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3 years ago

A brick is resting on a smooth, level surface. A student applies a force of

Chemistry

1 answer:

notsponge [240]3 years ago

8 0

Answer: C) The net force on the brick is 0 N, and the brick remains at rest.

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An aqueous KNO3 solution is made using 76.6 g of KNO3 diluted to a total solution volume of 1.84 L. (Assume a density of 1.05 g/

defon

Answer:

The answer to your question is Molarity = 0.41

Explanation:

Data

mass of KNO₃ = 76.6 g

volume = 1.84 l

density = 1.05 g/ml

Process

1.- Calculate the molecular mass of KNO₃

molecular mass = 39 + 14 + (16 x 3) = 101 g

2.- Calculate the number of moles

101 g of KNO₃ --------------- 1 mol

76.6 g of KNO₃ ------------ x

x = (76.6 x 1) / 101

x = 0.76 moles

3.- Calculate molarity

Molarity = $\frac{number of moles}{volume}$

Substitution

Molarity = $\frac{0.76}{1.84}$

Result

Molarity = 0.41

7 0

3 years ago

This is a material that allows heat/electricity to transfer.

Anvisha [2.4K]

A material that allows heat/electricity to transfer is called a conductor.

5 0

3 years ago

Read 2 more answers

The vapor pressure of substance X is 100. mm Hg at 1080.°C. The vapor pressure of substance X increases to 600. mm Hg at 1220.°C

artcher [175]

Explanation:

The given data is as follows.

$P_{1}$ = 100 mm Hg or $\frac{100}{760}atm$ = 0.13157 atm

$T_{1}$ = $1080 ^{o}C$ = (1080 + 273) K = 1357 K

$T_{2}$ = $1220 ^{o}C$ = (1220 + 273) K = 1493 K

$P_{2}$ = 600 mm Hg or $\frac{600}{760}atm$ = 0.7895 atm

R = 8.314 J/K mol

According to Clasius-Clapeyron equation,

$log(\frac{P_{2}}{P_{1}}) = \frac{\Delta H_{vap}}{2.303R}[\frac{1}{T_{1}} - \frac{1}{T_{2}}$

$log(\frac{0.7895}{0.13157}) = \frac{\Delta H_{vap}}{2.303 \times 8.314 J/mol K}[\frac{1}{1357 K} - \frac{1}{1493 K}]$

$log (6) = \frac{\Delta H_{vap}}{19.147}[\frac{(1493 - 1357) K}{1493 K \times 1357 K}]$

0.77815 = $\frac{\Delta H_{vap}}{19.147J/K mol} \times 6.713 \times 10^{-5} K$

$\Delta H_{vap}$ = $2.219 \times 10^{5}$ J/mol

= $2.219 \times 10^{5}J/mol \times 10^{-3}\frac{kJ}{1 J}$

= 221.9 kJ/mol

Thus, we can conclude that molar heat of vaporization of substance X is 221.9 kJ/mol.

4 0

3 years ago

How many grams are in 2.45E24 formula units of Fp3BZ2? The molar mass of Fp3Bz2. is 97.05 g/mol.

dlinn [17]

Answer:

394.99g

Explanation:

The number of moles of a substance can be calculated by dividing the number of atoms of such substance by Avagadro's number (6.02 × 10^23)

n = nA ÷ 6.02 × 10^23

The number of atoms of Fp3BZ2 in this question is 2.45E24 formula units i.e. 2.45 × 10^24

n = 2.45 × 10^24 ÷ 6.02 × 10^23

n = 2.45/6.02 × 10^(24-23)

n = 0.407 × 10¹

n = 4.07moles

Using mole = mass/molar mass

Where; molar mass of Fp3Bz2. is 97.05

g/mol.

mass = molar mass × mole

mass = 97.05 × 4.07

mass = 394.99g

4 0

3 years ago

1. the mass of one electron is

kupik [55]

Answer:

Mass of one electron is 9.1 × 10⁻³¹ kg

Mass of one proton is 1.673 × 10⁻²⁷ Kg

Mass of one neutron is 1.675 × 10⁻²⁷ Kg

-TheUnknownScientist 72

4 0

2 years ago