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3 years ago

Can someone help me please?

Chemistry

1 answer:

morpeh [17]3 years ago

8 0

This rock is balanced by roots on the ground that are very strong

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You have to make 500 mL of a 1.50 M BaCl2. You have 2.0 M barium chloride solution available. Determine how to make the needed d

Inga [223]

Answer:

You need 375 mL of BaCl2 solution.

Explanation:

M1V1=M2V2

Dilution formula. Substitute known values and solve for V1.

M1 = 2.0 M

M2 = 1.50 M

V2 = 500 mL

(2.0 M)(V1) = (1.50 M)(500 mL)

V1 = (1.50 M)(500 mL) / (2.0 M)

V1 = 375 mL

7 0

2 years ago

What do these two changes have in common? Shaking up salad dressing. Adding dish soap to water in a sink

Ghella [55]

I think it’s a maybe hope this helps

4 0

3 years ago

Compared to the nonmetals in Period 2, the metals in Period 2 generally have larger

konstantin123 [22]

The answer is atomic radii; the size or radii of an atom increases from left to right, versus the ionization energies and electronegativities of atoms which increase from right to left.

6 0

3 years ago

A 13.5g sample of gold is heated, then placed in a calorimeter containing 60g of water. The temperature of water increases from

sweet-ann [11.9K]

Answer:

$T_i~=163.1$ ºC

Explanation:

We have to start with the variables of the problem:

Mass of water = 60 g

Mass of gold = 13.5 g

Initial temperature of water= 19 ºC

Final temperature of water= 20 ºC

Initial temperature of gold= Unknow

Final temperature of gold= 20 ºC

Specific heat of gold = 0.13J/gºC

Specific heat of water = 4.186 J/g°C

Now if we remember the heat equation:

$Q_H_2_O=m_H_2_O*Cp_H_2_O*deltaT$

$Q_A_u=m_A_u*Cp_A_u*deltaT$

We can relate these equations if we take into account that all heat of gold is transfer to the water, so:

$m_H_2_O*Cp_H_2_O*deltaT=~-~m_A_u*Cp_A_u*deltaT$

Now we can put the values into the equation:

$60~g*4.186~J/g{\circ}C*(20-19)~{\circ}C=-(13.5~g*0.13~J/g{\circ}C*(20-T_i)~{\circ}C)$

Now we can solve for the initial temperature of gold, so:

$T_i~=(\frac{60~g*4.186~J/g{\circ}C*(20-19)~{\circ}C}{13.5~g*0.13~J/g{\circ}C})+20$

$T_i~=163.1$ ºC

I hope it helps!

5 0

3 years ago

Difference between emprical and molecular formula

Musya8 [376]

Answer:

The key difference between empirical and molecular formulas is that an empirical formula only gives the simplest ratio of atom whereas a molecular formula gives the exact number of each atom in a molecule.

3 0

2 years ago