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3 years ago

Density is mass per I'm it of volume. Which pair of lab instruments would a student use to measure the density of a seawater

Chemistry

1 answer:

kati45 [8]3 years ago

5 0

Density can be measured by D=Volume/mass

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Metals experience plastic deformation when _____.

xxTIMURxx [149]

Metals experience plastic deformation when a specific amount of pressure and temperature is applied to them. Most metals have low specific heat. Because of their low specific heat, they easily get hot when heat is applied to them.

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3 years ago

True or False: Helium gas will effuse more rapidly than carbon monoxide gas.

yulyashka [42]

Answer:

true

Explanation:

The lighter a gas is, the faster it will effuse; the heavier a gas is, the slower it will effuse. Of all the choices, helium (He) has the lowest molecular weight (atomic weight in this case), so it will have the highest rate of effusion.

6 0

4 years ago

Why are metals grouped together

Monica [59]

Because their atoms have the same number of electrons in the highest occupied energy level

5 0

4 years ago

How many grams of NaOH is needed to neutralize 90 mL of 1.5 N HCl?

AnnZ [28]

mol = conc × v

= 1.5 × 0.09

= 0.135 moles of HCl

HCl + NaOH > NaCl + H2O

1 mole HCl = 1 mole NaOH

0.135 mol HCl = x

x = 0.135 mol NaOH

mass = mol × molar mass

= 0.135 × 40

= 5.4 g

NaOH = 23 + 16 + 1 = 40 g/mol

I'm not a 100% sure if it's correct

6 0

3 years ago

Suppose 10.0 mL of 2.00 MNaOH is added to (a) 0.780 L of pure water and (b) 0.780 L of a buffer solution that is 0.682 Min butan

katrin2010 [14]

Answer:

a) pH will be 12.398

b) pH will be 4.82.

Explanation:

a) The moles of NaOH added = molarity X volume (L) = 2 X 0.01 = 0.02 moles

The total volume after addition of pure water = 0.780+0.01 = 0.79 L

The new concentration of /NaOH will be:

$molarity=\frac{molesofsolute}{volumeofsolution}=\frac{0.02}{0.79}=0.025M$

the [OH⁻] = 0.025

pOH = -log [OH⁻] = 1.602

pH = 14 -pOH = 12.398

b) The buffer has butanoic acid and butanoate ion.

i) Before addition of NaOH the pH will be calculated using Henderson Hassalbalch's equation:

$pH=pKa+log\frac{[salt]}{[acid]}$

pKa= $-logKa=-log(1.5X10^{-5})=4.82$

ii) on addition of base the pH will increase.

8 0

4 years ago