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3 years ago

True or false gasses can apply a force on an object

Chemistry

2 answers:

klasskru [66]3 years ago

6 0

Answer:

True, gasses can apply a force on an object.

WARRIOR [948]3 years ago

4 0

I think it’s false I’m not sure. I thought it was true

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Explain the reason according to bohr atomic model why atomic emission spectra

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contain only certain frequencies of light. Each s sublevel is related to a spherical s orbital. ... Bohr model: the electron is a particle; they hydrogen atomhas only certain allowable energy states. hope this helps

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3 years ago

Which statement accurately describe the two animals?

ANEK [815]

The elephant has more inertia than the dog so it takes more force to change the motion of the elephant.

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3 years ago

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Identify the polynomial. x3- y3 + z monomial binomial trinomial

Firdavs [7]

Answer: Option (c) is the correct answer.

Explanation:

A monomial is defined as an algebraic expression in which there is only one term present.

For example, 3x is a monomial.

A binomial is defined as an expression where there will be two terms present. For example, $(x^{2} + y^{2})$ is a binomial.

A polynomial is defined as an algebraic expression which contains more than two terms.

For example, $x^{3} - y^{3} + z$ is a polynomial.

Thus, we can conclude that the given expression $x^{3} - y^{3} + z$ is a polynomial.

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3 years ago

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

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3 years ago

USE YOUR OWN WORDS. Each question must have a three sentence long answer!

ivolga24 [154]

Answer:

I don't know zkjfn

Explanation:

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