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lorasvet [3.4K]
3 years ago
6

NH4, C032. OH", and PO43- are all examples of

Chemistry
2 answers:
LekaFEV [45]3 years ago
4 0

Answer:

the popultion of it

Explanation:

Dvinal [7]3 years ago
4 0

Answer:

polyatomic ions

Explanation:

:)

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Which statement best describes the oxidation numbers of the atoms found in magnesium chloride? A. Magnesium has a 2- oxidation n
In-s [12.5K]
B. is the correct answer of your question
6 0
4 years ago
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The British gold sovereign coin is an alloy of gold and copper having a total mass of 7.988 g, and is 22-karat gold.
azamat

Answer:

(a) m_{gold}=7.322g

(b)

V_{gold}=0.379cm^3

V_{copper}=0.122cm^3

(c) \rho _{coin}=15.94g/cm^3

Explanation:

Hello,

(a) In this case, with the given formula we easily compute the mass of gold contained in the sovereign  as shown below:

m_{gold}=\frac{m_{tota}*karats}{24}=\frac{7.988g*22}{24}=7.322g

(b) Now, by knowing the density of gold and copper, 19.32 and 8.94 g/cm³ respectively, we compute each volume, by also knowing that the rest of the coin contains copper:

V_{gold}=\frac{m_{gold}}{\rho_{gold}} =\frac{7.322g}{19.32g/cm^3}=0.379cm^3

m_{copper}=7.988g-7.322g=1.09g\\V_{copper}=\frac{m_{copper}}{\rho_{copper}}=\frac{1.09g}{8.94g/cm^3}  \\\\V_{copper}=0.122cm^3

(c) Finally, the volume is computed by dividing the total mass over the total volume containing both gold and copper:

\rho _{coin}=\frac{m_{total}}{V_{gold}+V_{copper}}=\frac{7.988 g}{0.379cm^3+0.122cm^3}\\  \\\rho _{coin}=15.94g/cm^3

Best regards.

3 0
4 years ago
Read 2 more answers
Materials expand when heated. Consider a metal rod of length L0 at temperature T0. If the temperature is changed by an amount ΔT
marysya [2.9K]

Answer:

(a) The length at temperature 180°C is 40.070 cm

(b) The length at temperature 90°C is 64.976 inches

(c) L(T, α) = 60·α·T - 9000·α + 60

Explanation:

(a) The given parameters are

The thermal expansion coefficient, α for steel = 1.24 × 10⁻⁵/°C

The initial length of the steel L₀ = 40 cm

The initial temperature, t₀ = 40°C

The length at temperature 180°C = L

Therefore, from the given relation, for change in length, ΔL, we have;

ΔL = α × L₀ × ΔT

The amount the temperature changed ΔT = 180°C - 40°C = 140°C

Therefore, the change in length, ΔL, is found as follows;

ΔL = α × L₀ × ΔT = 1.24 × 10⁻⁵/°C × 40 × 140°C = 0.07 cm

Therefore, L =  L₀ + ΔL = 40 + 0.07 = 40.07 cm

The length at temperature 180°C = 40.07 cm

(b) Given that the length at T = 120°C is 65 in., we have;

The temperature at which the new length is sought = 90°C

The amount the temperature changed ΔT = 90°C - 120°C = -30°C

ΔL = α × L₀ × ΔT = 1.24 × 10⁻⁵/°C × 65 × -30°C = -0.024375 inches

The length, L at 90°C is therefore, L = L₀ + ΔL = 65 - 0.024375 = 64.976 in.

The length at temperature 90°C = 64.976 inches

(c) L = L₀ + ΔL  = L₀ +  α × L₀ × ΔT = L₀ +  α × L₀ × (T - T₀)

Therefore;

L = 60 +  α × 60 × (T - 150°C)

L = 60 + α × 60 × T - 9000 × α

L(T, α) = 60·α·T - 9000·α + 60

7 0
4 years ago
Draw a schematic of the hydrogen atom with the single proton in the nucleus, and the n=1, n=2, n=3, and n=4 energy level options
max2010maxim [7]

Explanation:

According to Bohr's postulates, the electron in the present in the lower energy level can absorb energy and exits to higher energy level. Also, when this electron returns back to its orbit, it emits some energy.

Since the hydrogen consists of 1 electron and 1 proton. The lowest energy configuration of the hydrogen is when n =1 or, when the electron is present in the K-shell or the ground state.

The possible transition for the electron given in the question is :

n = 2, 3 and 4

The schematic diagram of the hydrogen atom consisting of these four quantum levels in which the electron can jump (Absorption) and comeback to from these energy levels (emission) .

3 0
4 years ago
Hydrazine (N 2 H 4 )) a rocket fuel reacts with oxygen to form nitrogen gas and water vapor . The reaction is represented with t
Drupady [299]

The mass of hydrazine (N₂H₄) required to produce 96 g of water (H₂O) is 85.4 g (Option C)

<h3>Balanced equation </h3>

N₂H₄ + O₂ —> N₂ + 2H₂O

Molar mass of N₂H₄ = (2×14) + (4×1) = 32 g/mol

Mass of N₂H₄ from the balanced equation = 1 × 32 = 32 g

Molar mass of H₂O = (2×1) + 16 = 18 g/mol

Mass of H₂O from the balanced equation = 2 × 18 = 36 g

SUMMARY

From the balanced equation above,

36 g of H₂O were produced by 32 g of N₂H₄

<h3>How to determine the mass of N₂H₄</h3>

From the balanced equation above,

36 g of H₂O were produced by 32 g of N₂H₄

Therefore,

96 g of H₂O will be produced by = (96 × 32) / 36 = 85.4 g of N₂H₄

Thus, 85.4 g of N₂H₄ is needed for the reaction

Learn more about stoichiometry:

brainly.com/question/14735801

8 0
3 years ago
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