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muminat
3 years ago
9

How do robots help scientists to perform experiments?

Chemistry
2 answers:
Andreas93 [3]3 years ago
8 0

Answer:

A. Robots go places humans are unable to go.

Arte-miy333 [17]3 years ago
3 0
The answer is number A
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Im hvaing a hard time getting the right answer
Tanya [424]

Answer:

V=23.9mL

Explanation:

Hello!

In this case for the solution you are given, we first use the mass to compute the moles of CuNO3:

n=2.49g*\frac{1mol}{125.55 g}=0.0198mol

Next, knowing that the molarity has units of moles over liters, we can solve for volume as follows:

M=\frac{n}{V}\\\\V=\frac{n}{M}

By plugging in the moles and molarity, we obtain:

V=\frac{0.0198mol}{0.830mol/L}=0.0239L

Which in mL is:

V=0.0239L*\frac{1000mL}{1L}\\\\V=23.9mL

Best regards!

6 0
3 years ago
Mercury, which is sometimes used in thermometers, has a density of 13.534 g/ml at room temperature. what volume of mercury conta
Sergeu [11.5K]
D - density: 13,534 g/ml
m - mass: 10g
V - volume: ??
_____________
d = m/V
V = m/d
V = 10/13,534
V = 0,7389 ml

:•)
6 0
3 years ago
1. Explain the relationship<br> between Polaris and Earth's tilt
Oduvanchick [21]

Answer:

that it i think

Explanation:

the earth revolves around the Sun once each year and spins on its axis of rotation once each day. This axis of rotation is tilted 23.5 degrees relative to its plane of orbit around the Sun. The axis of rotation is pointed toward Polaris the North Star. As the Earth orbits the Sun the tilt of Earth’s axis stays lined up with the North Star.

7 0
4 years ago
calculate the volume occupied by 10g of propane gas, under normal conditions of temperature and pressure
andriy [413]

Answer:

5.5 L

Explanation:

First we <u>convert 10 g of propane gas</u> (C₃H₈) to moles, using its <em>molar mass</em>:

  • 10 g ÷ 44 g/mol = 0.23 mol

Then we <u>use the PV=nRT formula</u>, where:

  • P = 1 atm & T = 293 K (This are normal conditions of T and P)
  • n = 0.23 mol
  • R = 0.082 atm·L·mol⁻¹·K⁻¹
  • V = ?

1 atm * V = 0.23 mol * 0.082 atm·L·mol⁻¹·K⁻¹ * 293 K

  • V = 5.5 L
3 0
3 years ago
The edge length of the unit cell of KCl (NaCl-like structure, FCC) is 6.28 Å. Assuming anion-cation contact along the cell edge,
Mila [183]

Answer:

1.33 Å

Explanation:

Given that the edge length , a of the KCl which forms the FCC lattice = 6.28 Å

Also,

For the FCC lattice in which the anion-cation contact along the cell edge , the ratio of the radius of the cation to that of anion is 0.731.

Thus,

\frac {r^+}{r^-}=0.731 .................1

Also, the sum of the radius of the cation and the anion in FCC is equal to half of the edge length.

Thus,

r^++r^-=\frac {a}{2}  ...................2

Given that:

Cl^-\ (r^-) = 1.82\ \dot{A}

To find,

K^+\ (r^+) = ? \dot{A}

Using 1 and 2 , we get:

1.731\ r^+=0.731\times \frac {6.28}{2}

<u>Size of the potassium ion = 1.33 Å</u>

4 0
3 years ago
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