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german
3 years ago
14

Does anyone know the anserw the symbol of the element

Chemistry
1 answer:
Komok [63]3 years ago
3 0

Cesium.

Groups are the vertical columns that run up and down while periods are the horizontal rows. So to find the answer to this, go to the first column (Group 1) and find the sixth period (row 6) which will land you on Cesium, element 55.

Hope this helps!

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What is the octet rule?
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The Octet rule is a general rule of thumb that applies to most atoms. Basically, it states that every atom wants to have eight valence electrons in its outermost electron shell.


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What is the function of a Mitochondrion​
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Answer:

Explanation:

Function. The mitochondrion is the site of ATP synthesis for the cell. The number of mitochondria found in a cell are therefore a good indicator of the cell's rate of metabolic activity; cells which are very metabolically active, such as hepatocytes, will have many mitochondria.

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What is the density of a liquid if it's volume is 125 mL and it's mass is 50g?
masha68 [24]

Answer:

0.4g/ml

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6 0
3 years ago
I want to know the steps.
Artyom0805 [142]

The answer for the following problem is described below.

<em><u> Therefore the standard enthalpy of combustion is -2800 kJ</u></em>

Explanation:

Given:

enthalpy of combustion of glucose(ΔH_{f} of C_{6}H_{12} O_{6}) =-1275.0

enthalpy of combustion of oxygen(ΔH_{f} of O_{2}) = zero

enthalpy of combustion of carbon dioxide(ΔH_{f} of CO_{2}) = -393.5

enthalpy of combustion of water(ΔH_{f} of H_{2} O) = -285.8

To solve :

standard enthalpy of combustion

We know;

ΔH_{f}  = ∈ΔH_{f} (products) - ∈ΔH_{f} (reactants)

C_{6}H_{12} O_{6} (s) +6 O_{2}(g) → 6 CO_{2} (g)+ 6 H_{2} O(l)

ΔH_{f} = [6 (-393.5) + 6(-285.8)] - [6 (0) + (-1275)]

ΔH_{f} = [6 (-393.5) + 6(-285.8)] - [0 - 1275]

ΔH_{f} = 6 (-393.5) + 6(-285.8)  - 0 + 1275

ΔH_{f} = -2361 - 1714 - 0 + 1275

ΔH_{f} =-2800 kJ

<em><u> Therefore the standard enthalpy of combustion is -2800 kJ</u></em>

7 0
3 years ago
The osmotic pressure of a solution formed by dissolving 35.0 mg of aspirin (c9h8o4) in 0.250 l of water at 25°c is __________ at
Lunna [17]
Mass of aspirin = 0.025 g
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moles of aspirin = .025g / 180.1583 g/mol = 0.000138767 moles
volume solution = .250 L
molarity of the solution = 0.000138767 moles / .250L =5.551 x 10 ^-04 Moles / liter
for aspirin i = Vant'Hoff factor = 1 particle in solution
T = 25 + 273 =298 K
osmotic pressure = M x R x T x i =
5.551 x 10 ^-04 mole L -1 x 0.08206 L atm K−1 mol−1 x 298 K x 1 = 0.0136 atmospheres
6 0
3 years ago
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