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3 years ago

What is the name of Se2Cl7?

Chemistry

1 answer:

Monica [59]3 years ago

4 0

Answer:

diselenium heptachloride

Explanation:

Se2Cl7

Se = selenium

Cl = chlorine

Don't forget the "-ide"

Diselenium heptachloride

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An element is a mixture of two isotopes.one isotope has an atomic mass of 34.969 amu and an abundance of 64.88

skad [1K]

An element is a mixture of two isotopes. One isotope has an atomic mass of 34.969 amu and an abundance of 64.88%. The other isotope has an atomic mass 36.966 amu.

5 0

4 years ago

What is the result at the end of meiosis II?

VMariaS [17]

Answer:

B.) two diploid cells

6 0

3 years ago

6) A 0.20 ml CO2 bubble in a cake batter is at 27°C. In the oven it gets

Nata [24]

Answer: The new volume of cake is 1.31 mL.

Explanation:

Given: $V_{1}$ = 0.20 mL, $T_{1} = 27^{o}C$

$V_{2}$ = ?, $T_{2} = 177^{o}C$

Formula used to calculate new volume is as follows.

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

Substitute the values into above formula as follows.

$\frac{V_{1}}{T_{1}} = \frac{V_{2}}{T_{2}}\\\frac{0.20 mL}{27^{o}C} = \frac{V_{2}}{177^{o}C}\\V_{2} = 1.31 mL$

Thus, we can conclude that the new volume of cake is 1.31 mL.

5 0

3 years ago

To determine the concentration of X in an unknown solution, 1.00 mL of 8.48 mM S was added to 3.00 mL of the unknown X solution

kogti [31]

Answer:

positif

Explanation:

$3.87169.843826 = y = x = .ion \: in \: cells = y = x. > \\ \geqslant {8}^{2} \times \frac{4}{3} | \geqslant | \times \frac{68.1 < }{3 = 8}$

6 0

3 years ago

Arizona was the site of a 400,000-acre wildfire in June 2002. How much carbon dioxide (CO2) was produced into the atmosphere by

kodGreya [7K]

Answer:

2.97 × 10¹³ g

Explanation:

First, we have to calculate the biomass the is burned. We can establish the following relations:

2.47 acre = 10,000 m²
10 kg of C occupy an area of 1 m²
50% of the biomass is burned

The biomass burned in the site of 400,000 acre is:

$400,000acre\times\frac{10,000m^{2} }{2.47acre} \times \frac{10kgC}{m^{2} } \times 50\% = 8.10 \times 10^{9} kgC$

Let's consider the combustion of carbon.

C(s) + O₂(g) ⇒ CO₂(g)

We can establish the following relations:

The molar mass of C is 12.01 g/mol
1 mole of C produces 1 mole of CO₂
The molar mass of CO₂ is 44.01 g/mol

The mass of produced is CO₂:

$8.10 \times 10^{12}gC \times \frac{1molC}{12.01gC} \times \frac{1molCO_{2}}{1molC} \times \frac{44.01gCO_{2}}{1molCO_{2}} =2.97 \times 10^{13} gCO_{2}$

4 0

3 years ago