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4 years ago

Which descriptions apply to eukaryotic cells? Check all that apply.

Chemistry

1 answer:

Anettt [7]4 years ago

3 0

Answer:

are usually multicellular

contain a true nucleus

contain membrane-bound organelles

Explanation:

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1. What is the mass of 22.4 L of H2 at STP?

Fed [463]

1.
n = m/2.016 g/mol = V/22.4 l
V = 22.4 l
Hence,
m = 2.016 g ≈ 2.02 g
B.) 2.02 g

2.
Balancing the equation, we get,
Mg + 2HCl → MgCl2 + H2
PV = nRT
V = nRT/P
V = 1.5 mol × 8.314 atm/(mol.K) × 298 K/(0.96 atm)
V = 3871.206
V ≈ 38.2 l

6 0

3 years ago

Read 2 more answers

An electromagnetic wave has a wavelength of 0.12 meters. what is that in nanometers?

tino4ka555 [31]

Answer:

1.2 × 10⁸ m

Explanation:

You want to convert metres to nanometres.

Recall that the multiplying prefix nano- means × 10⁻⁹, so

1 nm = 1 × 10⁻⁹ m

The conversion factor is either $\frac{\text{1 nm} }{10^{-9} \text{m}}$ or $\frac{10^{-9} \text{m} }{\text{1 nm}}$ .

You choose the one that has the desired units (nm) on top. Then

$\lambda = \text{0.12 m} \times \frac{\text{1 nm} }{10^{-9} \text{m}} = 1.2 \times 10^{8} \text{ nm}$

7 0

3 years ago

What mass of CDP (403 g mol−1) is in 10 mL of the buffered solution at the beginning of Experiment 1? 6.4 × 10−4 g 6.4 × 10−3 g

eduard

Compete Question:

What mass of CDP (403 g mol−1) is in 10 mL of the buffered solution at the beginning of Experiment 1?

Passage: "16 mmol of CDP in 1 L of buffer"

Answer:

6.4 × 10-2 g

Explanation:

$Mass = Mole × Molar Mass$

we are given from the question that 16 mmol of CDP is in 1 L of buffer

this mean that we have $16 × 10^-3$ moles of CDP in 1 liter of buffer.

so the mass of CDP in one liter of buffer will be calculate as,

mass of CDP = $16 × 10^-3$ × 403g mol−1

= $64 × 10^-1$

= 6.4 g/L

But because the question

asks us about the mass of CDP in 10 mL of solution, we will go further to calculate it like this:

6.4 g/L × 10 mL

6.4 g/L × 0.01 L = $6.4 × 10^-2$

8 0

4 years ago

1N2 + 3H2 -->

Hunter-Best [27]

Answer:

28.23 g NH₃

Explanation:

The balanced chemical equation is:

N₂(g) + 3 H₂(g) → 2 NH₃(g)

Thus, 1 mol of N₂ reacts with 2 moles of H₂ to produce 2 moles of NH₃. We convert the moles to mass (in grams) by using the molecular weight (MW) of each compound:

MW(N₂) = 2 x 14 g/mol = 28 g/mol

mass N₂= 1 mol x 28 g/mol = 28 g

MW(H₂) = 2 x 1 g/mol = 2 g/mol

mass H₂ = 3 mol x 2 g/mol = 6 g

MW(NH₃) = 14 g/mol + (3 x 1 g/mol) = 17 g/mol

mass NH₃= 2 moles x 17 g/mol = 34 g

Now, we have to figure out which is the limiting reactant. For this, we know that the stoichiometric ratio is 28 g N₂/6 g H₂. If we have 36.85 g of H₂, we need the following mass of N₂:

36.85 g H₂ x 28 g N₂/6 g H₂ = 171.97 g N₂

We have 23.15 g N₂ and we need 171.97 g. So, we have lesser N₂ than we need. Thus, the limiting reactant is N₂.

Now, we calculate the product (NH₃) by using the stoichiometric ratio 34 g NH₃/28 g N₂, with the mass of N₂ we have:

23.25 g N₂ x 34 g NH₃/28 g N₂ = 28.23 g NH₃

Therefore, the maximum amount of NH₃ that can be produced is 28.23 grams.

5 0

3 years ago

A certain element has a melting point over 700 ∘c and a density less than 2.00 g/cm3. give one possible identity for this elemen

Hunter-Best [27]

A melting point of over 700 C and a density of less than 2 g/cm3 can be observed for many group 2 elements. In this group, the density increases on moving down the group, whereas the melting point increases upto calcium and then starts decreasing.

Calcium, symbol Ca is the element with melting point around 840 C and density of 1.55 g/cm3 which is closest to the specified data range .

3 0

4 years ago