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Nadya [2.5K]
3 years ago
7

How many moles of sodium will react with water to form sodium hydroxide and 4.0 moles of hydrogen?

Chemistry
1 answer:
11Alexandr11 [23.1K]3 years ago
5 0
It is 4/10 of moles is this ane halp?
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Which sample of matter is classified as a substance?<br> (1) air (3) milk(2) ammonia (4) seawater
jeyben [28]
A chemical substance has the characteristics that it cannot be separated by physical  methods. Seawater and milk can be separated by sedimentation, and air has different components depending on other aspects (such as elevation). Only ammonia is a substance. (thus it can have a formula: NH<span>3)</span>
3 0
3 years ago
what volume of a 0.149 m potassium hydroxide solution is required to neutralize 17.0 ml of a 0.112 m hydrobromic acid solution?
IgorLugansk [536]

Answer: 12.78ml

Explanation:

Given that:

Volume of KOH Vb = ?

Concentration of KOH Cb = 0.149 m

Volume of HBr Va = 17.0 ml

Concentration of HBr Ca = 0.112 m

The equation is as follows

HBr(aq) + KOH(aq) --> KBr(aq) + H2O(l)

and the mole ratio of HBr to KOH is 1:1 (Na, Number of moles of HBr is 1; while Nb, number of moles of KOH is 1)

Then, to get the volume of a 0.149 m potassium hydroxide solution Vb, apply the formula (Ca x Va)/(Cb x Vb) = Na/Nb

(0.112 x 17.0)/(0.149 x Vb) = 1/1

(1.904)/(0.149Vb) = 1/1

cross multiply

1.904 x 1 = 0.149Vb x 1

1.904 = 0.149Vb

divide both sides by 0.149

1.904/0.149 = 0.149Vb/0.149

12.78ml = Vb

Thus, 12.78 ml of potassium hydroxide solution is required.

5 0
3 years ago
if 45.0 ml of 1.50 M Ca(OH)2 are needed to neutralize 25.0 ml of HI of unknown concentration, what is the molarity of the HI?
ipn [44]

Answer:

M of HI = 5.4 M.

Explanation:

  • We have the rule: at neutralization, the no. of millimoles of acid is equal to the no. of millimoles of the base.

<em>(XMV) acid = (XMV) base.</em>

where, X is the no. of (H) or (OH) reproducible in acid or base, respectively.

M is the molarity of the acid or base.

V is the volume of the acid or base.

<em>(XMV) HI = (XMV) Ca(OH)₂.</em>

For HI; X = 1, M = ??? M, V = 25.0 mL.

For Ca(OH)₂, X = 2, M = 1.5 M, V = 45.0 mL.

<em>∴ M of HI = (XMV) Ca(OH)₂ / (XV) HI</em> = (2)(1.5 M)(45.0 mL) / (1)(25.0 mL) = <em>5.4 M.</em>

6 0
3 years ago
Calculate the theoretical yield of alum expected from 0.9875 g of aluminum foil. assume the aluminum is the limiting reactant.
skelet666 [1.2K]

Answer: 17.34 grams of alum will be produced if 0.9875 g of Aluminium foil was used.

Explanation: Reaction to form alum from Aluminium is given as:

2Al(s)+2KOH(aq.)+2H_2O(l)+4H_2SO_4(aq.)\rightarrow 2KAl(SO_4)_2(s).12H_2O(l)+3H_2(g)

We are given Aluminium to be the limiting reactant, so the formation of alum will be dependent on Aluminium because it limits the formation of product.

By stoichiometry,

2 moles of Al is producing 2 moles of Alum

Mass of 2 moles of Aluminium = (2 × 27)g/mol = 54 g/mol

Mass of 2 moles of alum = (2 × 474)g/mol = 948 g/mol

54 g/mol of aluminium will produce 948 g/mol of alum, so

\text{0.9875 grams of aluminium will produce}=\frac{948g/mol}{54g/mol}\times 0.9875g

Amount of Alum produced = 17.34 grams

Theoretical yield of alum = 17.34 grams.

5 0
3 years ago
Plzz someone help me ASAP!
irina1246 [14]

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3 0
3 years ago
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