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xxMikexx [17]
3 years ago
13

How many moles are in 25g of NaCI?

Chemistry
1 answer:
emmainna [20.7K]3 years ago
4 0

Nacl = (. 23+35. 5)

= 58.5g

.

1 mol of Nacl = 58.5g

X mol Of Nacl = 25g

X Mol of Na Cl =25 ÷ 58.5

X mol of Nacl = 0.4 mol

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radioisotope actinium 225 has half life of 10 days, I begin with 16 kg of this isotope, how much will remain after 40 days?
Igoryamba
N(t)=N_{0}(\frac{1}{2})^\frac{t}{\tau_{_\frac{1}{2}}}}\\\\
N_{0}=16kg\\
t=40days\\
\tau_{_\frac{1}{2}}}=10days\\\\\\
N(t)=16kg*(\frac{1}{2})^{\frac{40}{10}}=16kg*\frac{1}{16}=1kg
3 0
3 years ago
Explain the Anchoring Phenomenon (summary)
Virty [35]
An anchoring phenomenon anchors all of the learning within a unit. So, it is a unit level event that the classroom is trying to make sense of as they engage in a series of lessons.

Since the questions the students ask about the anchor drive the learning within the unit, the anchor should be complex and require an understanding of several big science ideas to explain.

At strategic moments, the class revisits the anchoring phenomenon to review their initial questions to see which they have answered, which they are making progress on, and what new questions they may have to help us continue learning about the phenomenon.

Throughout the unit, the classroom and each student should be given opportunities to share their thinking and how it relates to the anchoring phenomenon.
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3 0
3 years ago
A sample of gas has a volume of 75.0 mL at 30.0 psi. Determine the pressure of the gas if its volume is changed to 15 mL and its
Assoli18 [71]

Answer:

5.995 psi

Explanation:

30 psi = 2.04 atm

75 mL = 0.075 L

15 mL = 0.015 L

0.075 L/ 2.04 atm = 0.015 L/x

0.075x = 0.0306

x = 0.408

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4 0
3 years ago
OS-182 has a half-life of 3.6 days. If a sample started at 175 grams, how much would be led after 18 days?
hodyreva [135]
I think c is correct answer
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3 0
3 years ago
Read 2 more answers
A 25.0 g sample of an alloy was heated to 100.0 oC and dropped into a beaker containing 90 grams of water at 25.32 oC. The tempe
kaheart [24]

Answer:

The specific heat of the alloy C_{a} = 0.37 \frac{KJ}{Kg K}

Explanation:

Mass of an alloy m_{a} = 25 gm

Initial temperature T_{a} = 100°c = 373 K

Mass of water m_{w} = 90 gm

Initial temperature of water T_{w} = 25.32 °c = 298.32 K

Final temperature T_{f} = 27.18 °c = 300.18 K

From energy balance equation

Heat lost by alloy = Heat gain by water

m_{a} C_{a}  [T_{a} - T_{f}] = m_{w} C_w (T_{f} -T_{w} )

25 × C_{a} × ( 373 - 300.18 ) = 90 × 4.2 (300.18 - 298.32)

C_{a} = 0.37 \frac{KJ}{Kg K}

This is the specific heat of the alloy.

6 0
3 years ago
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