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KonstantinChe [14]
3 years ago
12

Can someone please help ill mark you as brainlest

Chemistry
2 answers:
dexar [7]3 years ago
5 0
Its the molecules have 20 electrons so the bear fainted.
Vilka [71]3 years ago
3 0
20 will be your answer
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What is the mole-to-mole ratio between 25 Oxygen and 18 Water
eduard

25:18 meaning for every 25 moles of oxygen there's 18 moles of water

7 0
3 years ago
The length of a desk is 3.5 feet. how many centimeters is the length
Delvig [45]
1 foot = 30.48 cm
3.5 ft * 30.48 cm = 106.68 cm

Answer = 106.68 cm

Hope it helps :)
Please mark as brainliest
4 0
3 years ago
Read 2 more answers
Please help! ill give you 90 points!
Luba_88 [7]

Answer:

1.d = 2.70 g/mL

2.d = 13.6 g/mL

4.d = 1896 g / 212.52 cm3 = 8.9 g/cm3

Explanation:If this helped subscribe to Amiredagoat Yt

3 0
3 years ago
1. How much heat is required to raise the temperature of 4.64kg of lead from 150°C to 219°C?
diamong [38]

Answer: An amount of 40980.48 J/g^{o}C heat is required to raise the temperature of 4.64kg of lead from 150°C to 219°C.

Explanation:

Given: mass of lead = 4.64 kg

Convert kg into grams as follows.

1 kg = 1000 g\\4.64 kg = 4.64 kg \times \frac{1000 g}{1 kg}\\= 4640 g

T_{1} = 150^{o}C

T_{2} = 219^{o}C

The standard value of specific heat of lead is 0.128 J/g^{o}C.

Formula used to calculate heat is as follows.

q = m \times C \times \Delta T

where,

q = heat energy

m = mass of substance

C = specific heat of substance

\Delta T = change in temperature

Substitute the value into above formula as follows.

q = m \times C \times \Delta T\\= 4640 g \times 0.128 J/g^{o}C \times (219 - 150)^{o}C\\= 40980.48 J/g^{o}C

Thus, we can conclude that 40980.48 J/g^{o}C heat is required to raise the temperature of 4.64kg of lead from 150°C to 219°C.

8 0
3 years ago
Methane gas is burned
prisoha [69]
Yes it is burned and I need to do that question as well for my class
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3 years ago
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