What really is the Ebola virus?

1 answer:

7 0

The Ebola virus belongs to a family of viruses termed Filoviridae. Filovirus particles form long sometimes branched filaments of varying shapes, as well as shorter filaments , and may measure up to 14,000 nanometers in length with diameter of 80 nanometers. Viral particles contain one molecule of single stranded RNA enveloped in a lipid membrane. New viral particle bud from the surface of their host cell. Although Ebola virus was only discovered in 1976, it is an ancient virus and is thought to have split from other viruses thousands of years ago.

You might be interested in

what mass of hgbr2 can be produced from the reaction of 5.00 ml mercury (density = 13.6 g/mL) and 5.00 ml bromine (density = 3.1

Katarina [22]

Jsnsbshhsbshdjjshsbshd

5 0

3 years ago

Pls help...

rjkz [21]

Answer: c

Explanation:

4 0

3 years ago

Read 2 more answers

How many outer-shell d electrons are there in each of the following transition-metal ions?

lesya692 [45]

Answer:

The answer to your question is:

Re (III) has 5 electrons

Sc(III) = has 1 electron

Ru(IV) = has 6 electrons

Hg(II) = has 10 electrons

Explanation:

75 Re(III) = 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p⁶ 5s² 4d¹⁰ 5p⁶ 6s² 4f¹⁴ 5d⁵

21 Sc(III) = 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹

44 Ru(IV) = 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p⁶ 5s² 4d⁶

80 Hg(II) = 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p⁶ 5s² 4d¹⁰ 5p⁶ 6s² 4f¹⁴ 5d¹⁰

4 0

3 years ago

At which electrode does oxidation occur in a

Charra [1.4K]

(3) the anode in both a voltaic cell and an <span>electrolytic cell is your answer.</span>

6 0

3 years ago

Read 2 more answers

What volume of 0.205 m k3po4 solution is necessary to completely react with 130 ml of 0.0116 m nicl2?

Alja [10]

Following chemical reaction is involved for the reaction between K3PO4 and NiCl2

2K3PO4 + 3NiCl2 → 6KCl + Ni3(PO4)2

Volume of solutions is converted to litres for present calculations (1litre = 1000ml)
number of moles of NiCl2 = 0.130 × 0.0116
=0.001508 mol

Now, $\frac{\text{mole of NiCl2 }}{\text{mole of K3PO4}} = \frac{3}{2}$
∴ mole of K3PO4 = $\frac{0.001508X2}{3}$
=0.001005 mol
given that, K3PO4 is 0.205 M
i.e. 0.205 mol of K3PO4 is present in 1 litre of solution
then 0.001005 mol is present in X litre of solution
∴X = $\frac{0.001005}{0.205}$ =0.004902 litre =4.902ml

5 0

3 years ago