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2 years ago

9

At archery practice, Vladimir tests a bow and arrow he has recently designed. He is able to accelerate a 24.0-g arrow at an aver

age rate of 70.0 m/s^2. What is the bows average force on the arrow

1 answer:

Whitepunk [10]2 years ago

7 0

Answer: A

1.68 N

Explanation:

F = ma = 0.024(70.0) = 1.68 N

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Calculate the numbers of mL in 0.603 L

Artist 52 [7]

Answer:

603 mL

Explanation:

A milliliter is a unit of volume equal to 1/1000th of a liter. It is the same as a cubic centimeter.

6 0

2 years ago

Convert 6.7 x 1024 molecules of nitrogen dioxide into grams.

BlackZzzverrR [31]

Answer:

510 g NO₂

General Formulas and Concepts:

Avogadro's Number - 6.022 × 10²³ atoms, molecules, formula units, etc.
Reading the Periodic Table
Writing Compounds
Using Dimensional Analysis

Explanation:

<u>Step 1: Define</u>

6.7 × 10²⁴ molecules NO₂ (Nitrogen dioxide)

<u>Step 2: Define conversions</u>

Avogadro's Number

Molar Mass of N - 14.01 g/mol

Molar Mass of O - 16.00 g/mol

Molar Mass of NO₂ - 14.01 + 2(16.00) = 46.01 g/mol

<u>Step 3: Use Dimensional Analysis</u>

<u /> $6.7 \cdot 10^{24} \ molecules \ NO_2(\frac{1 \ mol \ NO_2}{6.022 \cdot 10^{23} \ molecules \ NO_2} )(\frac{46.01 \ g \ NO_2}{1 \ mol \ NO_2} )$ = 511.901 g NO₂

<u>Step 4: Check</u>

<em>We are given 2 sig figs. Follow sig fig rules.</em>

511.901 g NO₂ ≈ 510 g NO₂

6 0

2 years ago

A lake in California contains about 8,500,000,000,000 gallons of water. How can this

aniked [119]

Answer:

8.5 * 10^11

Explanation:

8 0

2 years ago

Chemist A burns 10.000 grams of C₂H₆ in a constant-volume container at 25ºC. Chemist B burns an unknown mass of C₂H₆ in a consta

alexandr1967 [171]

Answer:

10.000 grams

Explanation:

For the first law of thermodynamics, the energy must be conserved, that means that the energy in form of heat (Q) must be equal to the sum of work (W) and internal energy(ΔU) :

Q = W + ΔU

ΔU depends on the temperature and W in the variation of pressure and volume. Q depends on the temperature, but also the mass. So, there is the same temperature, ΔU is equal for both reaction, if there is no work done, the heat must be equal for both of them. So the mass such be the same.

5 0

2 years ago

The pre-exponential constant and activation energy for the diffusion of iron in cobalt are 1.7 x 10-5 m2/s and 273,300 J/mol, re

Kitty [74]

<u>Answer:</u> The temperature of the system will be 1622 K

<u>Explanation:</u>

The equation relating the pre-exponential factor and activation energy follows:

$\log D=\log D_o-\frac{E_a}{2.303RT}$

where,

D = diffusion coefficient = $2.7\times 10^{-14}m^2/s$

$D_o$ = pre-exponential constant = $1.7\times 10^{-5}m^2/s$

$E_a$ = activation energy of iron in cobalt = 273,300 J/mol

R = Gas constant = 8.314 J/mol.K

T = temperature = ?

Putting values in above equation, we get:

$\log (2.7\times 10^{-14})=\log (1.7\times 10^{-5})-\frac{273,300}{2.303\times 8.314\times T}\\\\T=1622K$

Hence, the temperature of the system will be 1622 K

8 0

2 years ago