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2 years ago

My atomic number is 35.453.

Chemistry

1 answer:

ruslelena [56]2 years ago

5 0

CL
Chlorine<span>
</span>You wiil find in periodic table :)

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Plz help :(

ziro4ka [17]

<h2>Answer:0.5 moles,1 mole</h2>

Explanation:

$Sr(NO_{3})_{2}$ is readily soluble in water.

$Sr(NO_{3})_{2}$ dissolves in water to give $Sr^{2+}$ ions and $NO_{3}^{-}$ ions.

The reaction is

$Sr(NO_{3})_{2}$ → $Sr^{2+}$ $+2NO_{3}^{-}$

So, $1$ mole of $Sr(NO_{3})_{2}$ gives $1$ mole of $Sr^{2+}$ and $2$ moles of $NO_{3}^{-}$ .

So, $0.5$ moles of $Sr(NO_{3})_{2}$ gives $0.5$ moles of $Sr^{2+}$ and $1$ mole of $NO_{3}^{-}$ .

5 0

3 years ago

By titration, it is found that 28.5 mL of 0.183 M NaOH(aq) is needed to neutralize 25.0 mL of HCl(aq). Calculate the

maksim [4K]

Answer:

0.209M

Explanation:

M1V1=M2V2

(28.5 mL)(0.183M)=(25.0mL)(M)

M2= 0.209M

*Text me at 561-400-5105 for private tutoring if interested: I can do homework, labs, and other assignments :)

4 0

2 years ago

What is the MOLAR heat of combustion of methane(CH₄) if 64.00g of methane are burned to heat 75.0 ml of water from 25.00°C to 95

melamori03 [73]

Answer:

-5.51 kJ/mol

Explanation:

Step 1: Calculate the heat required to heat the water.

We use the following expression.

$Q = c \times m \times \Delta T$

where,

c: specific heat capacity
m: mass
ΔT: change in the temperature

The average density of water is 1 g/mL, so 75.0 mL ≅ 75.0 g.

$Q = 4.184J/g.\°C \times 75.0g \times (95.00\°C - 25.00\°C) = 2.20 \times 10^{3} J = 2.20 kJ$

Step 2: Calculate the heat released by the methane

According to the law of conservation of energy, the sum of the heat released by the combustion of methane (Qc) and the heat absorbed by the water (Qw) is zero

Qc + Qw = 0

Qc = -Qw = -22.0 kJ

Step 3: Calculate the molar heat of combustion of methane.

The molar mass of methane is 16.04 g/mol. We use this data to find the molar heat of combustion of methane, considering that 22.0 kJ are released by the combustion of 64.00 g of methane.

$\frac{-22.0kJ}{64.00g} \times \frac{16.04g}{mol} = -5.51 kJ/mol$