Vinegar is a solution of acetic acid in water. If certain vinegar has a concentration of 1.5% v / v.

Which statement supports Newton’s first law of motion?

fenix001 [56]

C. Planets can move at a varying speed due to forces exerted in space.

6 0

3 years ago

A 58.5 g sample of glass is put into a calorimeter (see sketch at right) that contains 250.0 g of water. The glass sample starts

Ket [755]

Answer:

The specific heat capacity of glass is 0.70J/g°C

Explanation:

Heat lost by glass = heat gained by water

Heat lost by glass = mass × specific heat capacity (c) × (final temperature - initial temperature) = 58.5×c×(91.2 - 21.7) = 4065.75c

Heat gained by water = mass × specific heat capacity × (final temperature - initial temperature) = 250×4.2×(21.7 - 19) = 2835

4065.75c = 2835

c = 2835/4065.75 = 0.70J/g°C

5 0

3 years ago

Read 2 more answers

If you feed 100 kg of N2 gas and 100 kg of H2 gas into a

torisob [31]

Answer : The mass of ammonia produced can be, 121.429 k

Solution : Given,

Mass of $N_2$ = 100 kg = 100000 g

Mass of $H_2$ = 100 kg = 100000 g

Molar mass of $N_2$ = 28 g/mole

Molar mass of $H_2$ = 2 g/mole

Molar mass of $NH_3$ = 17 g/mole

First we have to calculate the moles of $N_2$ and $H_2$ .

$\text{ Moles of }N_2=\frac{\text{ Mass of }N_2}{\text{ Molar mass of }N_2}=\frac{100000g}{28g/mole}=3571.43moles$

$\text{ Moles of }H_2=\frac{\text{ Mass of }H_2}{\text{ Molar mass of }H_2}=\frac{100000g}{2g/mole}=50000moles$

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

$N_2+3H_2\rightarrow 2NH_3$

From the balanced reaction we conclude that

As, 1 mole of $N_2$ react with 3 mole of $H_2$

So, 3571.43 moles of $N_2$ react with $3571.43\times 3=10714.29$ moles of $H_2$

From this we conclude that, $H_2$ is an excess reagent because the given moles are greater than the required moles and $N_2$ is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of $NH_3$

From the reaction, we conclude that

As, 1 mole of $N_2$ react to give 2 mole of $NH_3$

So, 3571.43 moles of $N_2$ react to give $3571.43\times 2=7142.86$ moles of $NH_3$

Now we have to calculate the mass of $NH_3$

$\text{ Mass of }NH_3=\text{ Moles of }NH_3\times \text{ Molar mass of }NH_3$

$\text{ Mass of }NH_3=(7142.86moles)\times (17g/mole)=121428.62g=121.429kg$

Therefore, the mass of ammonia produced can be, 121.429 kg

6 0

3 years ago

The products

alexandr1967 [171]

Answer:

The products of self-ionization of water are OH⁻ and H⁺.

Explanation:

The water is self ionized according to the equation:

H₂O → OH⁻ + H⁺.

The ionic product for water (Kw) = [OH⁻][H⁺] = 10⁻¹⁴.

Kw is also called "self-ionization constant" or "auto-ionization constant".

6 0

3 years ago

Using the Bohr model, determine the energy, in joules, necessary to ionize a ground-state hydrogen atom. Show your calculations.

lord [1]

Answer:

The energy required to ionize the ground-state hydrogen atom is 2.18 x 10^-18 J or 13.6 eV.

Explanation:

To find the energy required to ionize ground-state hydrogen atom first we calculate the wavelength of photon required for this operation.

It is given by Bohr's Theory as:

1/λ = Rh (1/n1² - 1/n2²)

where,

λ = wavelength of photon

n1 = initial state = 1 (ground-state of hydrogen)

n2 = final state = ∞ (since, electron goes far away from atom after ionization)

Rh = Rhydberg's Constant = 1.097 x 10^7 /m

Therefore,

1/λ = (1.097 x 10^7 /m)(1/1² - 1/∞²)

λ = 9.115 x 10^-8 m = 91.15 nm

Now, for energy (E) we know that:

E = hc/λ

where,

h = Plank's Constant = 6.625 x 10^-34 J.s

c = speed of light = 3 x 10^8 m/s

Therefore,

E = (6.625 x 10^-34 J.s)(3 x 10^8 m/s)/(9.115 x 10^-8 m)

E = 2.18 x 10^-18 J

E = (2.18 x 10^-18 J)(1 eV/1.6 x 10^-19 J)

E = 13.6 eV

5 0

3 years ago