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3 years ago

Which determine the charge of an atom?

Chemistry

1 answer:

Andre45 [30]3 years ago

8 0

Answer:

the answer is B its the number of protons and neutrons

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1.2 moles of (NH4)3PO3

Aleks04 [339]

1.2 moles of (nph4)3po3 is.......159.6 grams

3 0

3 years ago

1. Which substance has the highest boiling point?

Katyanochek1 [597]

Answer:

your answer would be the last one gold

5 0

2 years ago

PLEASE HELP WILL GIVE 20PTS AND WILL MARK BRAINLIEST FOR RIGHT ANSWER!!!

Marysya12 [62]

Answer:

C) 0.457

Explanation:

The ratio between O2 and H2O is 1:2 according to the balanced equation. You can find how many moles is O2 by : 5.12/22.4 = 0.22857 ( 1 mole = 22.4 litters)

Moles of H2O will be 0.22857 * 2 = 0.457142.

Therefore answer C)

7 0

3 years ago

What volume of O2 collected at 22.0 and 728 mmHg would be produce by the decomposition of 8.15 g KClO3?

adell [148]

Answer:

There is 2.52 L of O2 collected

Explanation:

Step 1: Data given:

Temperature = 22.0 °C

Pressure = 728 mmHg = 728 /760 = 0.958 atm

Mass of KClO3 = 8.15 grams

Molar mass of KClO3 = 122.55 g/mol

Step 2: The balanced equation

2KClO3(s) → 2KCl(s) + 3O2(g)

Step 3: Calculate moles of KClO3

Moles KClO3 = mass KClO3 / molar mass KClO3

Moles KClO3= 8.15 grams / 122.55 g/mol

Moles KClO3 = 0.0665 moles

Step 4: Calculate moles of O2

For 2 moles of KClO3 we'll have 2 moles of KCl and 3 moles of O2 produced

For 0.0665 moles of KClO3 we have 3/2 * 0.0665 = 0.09975 moles

Step 5: Calculate vlume of O2

p*V = n*R*T

V = (n*R*T)/p

⇒ with n = the number of moles O2 = 0.09975 moles

⇒ with R = the gas constant = 0.08206 L*atm/K*mol

⇒ with T = 22.0 °C = 273 +22 = 295 Kelvin

⇒ with p = 0.958 atm

V = (0.09975 * 0.08206 * 295) / 0.958

V = 2.52 L

There is 2.52 L of O2 collected

7 0

3 years ago

If 10.0 mL of a .600 M of HNO3 reacts with 31.0 mL of .700M Ba(OH)2 solution, what is the molarity of Ba(OH)2 after the reaction

Tasya [4]

Answer:

0.456M

Explanation:

1. Balanced molecular equation

$2HNO_3+Ba(OH)_2\rightarrow Ba(NO_3)_2+2H_2O$

2. Mole ratio

$\dfrac{2molHNO_3}{1molBa(OH)_2}$

3. Moles of HNO₃

Number of moles = Molarity × Volume in liters
n = 0.600M × 0.0100 liter = 0.00600 mol HNO₃

4. Moles Ba(OH)₂

n = 0.700M × 0.0310 liter = 0.0217 mol

5. Limiting reactant

Actual ratio:

$\dfrac{0.0600molHNO_3}{0.0217molBa(OH)_2}\approx0.28$

Since the ratio of the moles of HNO₃ available to the moles of Ba(OH)₂ available is less than the theoretical mole ratio, HNO₃ is the limiting reactant.

Thus, 0.006 moles of HNO₃ will react completely with 0.003 moles of Ba(OH)₂ and 0.0217 - 0.003 = 0.0187 moles will be left over.

6. Final molarity of Ba(OH)₂

Molarity = number of moles / volume in liters
Molarity = 0.0187 mol / (0.0100 + 0.0031) liter = 0.456M

5 0

3 years ago