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Sonbull [250]
2 years ago
14

Given a gas with an initial volume of 3.0 L and a temperature of 290 K, what is the final volume if the

Chemistry
1 answer:
statuscvo [17]2 years ago
6 0

Answer:

2.8l

Explanation:

just make v2 de subject

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What volume in mt, of 0.5a M1HCI solution is needed to neutralize 77 ml of 1.54 M NaOH solution?
Rainbow [258]

Answer:

237.2 mL.

Explanation:

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

(XMV) acid = (XMV) base.

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) HCl = (XMV) NaOH.</em>

<em></em>

For HCl; X = 1, M = 0.5 M, V = ??? mL.

For NaOH, X = 1, M = 1.54 M, V = 77.0 mL.

<em>∴ V of HCl = (XMV) NaOH / (XV) HCl = (</em>1)(1.54 M)(77.0 mL) / (1)(0.5 M) = <em>237.2 mL.</em>

8 0
3 years ago
How does an objects speed affect its kinetic energy?​
rewona [7]
Because it’s kinetic energy INCREASES the speed
8 0
2 years ago
Base changes phenolphthalein to pink it is true or false​
galben [10]

Answer:

It’s false

Explanation:

it could be true if the question mentioned alkaline solution

4 0
2 years ago
Read 2 more answers
How many grams of NaOH are produced from 1.20 x 10^2 of Na2O
oksian1 [2.3K]
Write an balance the equation

Na2O + H2O -> 2 NaOH

Calculate the molecular mass of Na2O and NaOH from the atomic mass from the periodic table.

Na = 23
O=16
H=1

Na2O = 23 * 2 + 16 = 62
NaOH = 23+16+1= 40

For the stoichiometry of the reaction one mole of Na2O = 62g produce two mol of NaOH = 2* 40= 80 g

120 g Na2O x 80g NaOH / 62g Na2O=

154.8 g NaOH
5 0
3 years ago
Naomi is investigating the properties of a solid material. It takes 120 joules to raise the temperature of 10 grams of the mater
MrRissso [65]
When heat energy is supplied to a material it can raise the temperature of mass of the material.
Specific heat is the amount of energy required by 1 g of material to raise the temperature by 1 °C.
equation is 
H = mcΔt
H - heat energy 
m - mass of material 
c - specific heat of the material 
Δt - change in temperature
substituting the values in the equation 
120 J = 10 g x c x 5 °C
c = 2.4 Jg⁻¹°C⁻¹
3 0
3 years ago
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