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Sever21 [200]
3 years ago
13

What two properties most affect the strength of the gravitational forces

Chemistry
2 answers:
goblinko [34]3 years ago
7 0

1) Masses of the object

2) Distance between them

Tema [17]3 years ago
4 0
The two properties would be distance and mass.
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Two students prepare lab reports one for a chemistry class and the other for a biology class which of the following items will b
AleksandrR [38]

Answer:

the results of an investigation thank me later

Explanation:

4 0
3 years ago
If you have 0.50 mol of ca, how many atoms are present?
Evgen [1.6K]
Answer is: there are 3.011·10²³ atoms of calcium.

n(Ca) = 0.50 mol; amount of substance(calcium).
Na = 6.022·10²³ 1/mol;  Avogadro's constant or number.
N(Ca) = n(Ca) · Na.
N(Ca) = 0.50 mol · 6.022·10²³ 1/mol.
N(Ca) = 3.011·10²³; number of calcium atoms.
The mole is an SI unit which measures the number of particles in substance. One mole is equal to <span><span>6.022</span></span>·<span><span><span>10</span></span></span>²³<span> atoms.</span>
6 0
3 years ago
How many kinds of atoms are in a pure substance?
stepan [7]
Single type of atom builds a single atom
3 0
3 years ago
Read 2 more answers
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Elza [17]

Answer: There are 21.08\times 10^{23} molecules in 63.00 g of H_2O

Explanation:

According to avogadro's law, 1 mole of every substance occupies 22.4 L at STP and contains avogadro's number 6.023\times 10^{23} of particles.

To calculate the moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text {Molar mass}}=\frac{63.00g}{18g/mol}=3.5moles

1 mole of H_2O contains =  6.023\times 10^{23} molecules

Thus 3.5 moles of H_2O contains =  \frac{6.023\times 10^{23}}{1}\times 3.5=21.08\times 10^{23} molecules.

There are 21.08\times 10^{23} molecules in 63.00 g of H_2O

3 0
3 years ago
Calcular la presión final de un gas que inicialmente está a 21°C y 0,98 atm. Sabiendo que su temperatura aumenta a 37°C.
KatRina [158]

Answer:

1.03 atm

Explanation:

Primero <u>convertimos 21 °C y 37 °C a K</u>:

  • 21 °C + 273.16 = 294.16 K
  • 37 °C + 273.16 = 310.16 K

Una vez tenemos las temperaturas absolutas, podemos resolver este problema usando la<em> ley de Gay-Lussac</em>:

  • T₁P₂ = T₂P₁

En este caso:

  • T₁ = 294.16 K
  • P₂ = ?
  • T₂ = 310.16 K
  • P₁ = 0.98 atm

Colocando los datos:

  • 294.16 K * P₂ = 310.16 K * 0.98 atm

Y <u>despejando P₂</u>:

  • P₂ = 1.03 atm
7 0
3 years ago
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