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2 years ago

What happened to your projectile if you changed the liquid

Chemistry

1 answer:

svp [43]2 years ago

5 0

Answer:

Actors affecting the flight path of a Projectile are:

Gravity.

Air Resistance.

Speed of Release.

Angle of Release.

Height of Release.

Spin

Hope it's Helpful

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7.no . <br> with steps please anyone <br> no spam

gulaghasi [49]

Given mass=1.3g
Atomic mass=65u

Molar mass=65g/mol

Now

$\boxed{\sf No\:of\:moles=\dfrac{Given\:Mass}{Molar\:Mass}}$

$\\ \rm\longmapsto No\:of\;Moles=\dfrac{1.3}{65}=0.02mol$

6 0

3 years ago

Two atoms that are isotopes of one another must have the same number of what? Electrons, All Particles, Protons, or Neutrons

emmainna [20.7K]

atoms are made of 3 types of sub atomic particles; neutrons, protons and electrons

atomic number is the number of protons which is characteristic for the element. Atoms of the same element have the same number of protons.

mass number is the sum of the neutrons and protons.

Isotopes are atoms of the same element with different mass numbers. since they are the same element number of protons are the same but number of neutrons vary.

therefore 2 isotopes are of the same element so they have the same number of protons.

5 0

3 years ago

Read 2 more answers

The following information is to be used for the next 2 questions. In order to analyze for Mg and Ca, a 24-hour urine sample was

Ainat [17]

Answer:

Explanation:

From the given information:

The concentration of metal ions are:

$[Ca^{2+}]= \dfrac{0.003474 \ M \times 20.49 \ mL}{10.0 \ mL}$

$[Ca^{2+}]=0.007118 \ M$

$[Mg^2+] = \dfrac{0.003474 \ M\times (26.23 - 20.49 )mL}{10.0 \ mL}$

$=0.001994 \ M$

Mass of Ca²⁺ in 2.00 L urine sample is:

$= 2.00 L \times 0.001994 \dfrac{mol}{L} \times \dfrac{40.08 \ g}{1 \ mol}$

= 0.1598 g

Mass of Ca²⁺ = 159.0 mg

Mass of Mg²⁺ in 2.00 L urine sample is:

$= 2.00 L \times 0.007118 \dfrac{mol}{L} \times \dfrac{24.31 \ g}{1 \ mol}$

= 0.3461 g

Mass of Mg²⁺ = 346.1 mg

5 0

3 years ago

Examine the system of equations.<br> y = 4x +8,<br> y = 4x-1

storchak [24]

Hope this helps with the answer to your question

5 0

3 years ago

5. What are the relative rates of diffusion for methane, CH, and oxygen, O2? If O2 la travels 1.00 m in a certain amount of time

11Alexandr11 [23.1K]

Answer:

The relative rates of diffusion for methane and oxygen is 1.4142.

Methane gas will be able to travel 1.4142 meter in the same conditions.

Explanation:

To calculate the rate of diffusion of gas, we use Graham's Law.

This law states that the rate of effusion or diffusion of gas is inversely proportional to the square root of the molar mass of the gas. Mathematically written as:

$\text{Rate of diffusion}\propto \frac{1}{\sqrt{\text{Molar mass of the gas}}}$