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Korvikt [17]
3 years ago
15

How many calories are required to raise the temperature of 105 g of water from 30.0°c to 70.0°c?

Chemistry
1 answer:
ioda3 years ago
5 0
<span>4.200 calories. The formula is rise in heat (H) = the specific heat capacity (c) * rise in heat in degrees Fahrenheit (T) and volume of liquid (m) . H= cTm. Water, the typical comparison, has a specific heat capacity of 1. So to raise the Fahrenheit temperature of 105 grams of water 40 degrees, you multiply 105 * 40 * 1 = 4,200.</span>
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Mass is the number of protons + number of neutrons. Because electrons don't have significant weight.
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What is the overall charge of the compound frfr
matrenka [14]

Answer:

As any molecular or ionic compound, 0 net charge, as a hypothetical diatomic cation, 2+

Explanation:

The question states that this is a compound, and any molecular or ionic compound would have a net charge of 0.

However, such a compound doesn't exist. Francium generally exists in nature having a 1+ charge in its cation form. Combining two francium ions would result in a Fr_2^{2+} species which is simply a diatomic ion but not a compound.

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3 years ago
A person riding a bike has a mass of 55kg. They are moving with a velocity of 2m/s. What is the Kinetic Energy of the person?
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8 0
3 years ago
How much energy would be absorbed as heat by 75 g of iron when heated from 295 k to 301 k?
Katen [24]

Answer:

Q=202.5J

Explanation:

Hello,

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Q=mCp(T_2-T_1)

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Q=75g*0.450\frac{J}{g*K}*(301-295)K\\\\Q=202.5J

In such a way, since the temperature is increased heat is added, that is why it is positive.

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4 0
4 years ago
Read 2 more answers
An unknown compound contains only C , H , and O . Combustion of 5.60 g of this compound produced 13.7 g CO2 and 5.60 g H2O . Wha
son4ous [18]

Answer:

Empirical formula of compound  is C₄H₈O

Explanation:

Given data:

Mass of compound = 5.60 g

Mass of CO₂ = 13.7 g

Mass of H₂O = 5.60 g

Empirical formula of compound = ?

Solution:

Percentage of C:

13.7 g/5.60 g × 12/44× 100

2.45×0.273× 100 = 66.9%

Percentage of H:

5.60 g/ 5.60 g × 2.016/18 × 100

11.2%

Percentage of O:

(66.9% + 11.2%) - 100 = 21.9%

Grams atom of  C , H, O

66.9/12 = 5.6

11.2 / 1.008 = 11.11

21.9 / 16 = 1.4

Atomic ratio:

           C                 :               H           :              O

           5.6/1.4         :            11.11/1.4    :            1.4/1.4

              4                :               8            :               1

Empirical formula:

C₄H₈O

7 0
4 years ago
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