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nasty-shy [4]
3 years ago
13

A chemist adds 405.0mL of a 0.79M sodium carbonate solution to a reaction flask. Calculate the millimoles of sodium carbonate th

e chemist has added to the flask. Be sure your answer has the correct number of significant digits. Answer must be in mmol
Chemistry
1 answer:
Sophie [7]3 years ago
4 0

Answer:

320 mmol

Explanation:

405.0 mL * 1L/1000 mL = 0.4050 L

0.79 M = 0.79 mol/L

0.79 mol/L * 0.4050 L=0.32 mol

0.32 mol * 10^3 mmol/1 mol = 320 mmol

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Las sustancias ionicas y los metale sson electricamente neutros. ¿como es esto posible si estan formados por iones, que posen ca
Anton [14]

Answer:

ver explicacion

Explanation:

Los iones se forman cuando las especies químicas pierden o ganan electrones.

Las sustancias iónicas consisten en un ión positivo y negativo cuyas cargas se equilibran exactamente entre sí, por lo que el compuesto iónico es neutro.

Los átomos de metal se mantienen unidos por el enlace metálico. Esto implica la interacción entre iones metálicos cargados positivamente y un mar de electrones negativos. Las cargas positivas de los iones metálicos están exactamente equilibradas por el mar de electrones cargados negativamente, por lo que el metal es neutro.

8 0
2 years ago
If 4.0 g of helium gas occupies a volume of 22.4 L at 0 o C and a pressure of 1.0 atm, what volume does 3.0 g of He occupy under
WINSTONCH [101]

Answer:

the volume occupied by 3.0 g of the gas is 16.8 L.

Explanation:

Given;

initial reacting mass of the helium gas, m₁ = 4.0 g

volume occupied by the helium gas, V = 22.4 L

pressure of the gas, P = 1 .0 atm

temperature of the gas, T = 0⁰C = 273 K

atomic mass of helium gas, M = 4.0 g/mol

initial number of moles of the gas is calculated as follows;

n_1 = \frac{m_1}{M} \\\\n_1 = \frac{4}{4} = 1

The number of moles of the gas when the reacting mass is 3.0 g;

m₂ = 3.0 g

n_2 = \frac{m_2}{M} \\\\n_2 = \frac{3}{4} \\\\n_2 = 0.75 \ mol

The volume of the gas at 0.75 mol is determined using ideal gas law;

PV = nRT

PV = nRT\\\\\frac{V}{n} = \frac{RT}{P} \\\\since, \ \frac{RT}{P} \ is \ constant,\  then;\\\frac{V_1}{n_1} = \frac{V_2}{n_2} \\\\V_2 = \frac{V_1n_2}{n_1} \\\\V_2 = \frac{22.4 \times 0.75}{1} \\\\V_2 = 16.8 \ L

Therefore, the volume occupied by 3.0 g of the gas is 16.8 L.

4 0
3 years ago
What is the value for ∆Soreaction for the following reaction, given the standard entropy values? 2H2S(g) + SO2(g) 3Srhombic(s) +
Mademuasel [1]

Answer: \Delta S^{0} for the reaction is -186.75 J/K

Explanation:

Change in entropy (\Delta S^{0}) for the given reaction under standard condition is given by-

\Delta S^{0}= [3\times S_{rhombic}^{0}_{(s)}]+[2\times S_{H_{2}O}^{0}_{(g)}]-[2\times S_{H_{2}S}^{0}_{(g)}]-[1\times S_{SO_{2}}^{0}_{(g)}]

So \Delta S^{0} = [3\times 31.8 J/K.mol]+[2\times 188.825 J/K.mol]-[2\times 205.79 J/K.mol]-[1\times 248.22 J/K.mol] = -186.75 J/K

5 0
2 years ago
How to do q solution, qrxn, moles of Mg , and delta Hrxn?
Helga [31]

Answer:

<em> 14, 508J/K</em>

ΔHrxn =q/n

where q = heat absorbed and n = moles

Explanation:

<em>m = mass of substance (g) = 0.1184g</em>

1 mole of Mg - 24g

<em>n</em> moles - 0.1184g

<em>n = 0.0049 moles.</em>

Also, q = m × c × ΔT

<em> Heat Capacity, C of MgCl2 = 71.09 J/(mol K)</em>

<em>∴ specific heat c of MgCL2 = 71.09/0.0049 (from the formula c = C/n)</em>

<em>= 14, 508 J/K/kg</em>

ΔT=  (final - initial) temp = 38.3 - 27.2

= 11.1 °C.

mass of MgCl2 = 95.211 × 0.1184 = 11.27

⇒ q = 11.27g × 11.1 °C × <em>14, 508 j/K/kg </em>

<em>= 1,7117.7472 J °C-1 g-1</em>

<em />

<em>∴ ΔHrxn = q/n</em>

<em>=1,7117.7472  ÷ 0.1184 </em>

<em>= 14, 508J/K</em>

6 0
3 years ago
*click image*
nadya68 [22]

Answer:

Its be 1,2 and 4.

Explanation:

Since changing the direction won't do anything to a electric magnet.

6 0
3 years ago
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