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MrRissso [65]
3 years ago
9

How many grams of calcium are in 2.65 g of CaCO3?

Chemistry
1 answer:
myrzilka [38]3 years ago
3 0

Answer:

2.56

Explanation:

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HELP ILL GIVE U FREE POINTS AND BRAINLEST !! Observe the plate boundary shown here. What type of plate boundary is this? Many ge
Studentka2010 [4]
The answer is B.
You can rule C out because divergent means moving away. Rule out A because there is an oceanic and continental plate, not 2 of the same type. Rule D out for the same reason.
3 0
3 years ago
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A researcher recorder that a chemical reaction require 2.90 gal water and raised the temperature of the reaction vessel by 57°F.
Ivan

Answer:

10.875L

Explanation:

The problem here is a simple conversion. The conversion is from gal to liters Liter is a SI unit for recording volume as stated in the problem.

Given that:

              1 gallon of water = 3.75L

              2.90 gallon will be 2.9 x 3.75; 10.875L

4 0
3 years ago
The half-life of radium-226 is 1590 years. (a) A sample of radium-226 has a mass of 50 mg. Find a formula for the mass of the sa
Maksim231197 [3]

Answer:

Explanation:

a )

m = m₀ e^{-\lambda t

m is mass after time t . original mass is m₀ , λ is disintegration constant

λ = .693 / half life

= .693 / 1590

= .0004358

m = m₀ e^{- 0.0004358 t}

b )

m = 50 x e^{-.0004358\times 500}

= 40.21 mg .

c )

40 = 50 e^{-.0004358t

.8 = e^{-.0004358t

e^{.0004358t = 1.25

.0004358 t = .22314

t = 512 years .

4 0
3 years ago
The decomposition of carbon disulfide to carbon monosulfide and sulfur is first order with k=2.8 ×10^-7 at 1000°C .What is the h
RoseWind [281]

Answer:

2.5×10⁶ s

Explanation:

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

Rate constant (K) = 2.8×10¯⁷ s¯¹

Half-life (t½) =?

The half-life of a first order reaction is given by:

Half-life (t½) = 0.693 / Rate constant (K)

t½ = 0.693 / K

With the above formula, we can obtain the half-life of the reaction as follow:

Rate constant (K) = 2.8×10¯⁷ s¯¹

Half-life (t½) =?

t½ = 0.693 / K

t½ = 0.693 / 2.8×10¯⁷

t½ = 2.5×10⁶ s

Therefore, the half-life of the reaction is 2.5×10⁶ s

6 0
3 years ago
Calculate the hydronium ion concentration in an aqueous solution with a poh of 4.33 at 25°c.
Jlenok [28]

Taking into account the definition of pH and pOH, the hydronium ion concentration in an aqueous solution with a pOH of 4.33 at 25°c is 2.138×10⁻¹⁰ M.

<h3>Definition of pH</h3>

pH is a measure of acidity or alkalinity that indicates the amount of hydrogen ions present in a solution or substance.

The pH is defined as the negative base 10 logarithm of the activity of hydrogen ions, that is, the concentration of hydrogen ions or  hydronium ion H₃O⁺:

pH= - log [H⁺]= - log [H₃O⁺]

<h3>Definition of pOH</h3>

Similarly, pOH is a measure of hydroxyl ions in a solution and is expressed as the logarithm of the concentration of OH⁻ ions, with the sign changed:

pOH= - log [OH⁻]

<h3>Relationship between pH and pOH</h3>

The following relationship can be established between pH and pOH:

pOH + pH= 14

<h3>Concentration of hydronium ions</h3>

Being pOH= 4.33, pH is calculated as:

pH + 4.33= 14

pH= 14 - 4.33

<u><em>pH= 9.67</em></u>

Replacing in the definition of pH the concentration of hydronium ions is obtained:

- log [H₃O⁺]= 9.67

Solving:

[H₃O⁺]= 10⁻⁹ ⁶⁷

<u><em>[H₃O⁺]= 2.138×10⁻¹⁰ M</em></u>

Finally, the hydronium ion concentration in an aqueous solution with a pOH of 4.33 at 25°c is 2.138×10⁻¹⁰ M.

Learn more about pH and pOH:

brainly.com/question/16032912

brainly.com/question/13557815

#SPJ1

4 0
2 years ago
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