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harina [27]
2 years ago
13

30 mL of 0.6 M HCl solution is neutralized with 90 mL NaOH solution. What is the concentration of the base? Show all work Please

:).
Chemistry
1 answer:
skad [1K]2 years ago
3 0

From

<em>n</em><em>a</em><em>=</em><em>1</em>

<em>n</em><em>b</em><em>=</em><em>1</em>

<em>V</em><em>a</em><em>=</em><em>3</em><em>0</em><em>m</em><em>l</em><em>s</em>

<em>V</em><em>b</em><em>=</em><em>9</em><em>0</em><em>m</em><em>l</em><em>s</em>

<em>M</em><em>a</em><em>=</em><em>0</em><em>.</em><em>6</em><em>M</em>

<em>M</em><em>b</em><em>=</em><em>?</em>

<em>f</em><em>r</em><em>o</em><em>m</em>

<em>Mb</em><em>=</em><em><u>M</u></em><em><u>a</u></em><em><u>×</u></em><em><u>V</u></em><em><u>a</u></em><em><u>×</u></em><em><u>n</u></em><em><u>b</u></em>

<em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em>V</em><em>b</em><em>×</em><em>n</em><em>a</em>

<em>M</em><em>b</em><em>=</em><em><u>0</u></em><em><u>.</u></em><em><u>6</u></em><em><u>×</u></em><em><u>3</u></em><em><u>0</u></em><em><u>×</u></em><em><u>1</u></em>

<em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em>9</em><em>0</em><em>×</em><em>1</em>

<em>m</em><em>b</em><em>=</em><em><u>1</u></em><em><u>8</u></em>

<em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em>9</em><em>0</em>

<em>M</em><em>b</em><em>=</em><em>0</em><em>.</em><em>2</em><em>M</em>

<em>NaOH</em><em>=</em><em>4</em><em>0</em>

<em>F</em><em>r</em><em>o</em><em>m</em>

<em>c</em><em>o</em><em>n</em><em>c</em><em>=</em><em>m</em><em>o</em><em>l</em><em>a</em><em>r</em><em>i</em><em>t</em><em>y</em><em>×</em><em>m</em><em>r</em>

<em>c</em><em>o</em><em>n</em><em>c</em><em>=</em><em>0</em><em>.</em><em>2</em><em>×</em><em>4</em><em>0</em>

<em>c</em><em>o</em><em>n</em><em>c</em><em>=</em><em>8</em><em>g</em><em>/</em><em>m</em><em>o</em><em>l</em>

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<span>only D-glucose is found in disaccharides and polysaccharides.</span>
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Calculate the atomic mass of lead. The four lead isotopes have atomic masses and relative abundances of 203.973 amu (1.4%), 205.
blondinia [14]
207.217 amu
Work:
203.973 amu *(0.014) = 2.855 amu
205.974 amu *(0.241) = 49.639 amu
206.976 amu *(0.221) = 45.741 amu
207.977 amu *(0.524) = 108.979 amu

2.855 + 49.639 + 45.741 + 108.979 = 207.217 amu
8 0
3 years ago
I just started learning about kinetic molecular theory, and I’m not sure how to answer the question circled below
lions [1.4K]

Answer : The value of 'R' is 0.0821\text{ L atm }mol^{-1}K^{-1}

Solution : Given,

At STP conditions,

Pressure = 1 atm

Temperature = 273 K

Number of moles = 1 mole

Volume = 22.4 L

Formula used :     R=\frac{PV}{nT}

where,

R = Gas constant

P = pressure of gas

T = temperature of gas

V = volume of gas

n = number of moles of gas

Now put all the given values in this formula, we get the values of 'R'.

R=\frac{(1atm)\times (22.4L)}{(1mole)\times (273K)}

R=0.0821\text{ L atm }mol^{-1}K^{-1}

Therefore, the value of 'R' is 0.0821\text{ L atm }mol^{-1}K^{-1}.

7 0
3 years ago
Match each transition metal ion with its condensed ground-state electron configuration. A [Ar]3d2 B [Ar]4s23d3 C [Kr]4d10 D [Xe]
madreJ [45]

Answer:

Mn^{2+} : - F . [Ar]3d^{5}

Hg^{2+} : - G. [Xe]4f^{14}5d^{10}

La^{3+} : - D. [Xe]

Fe^{3+} : - F. [Ar]3d^{5}

Ag^{+} : - C. [Kr]4d^{10}

Co^{3+} : - E. [Ar]3d^{6}

Explanation:

The electronic configuration of the element Mn is:-

[Ar]3d^{5}4s^2

For, Mn^{2+}, 2 electrons are lost, thus the configuration is:-

[Ar]3d^{5}

The electronic configuration of the element Hg is:-

[Xe]4f^{14}5d^{10}6s^2

For, Hg^{2+}, 2 electrons are lost, thus the configuration is:-

[Xe]4f^{14}5d^{10}

The electronic configuration of the element La is:-

[Xe]5d^{1}6s^2

For, La^{3+}, 3 electrons are lost, thus the configuration is:-

[Xe]

The electronic configuration of the element Fe is:-

[Ar]3d^{6}4s^2

For, Fe^{3+}, 3 electrons are lost, thus the configuration is:-

[Ar]3d^{5}

The electronic configuration of the element Ag is:-

[Kr]4d^{10}5s^1

For, Ag^{+}, 1 electron is lost, thus the configuration is:-

[Kr]4d^{10}

The electronic configuration of the element Co is:-

[Ar]3d^{7}4s^2

For, Co^{3+}, 3 electrons are lost, thus the configuration is:-

[Ar]3d^{6}

7 0
3 years ago
A marble is L 1.32 W 1.32 H 1.32, what is the volume? Show your work
cupoosta [38]

The volume of the marble is 2.3 cube units.

From the question given above, the following data were obtained:

  • Length (L) = 1.32
  • Width (W) = 1.32
  • Height (H) = 1.32
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The volume of the marble can be obtained as follow:

V = L × W × H

V = 1.32 × 1.32 × 1.32

V = 2.3 cube units

Thus, the volume of the marble is 2.3 cube units.

Learn more about volume:

brainly.com/question/20665757

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