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3 years ago

How many protons are present in an atom

Chemistry

1 answer:

kogti [31]3 years ago

3 0

Answer:

36 protons

Explanation:

Because atomic number Z is equal number of protons while mass number is protons plus neutrons

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IF SOMEONE HELP ME WITH THIS, I'll GIVE 100 POINTS! Create an organization structure for fictional elements. There are many ways

Vlad1618 [11]

Bruh you didn’t give 100 points tho

6 0

3 years ago

Calculate the mass of butane needed to produce 65.5 g of carbon dioxide

jeka94

You take the mass of carbon dioxide, 56.8g, divide by its molar mass, 44.01g/mol, to produce the moles of carbon dioxide. This is multiplied by the molar ratio of butane/CO2, (2/8) = 1/4, which gives the moles of butane required to produce the carbon dioxide.

6 0

3 years ago

1. All ___matter______ is made of atoms that cannot be ______________________ or ____________________. 2. All _______________ of

marin [14]

Answer:

All the matter that is made of atoms that cannot be broken or beaten into thin sheet .2All matters of a given book are in mass and mixture

4 0

2 years ago

50.0 g of NaNO3 are dissolved into enough water to make 250 mL of solution. What is the molarity of this solution?

d1i1m1o1n [39]

Answer:

2.35 M

Explanation:

Molarity is mol/L of solution. We have to convert the g to mol and the mL to L. G to mol uses the molar mass of the compound. The molar mass of NaNO₃ is 85.00g/mol.

$50.0gNaNO3*\frac{1molNaNO3}{85.00gNaNO3} = 0.588molNaNO3$

Then you have to convert mL to L.

$250mL*\frac{1L}{1000mL} = 0.250L$

Now divide the mol by the L.

$\frac{0.588mol NaNO3}{0.250L} = 2.352 M$

Round to the smallest number of significant figures = 2.35M

7 0

3 years ago

You add 100.0 g of water at 52.0 °C to 100.0 g of ice at 0.00 °C. Some of the ice melts and cools the water to 0.00 °C. When the

lara31 [8.8K]

Answer:

$m_{ice} = 65.336\,g$

Explanation:

Accoding to the First Law of Thermodynamics, the heat released by the water melts a portion of ice. That is to say:

$Q_{water} = Q_{ice}$

$(100\,g)\cdot \left(4.184\,\frac{J}{kg\cdot ^{\textdegree}C}\right)\cdot (52\,^{\textdegree}C - 0\,^{\textdegree}C) = m_{ice}\cdot \left(333\,\frac{J}{g} \right)$

The amount of ice that is melt is:

$m_{ice} = 65.336\,g$

5 0

3 years ago