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guapka [62]
4 years ago
9

What is the density of a rectangular block that has a mass of 500 grams and a volume of 50cm^3

Chemistry
1 answer:
oee [108]4 years ago
6 0
Density = mass / volume

D = 500 g / 50 cm³

D = 10 g/cm³

hope this helps!
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What is pH?
sergij07 [2.7K]
PH = -log[H+]
That is, pH is the negative logarithm of the hydrogen ion concentration.

pOH = -log[OH-]
pOH is the negative logarithm of the hydroxide ion concentration.

The correct answer to your question is a. The negative logarithm of the hydrogen ion concentration.
5 0
3 years ago
What products would form if chlorine gas was bubbled through a solution of sodium bromide?
Misha Larkins [42]
NaCl would form because it’s a single replacement reaction
4 0
3 years ago
The absolute temperature of a gas is increased four times while maintaining a constant volume. What happens
poizon [28]

Answer:

<u>It increases by a factor of four</u>

Explanation:

Boyle's Law : At constant temperature , the volume of fixed mass of a gas is inversely proportional to its pressure.

pV = K.......(1)

pV = constant

Charles law : The volume of the gas is directly proportional to temperature at constant pressure.

V = KT

or V/T = K = constant ....(2)

Applying equation (1) and (2)

\frac{PV}{T}=K

\frac{P_{1}V_{1}}{T_{1}}=\frac{P_{2}V_{2}}{T_{2}}

According to question ,

T2 = 4 (T1)

V2 = V1

Put the value of T2 and V2 , The P2 can be calculated,

\frac{P_{1}V_{1}}{T_{1}}=\frac{P_{2}V_{1}}{4T_{1}}

V1 and V1 cancel each other

T1 and T1 cancel each other

We get,

P1=\frac{P2}{4}

or

P2 = 4 P1

So pressure increased by the factor of four

5 0
4 years ago
14.A sample of fluorine gas has a density of _____
yarga [219]

Answer:

d = 0.793 g/L

Explanation:

Given data:

Density of fluorine gas = ?

Pressure of gas = 0.554 atm

Temperature of gas = 50 °C (50+273.15K = 323.15 K)

Solution:

Formula:

PM = dRT

M = molar mass of gas

P = pressure

R = general gas constant

T = temperature

d = PM/RT

d = 0.554 atm × 37.99 g/mol / 0.0821 atm.L /mol.K × 323.15 K

d = 21.05 atm.g/mol/26.53 atm.L /mol

d = 0.793 g/L

8 0
3 years ago
Calculate the ph of a solution formed by mixing 200.0 ml of 0.30 m hclo with 300.0 ml of 0.20 m kclo. the ka for hclo is 2.9 × 1
velikii [3]
C(HClO) = 0,3 M.
<span>V(HClO) = 200 mL = 0,2 L.
n(HClO) = </span>c(HClO) · V(HClO).
n(HClO) = 0,06 mol.<span>
c(KClO</span>) = 0,2 M.
<span>V(KClO) = 0,3 L.
n(KClO) = 0,06 mol.
V(buffer solution) = 0,2 L + 0,3 L = 0,5 L.
ck</span>(HClO) = 0,06 mol ÷ 0,5 L = 0,12 M.
cs(KClO) = 0,06 mol ÷ 0,5 L = 0,12 M.<span>
Ka(HClO</span>) = 2,9·10⁻⁸.<span>
This is buffer solution, so use Henderson–Hasselbalch equation: 
pH = pKa + log(cs</span> ÷ ck).<span>
pH = -log(</span>2,9·10⁻⁸) + log(0,12 M ÷ 0,12 M).<span>
pH = 7,54 + 0.
pH = 7,54</span>
6 0
4 years ago
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