Please hurry which statement is a scientist most likely to use in describing work

The diameter of a carbon atom is 0.000000000154 m. What is this number expressed in scientific notation?

Viktor [21]

1.54 x 10^-10 (ten to the negative tenth power)

8 0

3 years ago

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The density of a block of wood with a volume of 50 cubic centimeters and a mass of 100 grams is?

german

2 g/cm³ is the density of the block of wood

Explanation:

To determine density of the block of wood we use the following formula:

density = mass / volume

density of the wood block = 100 g / 50 cm³

density of the wood block = 2 g/cm³

Learn more about:

density

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7 0

3 years ago

The equation below shows a reaction that is second order in A and first order in B.

olya-2409 [2.1K]

It’s the second one because the order of it is used as the exponent for it in the rate law equation

4 0

3 years ago

Calculate the morality for the following<br> 10.5 kg of Na2 SO4. 10H2O in 18.60 L

Serggg [28]

Answer:

1.75M

Explanation:

The problem deals with finding the molarity of the given compound.

The compound is:

Na₂SO₄.10H₂O

Mass of the compound = 10.5kg = 10500g

Volume of the compound = 18.6L

Molarity is the number of moles of solute in a solution;

Molarity = $\frac{number of moles }{volume}$

Number of moles = $\frac{mass}{molar mass}$

Molar mass of Na₂SO₄.10H₂O = 2(23) + 32 + 4(16) + 10[2(1) + 16]

= 322g/mol

Now;

Number of moles = $\frac{10500}{322}$ = 32.6mole

So;

Molarity = $\frac{32.6}{18.6}$ = 1.75M

5 0

3 years ago

When heated a sample consisting of only CaCO3 and MgCO3 yields a mixture of CaO and Mgo. If the weight of the combined oxides is

ExtremeBDS [4]

Answer:

38.83 % of CaCO3

61.17 % of MgCO3

Explanation:

where Moles of CaCO3 is equals to x and MgCO3 is y we have that...

CaCO3 molar mass = 100.09 g / mol = 100.09 x

MgCO3 molar mass = 84.31 g / mol = 84.31 y

decomposition reactions :

CaCO3 ---> CaO + CO2

MgCO3 ---> MgO + CO2

So we have that , Moles of CaO = Moles of CaCO3 = x

and Moles of MgO = Moles of MgCO3 = y

CaO molar mass = 56.08 g / mol

MgO molar mass = 40.30 g / mol

CaO = 56.08 x

MgO = 40.30 y

"If the weight of the combined oxides is equal to 51.00% of the initial sample weight,"

total mass of MgO and CaO = 51.00 % of Total Mass of MgCO3 and CaCO3

thus

56.08 x + 40.30 y = 0.51 ( 100.09 x + 84.31 y )

56.08 x + 40.30 y = 51.04 x + 42.99y

5.04 x = 2.7 y

y = 1.87 x

CaCO3 % in the sample

= 100.09 x × 100 / ( 100.09 x + 84.31 y )

= 10009 x / ( 100.09 x + 84.31 × 1.87 x )

= 10009 x / ( x ( 100.09 + 157.66 ) )

= 10009 / 257.75

= 38.83 %

MgCO3 % in the sample

= 100 - 38.83

= 61.17 %

7 0

3 years ago