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2 years ago

How many millimeters are in 251 cm?

Chemistry

2 answers:

Serga [27]2 years ago

4 0

Answer:

2510 mm

Explanation:

1 cm = 10 mm

251 cm = 10mm×251 = 2510 mm

antoniya [11.8K]2 years ago

3 0

Answer:

2510

Explanation:

Since there are 10 milimetres in 1 centimetre, 1 cm = 10mm

There are 10 times as many milimetres as centimetres.

251cm X 10 = mm

251cm = 2510mm

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Which layer of the earth is the thinnest and thickest?

Xelga [282]

Answer:

crust

Discuss with the whole class what the relative thicknesses of the layers are — that the inner core and outer core together form the thickest layer of the Earth and that the crust is by far the thinnest layer.

Explanation:

6 0

3 years ago

Start with 100.00 mL of 0.10 M acetic acid, CH3COOH. The solution has a pH of 2.87 at 25 oC. a) Calculate the Ka of acetic acid

jasenka [17]

Answer: a) The $K_a$ of acetic acid at $25^0C$ is $1.82\times 10^{-5}$

b) The percent dissociation for the solution is $4.27\times 10^{-3}$

Explanation:

$CH_3COOH\rightarrow CH_3COO^-H^+$

cM 0 0

$c-c\alpha$ $c\alpha$ $c\alpha$

So dissociation constant will be:

$K_a=\frac{(c\alpha)^{2}}{c-c\alpha}$

Give c= 0.10 M and $\alpha$ = ?

Also $pH=-log[H^+]$

$2.87=-log[H^+]$

$[H^+]=1.35\times 10^{-3}M$

$[CH_3COO^-]=1.35\times 10^{-3}M$

$[CH_3COOH]=(0.10M-1.35\times 10^{-3}=0.09806M$

Putting in the values we get:

$K_a=\frac{(1.35\times 10^{-3})^2}{(0.09806)}$

$K_a=1.82\times 10^{-5}$

b) $\alpha=\sqrt\frac{K_a}{c}$

$\alpha=\sqrt\frac{1.82\times 10^{-5}}{0.10}$

$\alpha=4.27\times 10^{-5}$

$\% \alpha=4.27\times 10^{-5}\times 100=4.27\times 10^{-3}$

5 0

3 years ago

Calculate the solubilities of the following compounds in a 0.02 M solution of barium nitrate using molar concentrations, first i

Law Incorporation [45]

Answer:

a. 1.7 × 10⁻⁴ mol·L⁻¹; b. 5.5 × 10⁻⁹ mol·L⁻¹

c. 2.3 × 10⁻⁴ mol·L⁻¹; 5.5 × 10⁻⁸ mol·L⁻¹

Explanation:

a. Silver iodate

Let s = the molar solubility.

AgIO₃(s) ⇌ Ag⁺(aq) + IO₃⁻(aq); Ksp = 3.0 × 10⁻⁸

E/mol·L⁻¹: s s

$K_{sp} =\text{[Ag$^{+}$][IO$_{3}$$^{-}$]} = s\times s = s^{2} = 3.0\times 10^{-8}\\s = \sqrt{3.0\times 10^{-8}} \text{ mol/L} = 1.7 \times 10^{-4} \text{ mol/L}$

b. Barium sulfate

BaSO₄(s) ⇌ Ba²⁺(aq) + SO₄²⁻(aq); Ksp = 1.1 × 10⁻¹⁰

I/mol·L⁻¹: 0.02 0

C/mol·L⁻¹: +s +s

E/mol·L⁻¹: 0.02 + s s

$K_{sp} =\text{[Ba$^{2+}$][SO$_{4}$$^{2-}$]} = (0.02 + s) \times s \approx 0.02s = 1.1\times 10^{-10}\\s = \dfrac{1.1\times 10^{-10}}{0.02} \text{ mol/L} = 5.5 \times 10^{-9} \text{ mol/L}$

c. Using ionic strength and activities

(i) Calculate the ionic strength of 0.02 mol·L⁻¹ Ba(NO₃)₂

The formula for ionic strength is

$\mu = \dfrac{1}{2} \sum_{i} {c_{i}z_{i}^{2}}\\\\\mu = \dfrac{1}{2} (\text{[Ba$^{2+}$]}\cdot (2+)^{2} + \text{[NO$_{3}$$^{-}$]}\times(-1)^{2}) = \dfrac{1}{2} (\text{0.02}\times 4 + \text{0.04}\times1)= \dfrac{1}{2} (0.08 + 0.04)\\\\= \dfrac{1}{2} \times0.12 = 0.06$

(ii) Silver iodate

a. Calculate the activity coefficients of the ions

$\log \gamma = -0.51z^{2}\sqrt{I} = -0.051(1)^{2}\sqrt{0.06} = -0.51\times 0.24 = -0.12\\\gamma = 10^{-0.12} = 0.75$

b. Calculate the solubility

AgIO₃(s) ⇌ Ag⁺(aq) + IO₃⁻(aq)

$K_{sp} =\text{[Ag$^{+}$]$\gamma_{Ag^{+}}$[IO$_{3}$$^{-}$]$\gamma_{IO_{3}^{-}}$} = s\times0.75\times s \times 0.75 =0.56s^{2}= 3.0 \times 10^{-8}\\s^{2} = \dfrac{3.0 \times 10^{-8}}{0.56} = 5.3 \times 10^{-8}\\\\s =2.3 \times 10^{-4}\text{ mol/L}$

(iii) Barium sulfate

a. Calculate the activity coefficients of the ions

$\log \gamma = -0.51z^{2}\sqrt{I} = -0.051(2)^{2}\sqrt{0.06} = -0.51\times16\times 0.24 = -0.50\\\gamma = 10^{-0.50} = 0.32$

b. Calculate the solubility

BaSO₄(s) ⇌ Ba²⁺(aq) + SO₄²⁻(aq

$K_{sp} =\text{[Ba$^{2+}$]$\gamma_{ Ba^{2+}}$[SO$_{4}$$^{2-}$]$\gamma_{ SO_{4}^{2-}}$} = (0.02 + s) \times 0.32\times s\times 0.32 \approx 0.02\times0.10s\\2.0\times 10^{-3}s = 1.1 \times 10^{-10}\\s = \dfrac{1.1\times 10^{-10}}{2.0 \times 10^{-3}} \text{ mol/L} = 5.5 \times 10^{-8} \text{ mol/L}$

7 0

2 years ago

How many atoms of each element does sodium carbonate, Na2CO3, contain?

vfiekz [6]

Sodium carbonate contains 3 elements: Na, C, and O. Step 2: Determine how many atoms of each element are in the formula. Na, Co, is made up of Na, C, and O3, so Na.co, has 2 atoms of sodium (Na) 1 atom of carbon (C) and 3 atoms of oxygen (O).

7 0

2 years ago

Read 2 more answers

2. What would be the outcome of a unicellular organism that cannot maintain homeostasis? (1 point)

Juli2301 [7.4K]

Answer:

Option A = it dies

Explanation:

Homeostasis:

It is the ability of organisms to maintain the equilibrium in its internal environment in order to deal with external environmental changes.

If the unicellular organism cannot maintain the homeostasis it would die.

Importance of homeostasis:

It is constant chemical and physical condition of cell. The unicellular organisms in maintained homeostasis condition can grow and reproduce. They respond to the environment and transfer energy. It is the primary objective of unicellular organism other wise they would not survive.

It is very important for them because they totally rely on it and can perform the necessary functions of life.

While in case of multi cellular organisms all cells communicate with each other to maintain the homeostasis.

3 0

3 years ago