All categories

3 years ago

Please don't give me a wedsite

Chemistry

1 answer:

Basile [38]3 years ago

8 0

Answer: The size of actual bed is 75 inches

Explanation: Scale refers to the ratio of the original dimension to the dimension on paper (miniaturised dimension). Since it is a ratio, it has no dimension.

Scale= original dimension/miniaturised dimension

Given that scale is 1:25

It means that every 1 inch on paper would correspond to 25 inches in reality.

The length of the model bed= 3 inches

Hence, the length of the original bed is 25*3 = 75 inches

You might be interested in

How long would it take a radio wave with a frequency of 6.50 ✕ 108 Hz to travel from Mars to Earth if the distance between the t

olga55 [171]

The radio wave travels at the speed of light so divide 8.00 x 10^7 by the speed of light.<span>

</span>

4 0

4 years ago

Read 2 more answers

True or False: When it is spring in the Northern Hemisphere, it is autumn in the Southern Hemisphere.

svlad2 [7]

Answer:

That is true

Explanation:

3 0

3 years ago

Read 2 more answers

A student mixes in a test tube 3.00mL of 0.050M CuSO4with 7.00mL of 0.20M NH3/NH41 . The solution becomes a deep blue color. Ass

valkas [14]

Answer:

$\large \boxed{\text{0.0035 mol/L}}$

Explanation:

We are given the volumes and concentrations of two reactants, so this is a limiting reactant problem.

We know that we will need moles, so, lets assemble all the data in one place.

Cu²⁺ + 4NH₃ ⟶ Cu(NH₃)₄²⁺

V/mL: 3.00 7.00

c/mol·L⁻¹: 0.050 0.20

1. Identify the limiting reactant

(a) Calculate the moles of each reactant

$\text{Moles of Cu}^{2+}= \text{3.00 mL solution} \times \dfrac{\text{0.050 mmol Cu}^{2+}}{\text{1 mL solution}} = \text{0.150 mmol Cu}^{2+}\\\\\text{Moles of NH}_{3} = \text{7.00 mL solution} \times \dfrac{\text{0.20 mmol NH}_{3}}{\text{1 mL solution}} = \text{0.140 mmol NH}_{3}$

(b) Calculate the moles of Cu(NH₃)₄²⁺ that can be formed from each reactant

(i) From Cu²⁺

$\text{Moles of Cu(NH$_{3}$)$_{4}$$^{2+}$} = \text{0.150 mmol Cu}^{2+} \times \dfrac{\text{1 mmol Cu(NH$_{3}$)$_{4}$$^{2+}$}}{\text{1 mmol Cu}^{2+}}\\\\= \text{0.150 mmol Cu(NH$_{3}$)$_{4}$$^{2+}$}$

(ii) From NH₃

$\text{Moles of Cu(NH$_{3}$)$_{4}$$^{2+}$} = \text{0.140 mmol NH}_{3} \times \dfrac{\text{1 mmol Cu(NH$_{3}$)$_{4}$$^{2+}$}}{\text{4 mmol NH}_{3}}\\\\= \text{0.0350 mmol Cu(NH$_{3}$)$_{4}$$^{2+}$}$

NH₃ is the limiting reactant, because it forms fewer moles of the complex ion.

(c) Concentration of the complex ion

$\text{The reaction forms 0.0350 mmol Cu(NH$_{3}$)$_{4}$$^{2+}$ in a total volume of 10.00 mL.}\\c = \dfrac{\text{moles}}{\text{litres}} = \dfrac{\text{0.0350 mmol}}{\text{10.00 mL}} = \textbf{0.0035 mol/L}\\\\\text{The concentration of the complex ion is $\large \boxed{\textbf{0.0035 mol/L}}$}$

7 0

3 years ago

Which one of the following reactions IS NOT balanced? 1. 2 KNO3 + 10 K → 5 K2O + N2 2. 2 SO2 + O2 → 2 SO3 3. SF4 + 3 H2O → H2SO3

Delicious77 [7]

Answer:

2 KNO₃ + 10 K → 5 K₂O + N₂ (option 1)

Explanation:

1. 2 KNO₃ + 10 K → 5 K₂O + N₂

We have in reactant side:

2 K, 2N, 6O and 10 K. In conclussion, 12 K, 2N and 6 O

We have in product side.

10 K, 5O and 2 N

This equation is unbalanced

2. 2 SO₂ + O₂ → 2 SO₃

In reactant side we have 2 S and 6 O

In product side we have 2 S and 6 O

3. SF₄ + 3 H₂O → H₂SO₃ + 4 HF

In reactant side we have 1 S, 4F, 6 H and 3 O

In product side we have 1 S, 4F, 6 H and 3 O

6 0

3 years ago

The ideal gas law is a combination of all the gas laws and states that PV = nRT. The value of n stands for

dimaraw [331]

Answer:

Explanation:

p is pressure, v is vo;lume, t is temperature, r is ideal gas contstant, therefore n is the moles

7 0

3 years ago

Read 2 more answers