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nignag [31]
3 years ago
5

How do scientist classify newly discovered organisms into domains and kingdoms? (6th Grade)

Chemistry
1 answer:
olga55 [171]3 years ago
5 0

Answer: Scientists look for cell types, the ability to make food, number of cells in their bodies.

Explanation: Scientists look for cell types, the ability to make food, number of cells in their bodies when placing newly-discovered organisms.

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Give the name and symbol of the prefixes used with SI units to indicate multiplication by the following exact quantites.
Bingel [31]

Answer:

A. Kilo , K, multiplication by 1000

B. Centi, c

C. Deci, d

D. Mili, m

E. Mega, M

F. Micro, u

7 0
3 years ago
Does this sound catchy or invite you to read more? "GMOs and pesticides are general amongst farmers for generations and should n
siniylev [52]

Answer:

It sounds fine, but it may be a bit too long. It's difficult to shorten things like this, but getting more straight to the point would give it that "catchy" feel.

Explanation:

3 0
3 years ago
25.0 mL of a hydrofluoric acid solution of unknown concentration is titrated with 0.200 M NaOH. After 20.0 mL of the base soluti
lesantik [10]

Answer:

[HF]₀ = 0.125M

Explanation:

NaOH + HF => NaF + H₂O

Adding 20ml of 0.200M NaOH into 25ml of HF solution neutralizes 0.004 mole of HF leaving 0.004 mole NaF in 0.045L with 0.001M H⁺ at pH = 3.   This is 0.089M NaF and 0.001M HF remaining.

=> 45ml of solution with pH = 3 and contains 0.089M NaF from titration becomes a common ion problem.

                HF  ⇄    H⁺    +      F⁻

C(eq)       [HF]     10⁻³M      0.089M (<= soln after adding 20ml 0.200M NaOH)

Ka = [H⁺][F⁻]/[HF]₀ => [HF]₀ = [H⁺][F⁻]/Ka

[HF]₀ = (0.001)(0.089)/(7.1 x 10⁻⁴) M = 0.125M

6 0
3 years ago
The volume of a sample gas, initially at 25 C and 158 mL, increased to 450 mL. What is the final temperature of the sample of ga
Rashid [163]

Answer:

Final temperature of the gas is  576 ^{0}\textrm{C}.

Explanation:

As the amount of gas and pressure of the gas remains constant therefore in accordance with Charles's law:

                                       \frac{V_{1}}{T_{1}}=\frac{V_{2}}{T_{2}}

where V_{1} and V_{2} are volume of gas at T_{1} and T_{2} temperature (in kelvin scale) respectively.

Here V_{1}=158mL , T_{1}=(273+25)K=298K and V_{2}=450mL

So  T_{2}=\frac{V_{2}T_{1}}{V_{1}}=\frac{(450mL)\times (298K)}{(158mL)}=849K 

849 K = (849-273) ^{0}\textrm{C} = 576 ^{0}\textrm{C}

So final temperature of the gas is  576 ^{0}\textrm{C}.

3 0
3 years ago
The molecular weight of a gas that has a density of 5.75 g/l at stp is __________ g/mol.
Ira Lisetskai [31]

Answer : The molecular weight of a gas is, 128.9 g/mole

Explanation : Given,

Density of a gas = 5.75 g/L

First we have to calculate the moles of gas.

At STP,

As, 22.4 liter volume of gas present in 1 mole of gas

So, 1 liter volume of gas present in \frac{1}{22.4}=0.0446 mole of gas

Now we have to calculate the molecular weight of a gas.

Formula used :

\text{Moles of gas}=\frac{\text{Mass of a gas}}{\text{Molecular weight of a gas}}

Now put all the given values in this formula, we get the molecular weight of a gas.

0.0446mole=\frac{5.75g}{\text{Molecular weight of a gas}}

\text{Molecular weight of a gas}=128.9g/mole

Therefore, the molecular weight of a gas is, 128.9 g/mole

8 0
3 years ago
Read 2 more answers
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