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Elza [17]
2 years ago
15

If a young girl learned that it would be 0° C, or 32° F, the next

Chemistry
1 answer:
Ad libitum [116K]2 years ago
7 0

Answer:

Cold

Explanation:

Because 0°C And below or 32°F Refers To Cold Weather

You might be interested in
1. How much heat (in calories) is needed to raise 20 g of H2O from 5°C to 40°C? (c = 1.0 cal/g °C)
KIM [24]

Answer:

700 calories

Explanation:

Using the formula below:

Q = m × c × ∆T

Where;

Q = amount of heat required (calories)

m = mass of substance (g)

c = specific heat of substance (cal/g°C)

∆T = change in temperature (°C)

According to this question, the following information was provided;

Q = ?

m = 20g

c = 1.0 cal/g °C

∆T = 40°C - 5°C = 35°C

Using the formula; Q = m × c × ∆T

Q = 20 × 1 × 35

Q = 700 calories

Hence, 700 cal of heat energy is needed to raise 20 g of H2O from 5°C to 40°C.

6 0
2 years ago
What is the volume of an 2.3 solution with 212 grams of calcium chloride dissolved in it
Travka [436]

830 mL. A 2.3 mol/L solution of CaCl2 has a volume of 830 mL

I am guessing that the concentration of your solution is 2.3 mol/L.

a) Moles of CaCl2

MM of CaCl2 = 110.98 g/mol

Moles of CaCl2 = 212 g CaCl2 x (1 mol CaCl2/110.98 g CaCl2)

= 1.910 mol CaCl2

b) Volume of solution

V = 1.910 mol CaCl2 x (1 L solution/2.3 mol CaCl2) = 0.83 L solution

= 830 mL solution


5 0
3 years ago
Help me please) attached the screen below. Thanks
Vitek1552 [10]

Answer:

1) d = 2.4 g/cm³

2) m = 25 g

3) v = 126.7 cm³

Explanation:

Given data:

Mass of material = 24 g

Volume of material = 10 cm³

Density of material = ?

Solution:

Formula:

d = m/v

by putting value,

d = 24 g / 10 cm³

d = 2.4 g/cm³

2) Given data:

Density of material = 5 g/cm³

Volume of material = 5 cm³

Mass of material = ?

Solution:

Formula:

d = m/v

5 g/cm³ = m / 5 cm³

m = 5 g/cm³×5 cm³

m = 25 g

3)Given data:

Density of material = 3 g/cm³

Mass of material = 380 g

Volume of material =  ?

Solution:

Formula:

d = m/v

3 g/cm³ = 380 g / v

v = 380 g /3 g/cm³

v = 126.7 cm³

7 0
3 years ago
The radioactivity of U-235 is of low intensity. Why then were the people of Hiroshima exposed to high intensity radiation in the
Marina86 [1]

Answer:

Explanation:

What occurred then is as a result of nuclear fission. This occurs as the Uranium-235 split into two smaller nuclei while releasing high energy neutrons. These neutrons bombard existing U-235 in the atmosphere and this reaction continue in a spontaneous manner until a chain reaction is formed of U-235, whose fall out fills the environment. This process was what led to people been exposed to high intensity radiation in the days and months after the atomic bomb was dropped.

7 0
3 years ago
What is the [OH-] of a substance that has a pH of 11?
Deffense [45]

Answer:

0.001 M OH-

Explanation:

[OH-] = 10^-pOH, so

pOH + pH = 14 and 14 - pH = pOH

14 - 11 = 3

[OH⁻] = 10⁻³ ; [OH-] = 0.001 M OH-

6 0
2 years ago
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