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yaroslaw [1]
3 years ago
12

What is the mass of 4.42 cm3 of platinum? The density of platinum is 22.5 g/cm3. (Sig Fig rules + units)

Chemistry
1 answer:
fgiga [73]3 years ago
7 0

Answer: 99.45 g

Explanation: we have 22.5 g of Platinum in every 1cm3 of Platinum so if we take 4.42 cm3 of Platinum we will have 4.42 × 22.5 g = 99.45 g of Platinum

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a gas has a volume of 350 cubic centimeters at 740 mmHg. how many cubic centimeters will the gas occupy at a pressure of 900 mmH
Harrizon [31]

Answer: 287.8 cm3

Explanation:

Given that:

Initial volume of gas V1 = 350 cm3

Initial pressure of gas P1 = 740 mmHg

New volume V2 = ?

New pressure P2 = 900 mmHg

Since, pressure and volume are involved while temperature is constant, apply the formula for Boyle's law

P1V1 = P2V2

740 mmHg x 350 cm3 = 900mmHg x V2

V2 = (740 mmHg x 350 cm3) /900mmHg

V2 = 259000 mmHg cm3 / 900mmHg

V2 = 287.8 cm3

Thus, the gas will occupy 287.8 cubic centimeters at the new pressure.

3 0
3 years ago
Srihari has a torch with him and 2 new batteries. However, the torch requires 3 batteries to work. Is there any way Srihari can
zloy xaker [14]

Answer:

go to the store to buy more batteries

Explanation:

6 0
3 years ago
Help PlS AFAP and thank you so much
Triss [41]

Answer:

B

Explanation:

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8 0
3 years ago
Read 2 more answers
For the reaction o(g) + o2(g) → o3(g) δh o = −107.2 kj/mol given that the bond enthalpy in o2(g) is 498.7 kj/mol, calculate the
Aneli [31]
The reaction;
O(g) +O2(g)→O3(g), ΔH = sum of bond enthalpy of reactants-sum of food enthalpy of products.
ΔH = ( bond enthalpy of O(g)+bond enthalpy of O2 (g) - bond enthalpy of O3(g)
-107.2 kJ/mol = O+487.7kJ/mol =O+487.7 kJ/mol +487.7kJ/mol =594.9 kJ/mol
Bond enthalpy (BE) of O3(g) is equals to 2× bond enthalpy of O3(g) because, O3(g) has two types of bonds from its lewis structure (0-0=0).
∴2BE of O3(g) = 594.9kJ/mol
Average bond enthalpy = 594.9kJ/mol/2
=297.45kJ/mol
∴ Averange bond enthalpy of O3(g) is 297.45kJ/mol.
8 0
3 years ago
A 65.0-gram sample of some unknown metal at 100.0° C is added to 100.8 grams of water at 22.0° C. The temperature of the water r
vampirchik [111]
We do a heat balance to solve this:
(m cp ΔT)water = -(m cp ΔT)metal
100.8 (4.18) (27 - 22) = -65 (cp)(27-100)
cp = 100.8 (4.18) (27 - 22) / (-65 (27-100))
cp = 0.44 J/ (°C × g)

The specific heat of the metal is 0.44 J/ (°C × g)
5 0
3 years ago
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