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3 years ago

Use complete sentences to describe how making an ‘and’ compound sentence effects your solution set.

Chemistry

1 answer:

Lyrx [107]3 years ago

5 0

Answer:

“And” indicates that both statements of the compound sentence are true at the same time. It is the overlap or intersection of the solution sets for the individual statements.

Explanation:

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The frequency of an x-ray wave is 3.0 x 1012 MHz. Its wave speed is 3.0x 108 m/s. Calculate the wavelength of the x-ray wave bel

elena55 [62]

Answer:

λ = 1*10⁻¹⁰m

Explanation:

Frequency (f) = 3.0*10¹²MHz = 3.0*10¹⁸Hz

Speed (v) = 3.0*10⁸m/s

Speed (v) of a wave = frequency (f) * wavelength (λ)

V = fλ

Solve for λ,

λ = v / f

λ = 3.0*10⁸ / 3.0*10¹⁸

λ = 1*10⁻¹⁰m

λ = 0.

6 0

3 years ago

Which of the following is an example of acceleration?

anzhelika [568]

Answer:

it's either B. or C.. hope this helps!

Explanation:

4 0

3 years ago

Read 2 more answers

20.00 g of aluminum (Al) reacts with 78.78 grams of molecular chlorine (Cl2), all of each reaction is completely consumed and as

shepuryov [24]

The reaction forms 98.76 g AlCl_3.

We have the masses of two reactants, so this is a limiting reactant problem.

We know that we will need a balanced equation with masses, moles, and molar masses of the compounds involved.

Step 1. Gather all the information in one place with molar masses above the formulas and everything else below them.

M_r: ___26.98 _70.91 __133.34

________2Al + 3Cl_2 → 2AlCl_3

Mass/g: 20.00 _78.78

Step 2. Calculate the moles of each reactant

Moles of Al = 20.00 g Al × (1 mol Al /26.98 g Al) = 0.741 29 mol Al

Moles of Cl_2 = 78.78 g Cl_2 × (1 mol Cl_2 /70.91 g Cl_2) = 1.11 10 mol Cl_2

Step 3. Identify the limiting reactant

Calculate the moles of AlCl_3 we can obtain from each reactant.

From Al: Moles of AlCl_3 = 0.741 29 mol Al × (2 mol AlCl_3/2 mol Al) = 0.741 29 mol AlCl_3

From Cl_2: Moles of AlCl_3 = 1.11 10 mol Cl_2 × (2 mol AlCl_3/3 mol Cl_2) = 0.740 66 mol AlCl_3

Cl_2 is the limiting reactant because it gives the smaller amount of AlCl_3.

Step 4. Calculate the mass of AlCl_3.

Mass = 0.740 66 mol AlCl_3 × 133.34 g/1 mol AlCl_3 = 98.76 g AlCl_3

The reaction produces 98.76 g AlCl_3.

4 0

3 years ago

The true absorbance for a 1.0 x 10 −5 M solution is 0.7526. If the percentage stray light for a spectrophotometer is 0.56%, calc

Korvikt [17]

Answer:

The percentage deviation is $\Delta M = 1.87$ %

Explanation:

From the question we are told that

The concentration is of the solution is $C = 1.0*10^{-5} M$

The true absorbance A = 0.7526

The percentage of transmittance due to stray light $z = 0.56$ % $=\frac{0.56}{100} = 0.0056$

Generally Absorbance is mathematically represented as

$A = -log T$

Where T is the percentage of true transmittance

Substituting value

$0.7526 = - log T$

$T = 10^{-0.7526}$

$= 0.177$

$= 17.7$ %

The Apparent absorbance is mathematically represented

$A_p = -log (T +z)$

Substituting values

$A_p = -log(0.177 + 0.0056)$

$= -log(0.1826)$

= 0.7385

The percentage by which apparent absorbance deviates from known absorbance is mathematically evaluated as

$\Delta A = \frac{A -A_p}{A} * \frac{100}{1}$

$= \frac{0.7526 - 0.7385}{0.7526} * \frac{100}{1}$

$\Delta A = 1.87$ %

Since Absorbance varies directly with concentration the percentage deviation of the apparent concentration from know concentration is

$\Delta M = 1.87$ %

6 0

4 years ago

How much liquid does this graduated cylinder contain?

nikklg [1K]

Answer:

I think 44 mL

Explanation:

I hope this helps u :D

8 0

3 years ago