanation:

i took the test on edg.                  sorry it is so late:(

7 0
3 years ago
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