Do anyone know the answer please

Chemistry

1 answer:

Nadya [2.5K]2 years ago

5 0

A) BaCl2
b)Li2S
hope this helps...

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What is the volume of 88 grams of CO2 gas at STP?

I am Lyosha [343]

! mole of CO2 at STP has a volume of 22.4 liters
88 grams = 2 moles

so the required volume = 2*22.4 = 44.8 liters

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2 years ago

What is the mass of 2.70 ×10^22 molecules of NaOH (Molar mass = 40.0 g/mol)?

Viefleur [7K]

Data:
Molar Mass of NaOH = 40 g/mol

Solving: <span>According to the Law Avogradro, we have in 1 mole of a substance, 6.02x10²³ atoms/mol or molecules
</span>
1 mol -------------------- 6.02*10²³ molecules
y mol -------------------- 2.70*10²² molecules

6.02*10²³y = 0.270*10²³
$y = \frac{0.270*\diagup\!\!\!\!\!\!10^{\diagup\!\!\!\!\!\!23}}{6.02*\diagup\!\!\!\!\!\!10^{\diagup\!\!\!\!\!\!23}}$
$\boxed{y \approx 0.04\:mol}$

Solving: <span>Find the mass value now
</span>
40 g ----------------- 1 mol of NaOH
x g ------------- 0.04 mol of NaOH

$x = 40*0.04$
$\boxed{\boxed{x = 1.6\:g}}\end{array}}\qquad\quad\checkmark$

Answer:
The mass is 1.6 grams

6 0

3 years ago

what volume of a 0.138 m potassium hydroxide solution is required to neutralize 26.0 ml of a 0.205 m nitric acid solution?

Law Incorporation [45]

A neutralization titration is a chemical response this is used to decide the composition of an answer and what kind of acid or base is in it. This is a way of volumetric analysis and the formula is ( $M1V1 = M2V2$ ).

Utilize the titration method of $M1V1 = M2V2$ in view that we're given the concentrations of every compound and the quantity of $KOH$ . Let: M1 = 0.138M, V1 = x, M2 = 0.205M, V2 = 26.0 ML.

M1 = initial mass
V1= initial volume
M2 = final mass
V2= final volume

$(M1V1 = M2V2)$
(0.138)(V1) = (0.205)x(26.0)
V2=(0.205)x(26.0)\ 0.138
V2 = 47.10 M/L
The final value of Volume needed for neutralization of nitric acid solution is V2 = 47.10 M/L