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2 years ago

How many sodium atoms does 3Na2S2O3 have?

Chemistry

1 answer:

zhuklara [117]2 years ago

4 0

Answer:

Element, Symbol, Atomic Mass, # of Atoms, Mass Percent. Sodium, Na, 137.9386 g/mol, 6, 29.0811%. Sulphur/Sulfur, S, 192.39 g/mol, 6, 40.5609%Explanation:

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What is the mass of solute dissolved in 50,0 g of 12.5% saline solution? A) 7.148 B) 6.258 C) 43.88 D) 4.00 g

Marizza181 [45]

Answer:

6.25 grams is the mass of solute dissolved.

Explanation:

w/w % : The percentage mass or fraction of mass of the of solute present in total mass of the solution.

$w/w\%=\frac{\text{Mass of solute}}{\text{Mass of solution}}\times 100$

Mass of the solution = 50.0 g

Mass of the solvent = x

w/w % = 12.5%

$12.5\%=\frac{x}{50.0 g}\times 100$

x = 6.25 g

6.25 grams is the mass of solute dissolved.

6 0

3 years ago

17 points Pls helpppppopppp

pentagon [3]

Answer:

Lipids

Explanation:

Lipids perform many different functions in a cell. Cells store energy for long-term use in the form of lipids called fats. Lipids also provide insulated from the environment for plants and animals

8 0

3 years ago

Given pH = 3.50 Find: [H3O+] and [OH-] Is this acidic, basic or neutral?

lys-0071 [83]

Answer:

Explanation:

Given parameters:

pH = 3.50

Unknown:

concentration of [H₃0⁺] = ?

concentration of [OH⁻] = ?

Solution:

In order to find the unknown, we use some simple expressions which best explains the pH scale and the equilibrium systems of aqueous solutions.

pH = -log₁₀[H₃O⁺]

[H₃O⁺] = inverse log₁₀ (-pH) = $10^{-pH}$ = $10^{-3.5}$

[H₃O⁺] = 3.2 x 10⁻⁴moldm⁻³

For the [OH⁻]:

we use : pOH = -log₁₀ [OH⁻]

Recall: pOH + pH = 14

pOH = 14 - pH = 14 - 3.5 = 10.5

Now we plug the value of pOH into pOH = -log₁₀ [OH⁻]

[OH⁻] = $10^{-pOH}$

[OH⁻] = $10^{-10.5}$ = 3.2 x 10⁻¹¹moldm⁻³

The solution is acidic as the concentration of H₃0⁺ is more than that of the OH⁻ ions.

8 0

3 years ago

What’s the answer ?

Romashka-Z-Leto [24]

Answer:

45.3°C

Explanation:

Step 1:

Data obtained from the question.

Initial pressure (P1) = 82KPa

Initial temperature (T1) = 26°C

Final pressure (P2) = 87.3KPa.

Final temperature (T2) =.?

Step 2:

Conversion of celsius temperature to Kelvin temperature.

This is illustrated below:

T(K) = T(°C) + 273

Initial temperature (T1) = 26°C

Initial temperature (T1) = 26°C + 273 = 299K.

Step 3:

Determination of the new temperature of the gas. This can be obtained as follow:

P1/T1 = P2/T2

82/299 = 87.3/T2

Cross multiply to express in linear form

82 x T2 = 299 x 87.3

Divide both side by 82

T2 = (299 x 87.3) /82

T2 = 318.3K

Step 4:

Conversion of 318.3K to celsius temperature. This is illustrated below:

T(°C) = T(K) – 273

T(K) = 318.3K

T(°C) = 318.3 – 273

T(°C) = 45.3°C.

Therefore, the new temperature of the gas in th tire is 45.3°C

6 0

3 years ago

Write a balanced equation for the complete combustion of each compound. a. formaldehyde (CH2O(g)) b. heptane (C7H17(l)) c. benze

Yuri [45]

Answer:

a. formaldehyde (CH₂O(g)): CH₂O(g) + O₂(g) → CO₂(g) + H₂O(g)

b. heptane (C₇H₁₇(l)): 4C₇H₁₇(l) + 45O₂(g) → 28CO₂(g) + 34H₂O(g)

c. benzene (C6H6(l)): C₆H₆(l) + 15/2 O₂(g) → 6CO₂(g) + 3 H₂O(g)

Explanation:

In a reaction of combustion, a hydrocarbon compound (composed of C, H and O) reacts with oxygen gas (O₂). The <u>complete</u> combustion of a hydrocarbon - such as formaldehyde, heptane, and benzene - produces carbon dioxide (CO₂) and water (H₂O).

Thus, we write the reactants and products for each combustion reaction and then we balance the atoms: C, H, and O.

a. formaldehyde (CH₂O(g)):

CH₂O(g) + O₂(g) → CO₂(g) + H₂O(g)

The chemical equation is already balanced with the coefficients 1: we have the same number of C atoms (1), H (2) and O (3) on both sides of the equation.

b. heptane (C₇H₁₇(l)):

C₇H₁₇(l) + O₂(g) → CO₂(g) + H₂O(g)

Here we have to write a coefficient 28 for CO₂ to balance the C atoms in the products side, and for C₇H₁₇ we write 28/4 = 7. With similar reasoning we found the coefficients for O₂ and H₂O:

4C₇H₁₇(l) + 45O₂(g) → 28CO₂(g) + 34H₂O(g)

c. benzene (C6H6(l)):

C₆H₆(l) + O₂(g) → CO₂(g) + H₂O(g)

First, we write a coefficient of 6 for CO₂ to balance the C atoms. Then, we have to balance H atoms: we write a coefficient 3 in H₂O. Now, we have 12 + 3 = 15 atoms of O on the reactants side. So, we write a half of these number of atoms in the coefficient for O₂: 15/2. We obtain the balanced equation:

C₆H₆(l) + 15/2 O₂(g) → 6CO₂(g) + 3 H₂O(g)

5 0

3 years ago