3 years ago

A B C D ............

Chemistry

1 answer:

Deffense [45]3 years ago

3 0

Answer:

I think gamma, I'm not sure tho

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If Sally has 250 grams of cement, how many kilograms does she have? How many milligrams?

MissTica

Answer:

0.250 kg

2500 g

Explanation:

1 kg = 1000g

1g = 10 mg

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2 years ago

Molecules are the smallest units of matter with the characteristic properties of a substance.

Vilka [71]

False, Atoms are the smallest units of matter that display both the chemical and physical properties of it, based on the structure of the atom.

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4 years ago

What is the change in density if a sample goes from 3.21 g/L to 5.43 g/mL?

Step2247 [10]

Answer:

$\Delta \rho =2.22 g/mL$

Explanation:

Hello,

In this case, since a change in science is widely known to be considered as a subtraction between the the final and initial values of two measured variables and is represented via Δ, here the final density is 5.43 g/mL and the initial one was 3.21 g/mL, therefore, the change in density is:

$\Delta \rho=\rho _f-\rho _i\\ \\\Delta \rho=5.43g/mL-3.21g/mL\\\\\Delta \rho =2.22 g/mL$

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3 years ago

Determine how many atoms of pure silver will be created when 19.83 x 1023 atoms of copper are used in the following reaction:

Lubov Fominskaja [6]

Answer:

$\boxed{3.966 \times 10^{24}\text{ atoms of Ag}}$

Explanation:

(a) Balanced equation

Cu + 2AgNO₃ ⟶ Cu(NO₃)₂+ 2Ag

(b) Calculation

You want to convert atoms of Cu to atoms of Ag.

The atomic ratio is ratio is 2 atoms Ag:1 atom Cu

$\text{Atoms of Ag} = 19.83 \times 10^{23}\text{atoms Cu} \times \dfrac{\text{2 atoms Ag}}{\text{1 atom Cu}}\\\\= 3.966 \times 10^{24}\text{ atoms of Ag}\\\\\text{The reaction will produce }\boxed{\mathbf{3.966 \times 10^{24}}\textbf{ atoms of Ag}}$

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3 years ago

Read 2 more answers

What is the relationship between the number of protons and the number of electrons in an atom?

aleksley [76]

In a neutral atom they are both equal, and their even quantities makes the atom neutral...

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3 years ago

A B C D ............​

A B C D ............