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djverab [1.8K]
3 years ago
14

Test (Speed, Velocity and Acceleration)

Chemistry
1 answer:
Troyanec [42]3 years ago
5 0

Answer:

1.5 miles

Explanation:

Given that,

Speed of dog, v = 15 mph

Time taken by the dog, t = 0.1 h

We need to find the distance travelled by the dog. We know that, distance is the product of speed and time. So,

d=vt

d=15\times 0.1\\\\d=1.5\ \text{miles}

Hence, the dog will cover a distance of 1.5 miles.

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Calculate the equilibrium number of vacancies per cubic meter for copper at 1000K. The energy for vacancy formation is 0.9eV/ato
nexus9112 [7]

Answer:

Therefore the equilibrium number of vacancies per unit cubic meter =2.34×10²⁴ vacancies/ mole

Explanation:

The equilibrium number of of vacancies is denoted by N_v.

It is depends on

  • total no. of atomic number(N)
  • energy required for vacancy
  • Boltzmann's constant (k)= 8.62×10⁻⁵ev K⁻¹
  • temperature (T).

N_v=Ne^{-\frac{Q_v}{kT} }

To find  equilibrium number of of vacancies we have find N.

N=\frac{N_A\ \rho}{A_{cu}}

Here ρ= 8.45 g/cm³  =8.45 ×10⁶m³

N_A= Avogadro Number = 6.023×10²³

A_{Cu}= 63.5 g/mole

N=\frac{6.023\times 10^{23}\times 8.45\times 10^{6}}{63.5}

   =8.01\times 10^{28 g/mole

Here Q_v=0.9 ev/atom , T= 1000k

Therefore the equilibrium number of vacancies per unit cubic meter,

N_v=( 8.01\times 10^{28}) e^{-(\frac{0.9}{8.62\times10^{-5}\times 1000})

   =2.34×10²⁴ vacancies/ mole

3 0
3 years ago
To test the effectiveness of a gunpowder mixture, 1 gram was exploded under controlled STP conditions, and the reaction chamber
Alekssandra [29.7K]

Answer:

1.3 × 10³ cm³

Explanation:

The gas occupies a volume of V₁ = 310 cm³ under standard temperature and pressure (STP), that is, T₁ = 273.15 K and P₁ = 1.0 atm. In order to find the volume V₂ under different conditions we can use the combined gas law formula.

\frac{P_{1}.V_{1}}{T_{1}} =\frac{P_{2}.V_{2}}{T_{2}} \\V_{2}=\frac{P_{1}.V_{1}.T_{2}}{T_{1}.P_{2}}=\frac{1.0atm\times 310cm^{3} \times 2473K }{273.15K \times 2.1atm} =1.3 \times 10^{3} cm^{3}

7 0
3 years ago
How many moles of oxygen are required to produce 37.15 g CO2?
denpristay [2]
Carbon = 12.010. Oxygen = 15.999 x 2 15.999 x 2 = 31.998 + 12.010 = 44.008 \frac{37.15 grams * 1 mole CO2}{44.008 grams}
8 0
4 years ago
Read 2 more answers
Matthew was working with different concentrations of hydrochloric acid in the lab. Which of these would BEST describe the result
CaHeK987 [17]
<span>The question says,'Mathew was working with different concentrations of hydrochloric acid in the lab. Which of these would best describe the resulsts Mathew would see if he was using a conductivity apparatus in each of the different acid concentration. The correct answer is C. This is because, acids conduct electricity, the stronger the acid, the brighter the electricity that will be produced while the weaker the acid, the weaker the electricity that will be produced. Thus, higher concentration equals tronger electricity.</span>
4 0
3 years ago
Read 2 more answers
During an experiment, the percent yield of calcium chloride from a reaction was 85.22%. Theoretically, the expected amount shoul
devlian [24]

The formula to find yield is

(Actual Yield)/(Theorectical Yield) x100

Just do the math.

85.22% x 113 = 96.2986

Convert it to 3 significant figures

Ans: 96.3g

7 0
3 years ago
Read 2 more answers
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