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2 years ago

What is the length of the x - component of the vector plotted below ?

Chemistry

1 answer:

MAVERICK [17]2 years ago

4 0

Answer:

Explanation:

You can make that into a triangle and find the hypotenuse by using the pythagorean theorem a^2 + b^2 = c^2

Your hypotenuse is your vector that you're looking for

Heres the work:

3^2 + 3^2= r^2

r=3

Please let me know if this helped, and rank it the brainliest if possible!

You might be interested in

Convert 2.1 mole of Al2(SO4)3 ionic units to a number of particles.

Semenov [28]

1.26 x 10²⁴particles

Explanation:

Given parameters:

Number of moles of the compound = 2.1 moles

Unknown:

Number of particles contained in the compound = ?

Solution:

A mole can be defined as the amount of a substance that contains Avogadro's number of particles 6.02 x 10²³.

The particles mentioned here can be atoms, molecules, formula units, ions, electrons, protons, neutrons and so forth.

To solve for the number of moles:

Number of particles = number of moles x 6.02 x 10²³

Number of particles = 2.1 x 6.02 x 10²³ = 12.64 x 10²³ = 1.26 x 10²⁴particles

Learn more:

mole calculation brainly.com/question/1841136

#learnwithBrainly

8 0

2 years ago

A white powder is known to be a mixture of magnesium oxide and aluminum oxide. 100cm3 of 2moldm-3 NaOH(aq) is just sufficient to

Effectus [21]

Answer: D.Aluminium Oxide 0.10, Magnesium Oxide 0.50

Explanation:

Number of moles of NaOH= number of moles × volume

Number of moles= 100/1000 × 2 = 0.2 moles

Since;

2 moles of NaOH yield 1 mole of Al2O3

0.2 moles of NaOH will yield 0.2 × 1/2 = 0.1 moles of Al2O3.

Number of moles of HCl= 800/1000 × 2 = 1.6 moles

If 1 mole of Al2O3 requires 6 moles of HCl

0.1 moles of Al2O3 requires 0.1 × 6 = 0.6 moles of HCl.

Number of moles of HCl left after reaction with Al2O3 = 1.6- 0.6 = 1 mole

This leftover reacts with MgO

But;

1 mole of MgO reacts with 2 moles of HCl

x moles of MgO reacts with 1 mole of HCl

Thus; x= 0.5 moles of MgO

8 0

2 years ago

Determine how many grams of silver would be produced, if 12.83 x 10^23 atoms of copper react with an excess of silver nitrate. G

AnnyKZ [126]

1) Chemical equation

Cu + 2AgNO3 ---> Cu (NO3)2 + 2Ag

2) molar ratios

1 mol Cu: 2 moles AgNO3 : 1 mol Cu (NO3)2 : 2 mol Ag

3) Convert 12. 83 * 10^23 atoms of Cu in moles

12.83 * 10 ^ 23 atoms / (6.02 * 10^23 atoms / mol) = 2.131 mol Cu

4) Use the proportions

2.131 mol Cu * 2 mol Ag / 1 mol Cu = 4.262 mol Ag

5) Use the atomic mass of silver to convert 4.262 mol in grams

mass = number of moles * atomic mass = 4.262 mol * 107.9 g / mol = 459.9 grams

Answer: 459.9 g

5 0

2 years ago

why does every human measurement have uncertainty associated with it? Why must we manage the uncertainty during calculations?

dimulka [17.4K]

Because there are many numbers of the human measurement . we must manage the uncertainly doing calculations because we can know what we are calculating.

3 0

2 years ago

Describe how you would prepare exactly 100 mL of 0.109 M picolinate buffer, pH 5.61. Possible starting materials are pure picoli

Pepsi [2]

Answer:

1.342g of picolinic acid and 6.743mL of 1.0M NaOH diluting the mixture to 100.0mL

Explanation:

<em>The pKa of the picolinic acid is 5.4.</em>

Using Henderson-Hasselbalch formula for picolinic-picolinate buffer:

pH = pKa + log [Picolinate] / [Picolinic]

<em>Where [] could be taken as moles of each species</em>

<em />

5.61 = 5.4 + log [Picolinate] / [Picolinic]

0.21 = log [Picolinate] / [Picolinic]

1.62181 = [Picolinate] / [Picolinic] <em>(1)</em>

Now, both picolinate and picolinic acid will be:

0.100L * (0.109mol / L) =

0.0109 moles = [Picolinate] + [Picolinic] <em>(2)</em>

First, as we will start with picolinic acid, we need add:

0.0109 moles picolinic acid * (123.10g/mol) = 1.342g of picolinic acid

Now, replacing (2) in (1):

1.62181 = 0.0109 moles - [Picolinic] / [Picolinic]

1.62181 [Picolinic] = 0.0109 moles - [Picolinic]

2.62181 [Picolinic] = 0.0109 moles

[Picolinic] = 4.157x10⁻³ moles

And:

[Picolinate] = 0.0109 - 4.157x10⁻³ moles =

<h3>6.743x10⁻³ moles</h3><h3 />

To obtain these moles of picolinate ion we need to make the reaction of the picolinic acid with NaOH:

Picolinic acid + NaOH → Picolinate + Water

<em>That means to obtain 6.743x10⁻³ moles of picolinate ion we need to add 6.743x10⁻³ moles of NaOH</em>

<em />

6.743x10⁻³ moles of NaOH that is 1.0M are, in mL:

6.743x10⁻³ moles * (1L / 1mol) = 6.743x10⁻³L * 1000 =

<h3>6.743mL of the 1.0M NaOH must be added</h3><h3 />

Thus, we obtain the desire moles of picolinate and picolinic acid to obtain the buffer we want, the last step is:

<h3>Dilute the mixture to 100mL, the volume we need to prepare</h3>

3 0

2 years ago