If 50.75 g of a gas occupies 10.0 L at STP, 129.3 g of the gas will occupy 25.48 L at STP.
<h3>How to calculate volume?</h3>
The volume of a gas at STP can be calculated using the direct proportion method.
According to this question, 50.75 g of a gas occupies 10.0 L at STP, then 129.3g of the same gas will occupy the following:
= 129.3 × 10/50.75
= 25.48L
Therefore, if 50.75 g of a gas occupies 10.0 L at STP, 129.3 g of the gas will occupy 25.48 L at STP.
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Answer:
Explanation:
Unit 10 - Acid/Base ... (a) Mg(OH. 2. ) (b) Mg(OH). 2. (c) Mg. 2. OH. (d) MgOH. 2. Standard: ... balanced equation for these neutralization reactions: 3. HCl + NaOH → ... H2CO3 + Ca(OH)2 → ... C5.7B Predict products of an acid-base neutralization. 8. 2 NH4OH + H2S ...An Arrhenius base is a compound that increases the OH − ion concentration in ... and a base is called a neutralization reaction and can be represented as follows: ... chemical equation for the neutralization reaction between HCl and Mg(OH) 2. ... acid, an Arrhenius base, or neither. a) NaOH. b) C 2H 5OH. c) H 3PO 4. 6
Answer:
V₁ = 10 mL
Explanation:
Given data:
Initial volume of HCl = ?
Initial molarity = 3.0 M
Final molarity = 0.10 M
Final volume = 300.0 mL
Solution:
Formula:
M₁V₁ = M₂V₂
M₁ = Initial molarity
V₁ = Initial volume of HCl
M₂ =Final molarity
V₂ = Final volume
Now we will put the values.
3.0 M ×V₁ = 0.10 M×300.0 mL
3.0 M ×V₁ = 30 M.mL
V₁ = 30 M.mL /3.0 M
V₁ = 10 mL
Answer:
–36 KJ.
Explanation:
The equation for the reaction is given below:
2B + C —› D + E. ΔH = – 24 KJ
From the equation above,
1 mole of D required – 24 KJ of energy.
Now, we shall determine the energy change associated with 1.5 moles of D.
This can be obtained as illustrated below:
From the equation above,
1 mole of D required – 24 KJ of energy
Therefore,
1.5 moles of D will require = 1.5 × – 24 = –36 KJ.
Therefore, –36 KJ of energy is associated with 1.5 moles of D.
