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Papessa [141]
3 years ago
12

MgCI2 how many element ?

Chemistry
1 answer:
polet [3.4K]3 years ago
7 0

Answer:

2 elements magnesium(Mg) and chlorine (Cl)

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A silver rod and a SHE are dipped into a saturated aqueous solution of silver oxalate, Ag2C2O4, at 25°C. The measured potential
nika2105 [10]

Answer:

3.50*10^-11 mol3 dm-9

Explanation:

A silver rod and a SHE are dipped into a saturated aqueous solution of silver oxalate, Ag2C2O4, at 25°C. The measured potential difference between the rod and the SHE is 0.5812 V, the rod being positive. Calculate the solubility product constant for silver oxalate.

Ag2C2O4 -->  2Ag+  +  C2O4 2-

So Ksp = [Ag+]^2 * [C2O42-]

In 1 L, 2.06*10^-4 mol of silver oxalate dissolve, giving, the same number of mol of oxalate ions, and twice the number of mol (4.12*10^-4) of silver ions.

So Ksp = (4.12*10^-4)^2 * (2.06*10^-4)

= 3.50*10^-11 mol3 dm-9

7 0
3 years ago
How many molecules of sugar are there in 1 can of soda?
Korvikt [17]

Answer:

5.8\cdot 10^{22}

Explanation:

Since no information was provided about the can of soda, let's take an example of 1 can of 12 fl oz (368 g) Coca-Cola.

The nutrition facts state that it contains 33 g of sugar. In order to calculate the number of molecules, we firstly need to know the molecular formula of sugar. Sugar can be represented by C_{12}H_{22}O_{11}.

Our first step is to find the molar mass of sugar:

M_{C_{12}H_{22}O_{11}} = 342.3 g/mol

Secondly, dividing mass of sugar, m = 33 g, by the molar mass will yield the number of moles of sugar:

n_{C_{12}H_{22}O_{11}}=\frac{m}{M}

Finally, multiplying moles by the Avogadro's constant will yield the number of molecules:

N_{C_{12}H_{22}O_{11}}=n_{C_{12}H_{22}O_{11}}\cdot N_A =\frac{m}{M} \cdot N_A = \frac{33 g}{342.3 g/mol} \cdot 6.022 \cdot 10^{23} mol^{-1} = 5.8\cdot 10^{22}

8 0
4 years ago
Gifblaar is a small South African shrub and one of the most poisonous plants known because it contains fluoroacetic acid (FCH2CO
PSYCHO15rus [73]

[H_{3}O^{+}] = 0.00770 M

The equilibrium equation representing the dissociation of FCH_{2}COOH

FCH_{2}COOH(aq) + H_{2}O (l)    FCH_{2}COO^{-}(aq)+ H_{3}O^{+}(aq)

Given [H_{3}O^{+}] = 0.00770 M

Let the initial concentration of acid be x and change y

So y = [H_{3}O^{+}] =[FCH_{2}COO^{-}] = 0.00770 M

pK_{a} = 2.59K_{a} = 10^{-2.59}   = 0.00257 M

K_{a} = \frac{(0.00770 M)(0.00770 M)}{x - 0.00770}

0.00257 = \frac{0.00005929}{x - 0.00770}

0.00257 x - 0.00001979 = 0.00005929

x = 0.031 M

Therefore, initial concentration of the weak acid is <u>0.031 M</u>

4 0
4 years ago
Do Metals or Nonmetals have larger ion size compared to the size of the neutral atom?
Stella [2.4K]

Answer:

b

Explanation:

3 0
3 years ago
Read 2 more answers
Can someone please help its science
anastassius [24]

Answer:

c

Explanation:

6 0
3 years ago
Read 2 more answers
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