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Llana [10]
3 years ago
11

15 g de nitrato de amonio, NH4NO3 se disuelven en 135 g de agua. La densidad de la

Chemistry
1 answer:
MrRa [10]3 years ago
7 0

Answer:

δ sln = 0.111 g/mL

Explanation:

density (δ): mass/volume

⇒ δ ≡ g solute/mL H2O

∴ m NH4NO3 = 15g

∴ δH2O ≅ 1 g/mL

∴ V H2O = (135g H2O).( mL/1gH2O) = 135 mL H2O

⇒ δ sln = 15 g NH4NO3 / 135 mL H2O = 0.111 g/mL

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A sample of gas at a pressure of 121.59 kPa, a volume of 31 L, and a temperature of 360 K contains how many moles of gas? A. 0.7
natali 33 [55]

Answer:

C. 1.3 mol

Explanation:

PV = nRT

where P is absolute pressure,

V is volume,

n is number of moles,

R is universal gas constant,

and T is absolute temperature.

Given:

P = 121.59 kPa

V = 31 L

T = 360 K

R = 8.3145 L kPa / mol / K

Find: n

n = PV / (RT)

n = (121.59 kPa × 31 L) / (8.3145 L kPa / mol / K × 360 K)

n = (3769.29 L kPa) / (2993.22 L kPa / mol)

n = 1.26 mol

Round to two significant figures, there are 1.3 moles of gas.

4 0
3 years ago
How can you tell the difference between non metals from metals on the periodic element chart with no colors
Leni [432]
You can tell the difference by How shiny it is and by how heavy it is
6 0
3 years ago
What is the molecular formula of a compound with the empirical formula SO and molecular weight 96.13?
anyanavicka [17]

S2O2 because the molecular formula of SO is just SO because they are both diatomic elements, but the empirical formula is just the but mot simplified, or S2O2.

4 0
3 years ago
True of false metals like copper are sometimes used to fill cavities in teeth
ElenaW [278]
I'd say false because copper is what they used to drink out of in the middle ages, and it caused them to get very sick, so I doubt they would use copper in your mouth.
7 0
3 years ago
Read 2 more answers
What mass of water could be warmed from 21.4 degrees celsius to 43.4 degrees celsius by the pellet dropped inside it? Heat capac
Artist 52 [7]

42.34 g of water could be warmed from 21.4°C to 43.4°C  by the pellet dropped inside it

Heat loss by the pellet is equal to the Heat gained by the water.

q_{w} = -q_{p} ….(1)

where, q_{w} is the heat gained by water

q_{p} is the heat loss by pellet

q_{w} = mCΔT

where m = mass of water

C = specific heat capacity of water = 4.184 J/g-°C

ΔT = Increase in temperature

ΔT for water = 43.4 - 21.4 = 22°C

q_{w} = m × 4.184 × 22 …. (2)

Now

q_{p} = H_{c} ×ΔT

where H_{c} = Heat capacity of pellet = 56J/°C

Δ T for pellet = 43.4 - 113 =- 69.6°C

q_{p} = 56 × -69.6 = -3897.6 J

From equation (1) and (2)

-m× 4.184 × 22 =-3897.6

m= 42.34 g

Hence, 42.34 g of water could be warmed from 21.4 degrees Celsius to 43.4 degrees Celsius by the pellet dropped inside it.

Learn more about specific heat here brainly.com/question/16559442

#SPJ1

6 0
2 years ago
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